Hello everybody, welcome to part seven of our reading of the Magdalen manuscript. If this is your first time here on the channel, welcome. I'm very glad to have you here, but you might want to start with the part one through six before moving into part seven. As those of you who have been following along, you know the first part of this book is allegedly a channeling from Mary Magdalen herself where we get into the practices of the priestess hood of Isis and the alchemies of Horace. Since we've finished reading through the manuscript, we're going through some other information on these practices which started last week in part six. Today we're going to be starting a section called internal alchemy, introducing the fundamentals of internal alchemy. Without an understanding of the alchemical terms used in the manuscript, it will be rather difficult to understand its essential concepts. In addition, some of the central ideas at the heart of the internal alchemy may not be familiar to many readers. For this reason, I have included a brief overview of internal alchemy in general and a larger overview of Egyptian alchemy in particular. It is my hope that these introductory sections will provide the reader with a deeper understanding and application of the manuscript and that's a personal note from Tom Kenyon who is the channeler for the Magdalen manuscript. I decided to make tea this morning. Rumbling through the cupboard half awake, I found a small kettle hidden in the corner. I poured some water in and sat on the stove. Lighting the burner, I went off to do other things like cleaning the counter from last night's dinner. It's so funny because I'm from the south when someone says they want to make tea. My initial thought is sweet tea, iced tea, not hot tea, iced tea. Soon I heard the familiar rumbling of boiling water. Sure enough, small clouds of hot mist were hovering over the stove. I turned off the gas and I poured hot water into an empty cup. The water hissed as it slid over the hot middle surface of the pot into the waiting container. I plopped a tea bag and finished cleaning the counter. While attending to the task at hand, invisibly, imperceptibly, the hot water found its way into the tea leaves. What had just been a cup of hot water was now a cup of tea. And if y'all want, maybe one day I'll share with you my grandmother's recipe for sweet tea, my grandmother from South Georgia, everybody's mama and grandmama has their own recipe for sweet tea. But my grandmother, Marianne, my dad's mom, who just recently passed away, she made a very specific type of sweet tea that also included pineapple juice. So if you guys ever want that recipe, just let me know and I will share it with you guys. Or if you ever come visit me down here in the south, I'll make it for you. What you may ask, does making tea have to do with alchemy? Plenty. The art of alchemy is simply changing one form into another. Most people think of alchemy as a medieval obsession for changing lead into gold. And while this is one form of alchemy, anything that causes a change in form is alchemy as well. Turning water into steam is alchemy. Changing a dry tea bag into tea is also alchemy. And we know if you follow along with my episodes with Shanti from Aquarius Rising Africa, we talk about that a lot, that's the power of spirituality is you being able to turn your own lead, your own darkness into that gold into something meaningful. While external alchemy like changing lead into gold is a fascinating enterprise, I am more intrigued by the internal alchemy, like the type discussed by Magdalene in the manuscript, like I was just mentioning with Shanti. But whether one is attempting to create enhanced abilities in oneself or to brew a cup of tea, some of the principles are the same. All successful alchemies must have three elements. One, a substance to be transformed. Two, a container to hold the alchemical reaction. And three, energy. Had I for instance stumbled into the kitchen this morning and poured water onto the burner instead of into the kettle, I would not have created alchemy. I would have created a mess. In external alchemies like making tea or nuclear fission, yes, atomic power is alchemy as well. The containers are pretty obvious. Different containers are required for different tasks. A tea cup for instance will do quite well for making tea, but is a very poor container for a nuclear reactor. For that, you will need a massive amount of concrete lead and a tremendous quantity of water. With internal alchemies, the containers are more abstract as are the goals. The purpose of internal alchemy is to change consciousness to accelerate one's own personal evolution. I do not, by the way, mean evolution in the usual sense. I doubt that anyone will physically sprout wings and fly as a result of practicing internal alchemy. However, the changes wrought by such practices can be profound that one feels as if one is flying above life and viewing it from a greatly expanded perspective. This is why many of the alchemical symbols from around the world involve flying beings. Like the Garyadas of the Balinese Hinduism, the Dragons of Daoism, or Horus, the hawk-like god of Egyptian alchemy to name a few. In all systems of internal alchemy, the container for the alchemical reaction is awareness itself. In other words, mental focus. As you read these words, you are holding them in the container of your awareness, and they are hopefully making sense to you. But if your awareness were to wander, say to a conversation in the next room, the container would shift and my words would not register. Even though you might be going through the motion of reading, the words would not be impacting you because they were outside the container of your awareness. There is a basic concept in all forms of internal alchemy, which can be stated quite simply, energy follows awareness. In this reading example above, for instance, the energy of your attention follows where you put your awareness. If you pay attention to what you are reading, the words will make an impression in your mind. If, however, you pay attention to that conversation you were talking about, the words of this book won't impress your mind, but the words of the conversation will. Yes. And this Thursday, I'm going to be releasing my video on misunderstandings of yoga. And focus is a big part of yoga. For example, in traditional yoga, we practice the Tristana method, which is breath, posture, and drishti-drishti is a focal point of your eyes. You have to keep your eyes focused, not wandering around the room or closed during your practice. That's bad to close your eyes during your practice. And that keeps your mind focused. Yoga, Tristana, Rodha, we're looking for that one pointed focus. That's what meditation is. It's getting the mind to focus. Once the mind shifts, that focus is lost, meditation is lost, etc., etc. So it's all the same thing here. The alchemical container of inner alchemy is awareness itself. The substance transformed the process of inner alchemy very according to the stream or lineage. Some of these substances include such things as physical neurotransmitters, hormones, saliva, and sexual fluids. There are also, however, the whole class of subtle substances that include such your focal things as chi, the Taoist stream, Purana, and the yogic stream, the winds in the Tibetan stream, and the netters in the Egyptian. It is this class of subtle substances that inner alchemy is the most concerned with. And for the average person, this is one of the most difficult concepts to understand. The reason I think has to do with our conditioning to the Newtonian world of everyday reality. We are not trained, generally speaking, to attend to these subtle energies behind the shadow play of physical reality. Let me give you an example of what I am talking about. And I actually bring this up a lot with my yoga courses too, where trying to eventually tap into subtle body, subtle responses, and he is correct. Most people, when they start practicing yoga or start practicing any type of spiritual discipline, don't have any type of connection to their subtle body. That is why in practices like yoga, we start with the asana, with the physical practice, with the gross body, which is just your basic body understanding sensation in your legs and your arms and your stomach. And the more and more and more you practice focusing in these postures on your drishti, on your breath, you start to tap into that subtle energy. Purana, which is what he was saying in yoga, what we call your life force, your product, that's your upward rising energy. So for an example, when I teach traditional led primary, I don't do this, but like in a workshop, Surya Namaskar, so your sun salutations, Surya is what is what is represented, represents Prana, because Prana is sun, it's solar energy, it's rising energy, the moon within represents the up Prana or the downward rising energy. But because Prana is upward rising, that means it's hot, like the sun is hot, right? So we start off with Surya Namaskar, Surya sun Namaskar is a greeting hello, so your sun salutations are basically warming up your body, creating heat, creating that light force, that movement. And so I will tell my students a lot in their downward facing dog, in a workshop or a course, after you, you're getting sweaty, you have that moment of pause and down dog, see if you can actually tap in and feel the circulation of that energy starting to emerge in your body. All right, so he goes on to say, remember that cup of tea I made this morning? Well, I never finished it, and now it sits cold on the counter, I got caught up with other things and forgot about it. For some reason, I didn't rinse out the old cup, but instead got a clean one from the cabinet, I put it next to the old one filled it with hot water and flopped in another tea bag. There were small billows of steam rising from the new cup of tea, but nothing was right as you out of the old cup except the smell of cold tea. How odd, I thought. I picked up both tea cups, one cold and one hot. I felt like Goldilocks and the three bears. This one's too cold, this one's too hot. I put down the hot one to let it cool off so that it might get to be just right. I was in the two side by side worlds of Newtonian and quantum reality. I could hold the tea cups in my hand. One was cold and one was hot. But the reason for this difference was from the realm much subtler than the two tea cups or the watery tea they contained. It had to do with the individual molecules of water. Although I could not see it, the molecules in the hot water were running about helter skelter like a mob at a shopping mall bumping into each other and in the process heating up things. In fact, when I had brought the water to boil on the stove, they were at their most riotous, a molecular body bra of extreme proportions. The cold tea cup, on the other hand, held a group of lethargic molecules. The energy that had heated them up had long gone past out of the cup in the form of radiant heat. And the once riotous molecules were now slow like dowagers in a nursing home. The only difference between the party animals in the hot cup of tea and the sleepy ones in the cold cup of tea was energy. I could not with my physical sense see the difference between the molecules of water since they were too small for me to see. All I could sense was the after effects of their energy or lack of it, namely heat or cold. In point of fact, that is all any of us experience in the physical world, the after effects of something happening in the subtle or quantum world. The task of the alchemist is to become aware of and sensitive to the subtle realms that escape most people. The reason for this is that the subtle realm of quantum world holds the best type of substances for our chemical transformation. For the most part, the substances in the Newtonian or dense physical world are too gross, too dense to be transformed the power of alchemical attention, but the subtle realm, the quantum world is very receptive to the focus of such attention. In my example of the two tea cups, the common element was energy or lack of it. In this case, the energy was applied from an external source and most internal alchemies. However, the source of energy is usually the energy of consciousness itself. What do I mean by this? Become aware of wanting your hands for a moment. Just shift your attention to it. Be aware of its position, its weight and the physical sensation in the hand. After a moment of this, become aware of your other hand. Which hand has more sensation in it? In other words, which one has more energy? For most people, the hand that has been focused on will have the most sensation. This is because an energy follows awareness. The alchemical container of awareness was focused on a particular hand and this resulted in an increase of perceived energy in that hand. The neurological reason for this are quite complex, but the practice of holding attention is quite simple. We do it every day. Sometimes the energy of the breath is used to power the alchemical reaction. At times, an external energy source might be used such as the sun or a ceremonial fire. In some rare forms of alchemy, other elements such as water or air are used. In the use of external energies to drive an alchemical reaction, the alchemist might focus on a flame while holding within the container of awareness the substance to be transformed or the energy of the sun might be used to power an internal alchemical reaction in the same way. To give you an example, I am writing this part of the book in the month of August on the island of Peros in Greece. Every afternoon around six when the sun is not blazing hot, I sit on the deck for about an hour and do an alchemical practice within the ancient Egyptian stream. To explain how I do this, I need to define a term from Egyptian alchemy, the Ka. The Ka body is sometimes called the etheric double for a spiritual twin. It is the same shape and size as the physical body, the cut, but it is made of energy and has very little mass or density. As I hold the subtle body in the container of my awareness, I sense the sun and I draw the subtle energies available to me through the sunlight into my Ka. Some of these energies from the sun have been scientifically documented such as ultraviolet radiation in the primary colors of full spectrum light. Studies have shown that full spectrum light has a very positive effect on health. It does. But there are whole classes of more subtle energies that science has not yet verified. These extremely subtle energies are presumably too subtle for our current methods of measurement. Daoists might refer to this type of subtle energy as Solarized Yang Chi. A yogi might refer to it as the Prana of Surya. Again, Surya is Prana, the sun. An alchemist within the Egyptian street might simply say that it is the Netter of Ra, the power of the sun god. But whatever you call it, there is another type of subtle energy in sunlight. In my practice of this technique, I have discovered what I call the armchair method. I used to use an elaborate system of standing movements to draw the sun's energy into my Ka. I still sometimes do this when I'm feeling particularly energetic, but I've also found that the method works just as well if I am sitting on my butt. In this case, around six in the afternoon, I sit in a canvas lounge chair provided by our benevolent landlord, Stefanos. I take off my shirt and lean back. Holding my Ka body in the container of my awareness, I simply become aware of my Ka, or you might say I imagine it as a body of luminous light. And then I breathe in in relaxed manner. On an inhale, I draw the subtle energy of the sun into my Ka, so Ka is like the energetic body, if that makes sense. Sometimes I draw the energy in through my navel and circulated around the Ka. Other times I draw the energy of the sun into my solar plexus, which is from the viewpoint of Egyptian alchemy and aspect of Ra. Yeah, your solar plexus, so that's Manipura. That's your third chakra. That's your Helios. That's where your heat comes from in your body. As I charge this miniature sun within my Ka, the excess energy spontaneously flows outward into my body. Sometimes I prefer to imagine my Ka as some kind of a magnet, which it is, and draw the energies from the sun directly into the entire Ka at once. While this method may seem odd or laxadaisical, it has all three elements of substance, container, and awareness. And so it is, in fact, a form of alchemy. As I focus on my Ka, I am holding it in the alchemical container of my awareness. This substance to be transformed is the Ka itself. In this case, I am engaging in something called an energy building practice. Why I would do such a thing will become clear in the next section when I talk about Egyptian alchemy. So there we have it, two of the essential elements, a substance to be transformed, my Ka, and a container to hold the alchemical reaction, my own awareness. The third element, energy is, of course, provided by the sun itself. When I do this practice correctly, example, hold all three elements together, there is a tremendous increase of energy and vibration within my Ka body. At times, however, my mind will wander into thoughts or fantasies. And at these moments, the building of the energy in the Ka decreases. If I don't bring my mind back to an awareness of the Ka, the building of energy will completely stop. This is of course, because I have lost the container for the reaction when my mind wanders. To use our tea analogy, wherever my mind wanders, it is like taking the tea kettle off the stove. The art, if you will, of internal alchemy is to hold all three elements of substance, energy and container together long enough for an alchemical reaction to occur. And the discipline required of an alchemical to accomplish such a feat can be considerable. All alchemists, regardless of their lineage, must attend to three elements, container, substance and energy. Each stream offers its own particular ways to strengthen the alchemical container. And by practicing the methods for strengthening the container of awareness, the alchemist can hold together ever increasingly intense alchemical reactions. The alchemical practitioner must also become keenly sensitive to the subtleties of substance. As awareness becomes refined, the alchemist is able to distinguish the most subtle characteristics of energetic substances. And a sense begins to unfold on how best to use these substances in an alchemical reaction and to what end. These substances range from actual physical materials such as saliva, sexual fluids, etc. to very subtle substances that are purely in the quantum realm of existence. As mentioned earlier, it is these quantum substances that are most easily transmuted during the process of inner alchemy. Finally, the alchemist must collect energy, since it is needed to drive alchemical reactions. There is a virtually unlimited field of possible energy sources for such inner work. And each alchemical tradition or stream offers its own suggestion on how to collect these types of energy. Indeed, the various streams of internal alchemy throughout the world offer a wide variety of clever and generous and sometimes remarkable methods for cultivation of energy. In the manuscript, Magdalene states that her task as an initiate was to assist Yashua in the building of his subtle body, the Ka, through specific energy building practices. This was accomplished according to the Magdalene through the masterful use of sexual energy. As an initiate of ISIS, she was using methods from one of the oldest alchemical streams on the planet. And it is to this that we now turn our attention. All right, that was a shorter section today, guys, but we are going to end it there. Next week, we're going to start with the Egyptian alchemies. I wanted to end it there because I want to give each of you a chance to kind of practice that idea of filling your Ka body, your subtle body. Again, you can do that through physical exercise, through yoga, through the exercise that he uses, whatever works for you. All right, guys, leave me your thoughts and your opinions down in the comment section below, and I will talk to you soon.


Hello and welcome to our Q&A. Well it is race day and it is Max Frick that we're talking to. So first of all, how are you doing? Yeah, I'm alright. Just trying to keep dry at the minute. Hopefully this rain holds off for the meeting later on. The sun's shining now. Don't you worry about a thing. Okay, well we've got all of your questions come in. So thank you so much. So Inga wants to know, what do you love about Malilla? About Malilla. It's such a nice track. There's a little bit of camber here. It's nice and smooth. So hopefully, whether the meeting, the track's going to be awesome today. Hopefully you can have some good starts and score a few points. Okay. And we've got Peter. He wants to know, who's your favourite other speedway rider? In the GPs. Yeah, go with that. Okay. Bartek's pretty exciting to watch. And then Doyle is another one who's quite exciting to watch. So I'd say one of those two. But then also obviously Ty's been pretty dominant the last few years. So you can have a combination of all of those guys. It's pretty awesome. Okay, favourite food? You can't be at a home cooked meal, I don't think. So we don't get to go home very often to Australia. So it's nice to go home and have a little bit of food. Some home cooked meal from mum, maybe. It's always nice. Are you just dropping hints in case she can come over and cook you some food? Yeah, maybe not come over, but I'm looking forward to going home at the end of the year and having some food, that's for sure. Okay. And favourite movie of all time ever? Of all time ever, the greatest movie is Forrest Gump. That's my favourite movie. I thought you were going to go for like a legendary one. You just went for Forrest Gump. Alright, okay. Taylor says, have you enjoyed riding in the Grand Prix so far? I'm going with a solid yes. Yeah, exactly. I've had a few good results. I'm just trying to keep consistent and pick up as much information as I can and learn as much as I can. And hopefully one day I can put that to good use in the future, being in the series maybe. Okay. Alan wants to know if you've got any lucky socks? No lucky socks. At one point I had a set of happy socks that are nice and colourful and I used them, but not anymore now. Okay, alright. And Richard says, I don't know why, I don't know what Richard's up to, but he wants to know if you've ever suffered a bad experience in an airport. Yes, everyone's flown through Heathrow. I've had a few close calls, that's for sure. Running through the airport, I've done a couple too many times, but nothing too bad I don't think. I've had nothing too major that I can remember. Alright, solid. Well, we've all lost luggage, so yeah, we'll go with that. Alright, now this is the groundbreaking, very serious and important question. We've left it for last. Would you rather fight 10 spider-sized kangaroos or one kangaroo-sized spider? Probably 10. What was the first one? Well, 10 spider-sized kangaroos, that's just disgusting, or one kangaroo-sized spider. Why would you? It's going to be the small 10 ones, isn't it? Yeah, it's got to be the small 10 ones, that's for sure. Would you ever box with a kangaroo? Have you seen them box? I've seen them box. I've kept my distance. I'm more of a lover, not a fighter, but yeah, it's pretty interesting. Alright, there you go. See, we went Australian. It's all exciting here at the Q&A. Alright, thanks so much, Max. Enjoy tonight. Right, thank you.


A, B, C, Roar! A is for Alligator. A, B is for Baboon. C is for Camel. D is for Donkey. E is for Elk. Is for Fox. G is for Goat. G, H is for Hummingbird. I is for Inchworm. J is for Jellyfish. J, K is for Koala. Is for Lion. Is for Monkey. Is for Newt. Is for Octopus. Is for Porcupine. Q is for Quail. Q is for Rhinoceros. Is for Seahorse. S, T is for Turtle. U is for Unicorn Beetle. V is for Vulture. W is for Wolf. W is for Xenops. Y is for Yak. Y, Z is for Zebu. Hey! What about me? Z.


Hello, I am Yumti Yong Longkomer from BA First Semester History Department and I will be giving a book review on Kafka on the Shore by Haruki Murakami. Haruki Murakami is a Japanese writer and he was born in Kyoto in 1949 and now lives near Tokyo. His other works include After the Quake, Dance Dance Dance, The Elephant Vanishes, Norwegian Wood and the Wind Up Bird Chronicle. So Kafka on the Shore has a dual perspective. It is powered by two prominent characters who are Kafka Tamura, who is a 15 year old boy who runs away from home because to escape a prophecy which is more like a curse. Then you have Nakata who is an aging simpleton and due to an incident he is able to talk to cats. Now although these two different characters seem to have a different journey, their parts eventually intertwine at the end of the novel. So Kafka on the Shore is my first Haruki Murakami book and personally I really enjoyed reading it and yeah it was really confusing at first so I had to read it a couple of times and to be honest I'm still confused and it has left me speculating about a lot of things. There are some parts that I didn't like reading and I won't be spilling any spoilers so let's just say that there are some really detailed bloody and questionable scenes. So my advice would be just read the novel with a very open mindset because it's very beautifully and well written. So the novel has themes of duality, fate and it is also a coming of age for Kafka the 15 year old boy and there is magical realism in the split realities like there are a lot of there is dream world and a lot of spirits and the ability to talk to cats and many more that you will find as you read the as you read the novel the author also makes a lot of references to writers classical music and even Greek mythology. So the magic about this novel is in the unexplained unexplained because the novel is actually very confusing and it has a lot of random events and some really bizarre characters as well but all these eventually make sense at the end and you know it leaves the reader a lot to speculate and a lot to work on and I think that's what's really interesting about this book and so yeah I would rate it a five out of five and I would definitely recommend you to read it. I really enjoyed reading it and if you have read the book I would really be happy to know your thoughts on the novel. That's all from me. Happy reading.


Good afternoon.  I was a bit sick. And for the same reason, I don't have new homework. But you will still get homework maybe tomorrow, and you will also get back your homework. OK. One week? What? Well, I don't know. Maybe you will just give some small team. OK, I still have to decide. But to just one, at least one more exercise. OK, I think we can start. OK, so we had been talking about relative homology. So I recall that we had, so A was a subspace of x, the topological spaces. Then we have, but in a natural way, the n chains, single n chains in A are contained in the n chains in x. And we can consider the relative n chains, which is just a quotient. And then in the usual way, one has the differential d. In the usual way, one associates to this, the relative homology. And we had seen that, say, writing i for the embedding from A to x, we have the long x-sector homology sequence. So this was something like you have some hn of A, hn x, hn of xA. And then we have this. This is the map given by the push forward by i. And here we have some boundary map to hn minus 1 of A. And it goes on like this, a long x-sector sequence. And one thing that is useful is to know how to compute these, how to simplify the computation of this, because it's given to us in this rather complicated way. And this is given by the excision theorem, which says that, again, if x is topological space and z and A are subspaces of x, such that the closure of z is contained in the interior of A. So we have here x, we have here A, and we have here z. Then we can cut out z from both x and A and get the same homology. So then the hn of xA is isomorphic, hn of x minus C, A minus C. And this uses this theorem of the cover, which I want to recall. So we had defined cn. So if, say, A and B are subsets of x, such that the closures, such that the interiors cover x. So interior of A union the interior of B is equal to x. Then we had seen, or then I had claimed, that so we can consider this what was called cn of A plus B. Which was supposed to be, was defined to be the set of all n chains, which are linear, such that every simplex is either contained in A or in B. So it's a set of all sum sigma, A sigma times omega, just like some i, ai, sigma i, i equals 1 to n for some n. And the ai are elements in z. And sigma i is a map from delta n to x with either sigma i of delta n is contained in A or sigma i of delta n is contained in B. And by the way, notice that by definition, we have that cn of A is contained in cn of x and cn of B is contained in cn of x. And if you think of what this means, we just have that cn of A plus B is just the sum. No, not the direct sum, the sum. So these are all formal sums, which can be written as sum of a formal sum with things in A and with things in B. And then the statement was that the relative homology. So the homology, so I can just say it like that. If I take the homology of the chain complex of x with d, which is just the homology of x, is equal to the homology of a k. So this was the theorem of the cover. So you can compute the homology of a space by looking at chains such that each of the simplices is contained either in A or in B. So there are no simplices which go over both. OK. And then I wanted to say one more thing, which we had maybe I think briefly said at the end. So if we had introduced an exercise, the reduced homology, I will not repeat. One thing that we had seen is that if x is path connected, the reduced homology is 0. And then it's a remark and not difficult to prove that the long exact homology sequence also holds in reduced homology. So recall that for n bigger than 0, the reduced homology is the same as the usual homology. So there's nothing there. But as I said, if the space is path connected, then the 0's reduced homology is just 0. And that makes it easier to apply the long exact sequence because most of the spaces you look at are path connected and then it starts with 0 in the beginning. So you don't have to have so complicated arguments. OK. And now we want to be more or less, I will talk a bit long. But anyway, now we are at the point of continuing where we were. So I wanted to prove this invariance of dimension for this. I actually will have to use a theorem I will only prove later. So we'll prove below the following that I don't know whether or not. So this is, I don't know, I call it theorem, but that's maybe nonsense, that if I take hk of Sn, this is equal to 0. So Sn is the n-sphere, the usual n-sphere. So this will be 0 if k is different from 0 and n, and z if k is equal to 0 or n. And for the reduced homology, it's clear so that this thing is connected. So reduced homology will be 0 also if k is equal to 0. So we are going in two cases. So this I will prove later. But now I want to use it to prove the invariance of dimension, so which says that if you have open subsets of Rn and Rm, which are homeomorphic, then n must be equal to m. So the dimension of open subsets is preserved by homeomorphism, which is something which maybe seems obvious to you, but it's not so clear how y should be true and you can now prove it using homology. So let u subset Rn and v subset Rm be open and non-empty and u homomorphic to v. Then it follows that n is equal to m. You know, I think there are examples of continuous surjective maps from, say, R to R2 or things like that. So you can see that this is not a, but if they are homeomorphisms, then this cannot happen. The dimension is preserved. And this we do using this relative homology and the excision theorem. So we take a point. So we have assumed that x and that u and v are non-empty. Obviously, otherwise, there's also no statement. You cannot certainly conclude anything from talking about the empty set. So let x be a point in u. And now we look at the relative homology, hk of u, u minus x. We can use excision. What is it? I claim this is equal to hk of Rn. So we were u was in Rn, Rn without x. So we just take, in the excision theorem, we take z equal to Rn minus u. And then you see that the excision theorem applies because this is contained, the closure of this, which is actually closed, is contained in the interior of both sets of Rn minus x. So then we can apply the long x-sector homology sequence. And I will apply it to the reduced homology. I mean, one can also do otherwise, but it gets slightly simpler if one is reduced homology. So I take hk of Rn, maps to hk of Rn comma Rn without x, goes to now the boundary map, hk minus 1 of Rn minus x, goes to hk minus 1 of Rn. And then it goes on. So if you look at this, Rn is contractible. So all the homology, except the 0th homology, is 0. The 0th homology is z. But you can also, if you take the reduced homology, then just all the reduced homology is 0. Because it's path connected and the higher homology is 0. So this is 0 for all k. And this is also 0 for all k. So we find that this relative homology is the same as this. So for all k, we have that hk of the relative homology is equal to the homology of this. But now you have that Sn minus 1 is a deformation retract of Rn minus x. I mean, you could assume you just, or at least is a, I mean, you can assume that the point x is 0. I mean, it's just by shifting. So assume that, then you just have here the Sn minus 1. You have here the point x. And you just can, if you just make the homotopy from the identity on Sn minus 1 to the identity on Rn minus the point, if you are inside, you kind of pull here with a certain speed so that at 0 you are at 1 you arrive here. And if you're outside, you do it here, all along straight lines going to infinity. And you find that you have a, you get in this way a homotopy equivalence between the identity on Sn minus 1 and the identity on Rn minus x. So it follows that this here is actually equal to hk minus 1 of Sn minus 1, which we know. So we find that hk, what? What's the problem? So we find that hk of, say, now I go back to where we were, u without x is equal to hk minus 1 of Sn minus 1, the reduced homology, which is equal to, according to what we have written here, to 0 when k is different from n and z if k is equal to n. So if we take this relative homology, it will remember the number n, because it's the only part where the homology is not n.  And as you see, according to what we just proved, this is 0. I mean, both sides are 0 unless n is equal to the, so this one is 0 and z. And this is if k is equal to n. And this one is 0 or z if k is equal to m. And these things are always supposed to be equal, so it means n is equal to m. OK, so this was this application. So I hope that's clear. So now we want to, so this was one small application of the excision theorem and the long exact homology sequence. So now we want to come to a kind of even somewhat more useful tool to directly find an exact sequence for the homology without talking about relative homology. And this is the Meier-Vietore sequence. And this is in some sense maybe the most useful thing that we will introduce in order to compute homology. So this is the Meier-Vietore sequence. So this is after two mathematicians, one called Meier, the other one, Vietores. I don't know very much about Meier, but Vietores died very recently at the age of 105 or something years. And I think he still was publishing papers until very soon before, so he can have a long career. But this I think he did when he was rather young. So I mean, this is the Meier-Vietore sequence. So let x be a topological space. A and B open subsets of x.  So x is equal to A union B. And I just for security, I mean, to have the notation, I introduce all the inclusions. So I take EA from A intersected B to A, the inclusion, IB from A intersected B to B. And then we also have, say, GA from A to x, the inclusion, and JB from B to x, the inclusion. So that's that we can see what all the maps are. So then we have a long exact sequence. So it somehow goes on in the usual way, Hn of A intersected B. So it's always, so all the maps, except for the boundary map, are given by inclusions. So here this will be, this goes to Hn of A plus Hn of B. And this is just by the two inclusions. So IA, IB. So the map from this to this one is the push forward by IA. And the one to the other one is the push forward by IB. And then we have the map to x. And here we basically take the difference. So we take, again, the push forwards by these two inclusion, but we take the difference. So we take, say, JB star minus GA star. So if you have a class alpha here, we send it to JB star of alpha. So the push forward minus JB star of alpha. And this would be a class here. And then we have the boundary map delta, which will, again, as usual, bring us to Hn minus 1 of A intersected B. And it goes on like this. So now, in this case, we want to give some kind of a proof. This, again, follows from the theorem of the cover that we had. That's why I repeated it. So the proof, we have a short exact sequence of chain complexes. So 0 goes to D star of A intersected B. Goes to C star of A plus C star of B goes to C star of A plus B goes to 0, where this map, I call phi in this psi and phi, is just IA star IB star. So if I have a, if I take class alpha here, so phi of alpha, if I have a chain, so linear combination of singular simplices in A intersected B, I can view them as singular simplices in A and also singular simplices in B. They just take the sum. I mean, I take it here and here. So this is equal to alpha. And here, I can, for psi, it's the same thing. I mean, again, just translate this J. So that means if I take psi of alpha, if I have a singular chain in A, I can view it as a singular chain in X.  So I have now, here are two things, alpha better. And I take the difference, beta minus alpha. And why is that an exact sequence? So it's kind of clear that these are all maps of chain complexes. It's just the D is always the same D as we had before. This is the restriction of the D in X on C star X. And the D on C star B is the restriction of that to B. The D is always the same. So it's certainly, these are all chain maps. So the first step, so phi is certainly injective because we know that according to the definition, we just have that C star of A intersected B is just a subset of C star of A and also of C star of B. And the map here is just the inclusion. I mean, the inclusion is called I A lower star. But it is just the inclusion. And now, what else? It's obviously injective. So the composition is obviously 0. If I take psi composed with phi, so psi was with phi is 0. Because if I take psi composed with phi and apply it to some class alpha. So what do we do? I mean, we first send it to alpha, alpha. And then we subtract it from itself. So it cannot really be more obvious. And now, but we want to say that the image is precisely equal to the kernel. So if we have alpha beta, which lies in C NA plus CNB, if this is in the kernel of psi, what does it mean? It means that beta minus alpha is equal to 0. So it means beta is equal to alpha. But alpha likes in CN of A, beta likes in CN of B. So it lies in the intersection of the two. The section of the two is CN of A intersected B, if you look at the definition. And so we see that then it is equal to phi. And then finally, the subjectivity is by definition. If you look at the, we have just observed that. We had seen that CN of A and CN of B are both sub-modules, I mean, sub-rings of CN of X. These are all the simplices where this is contained here. And by definition, we have CN of A plus B is equal to CN of A plus CN of B, inside CN of X. Or if you want to, and so therefore, and this is precisely the image, which by definition is the image of this map. But to see it more explicitly, we can also say, so let more explicitly, if we take an element alpha in CN of A plus B, then we can write it as alpha is equal to sum over sum i, alpha i sigma i plus sum over sum j, beta j tau j, where alpha i and beta j are some integers. And sigma i is a map from delta n to A. And tau j is a map from delta n to B. Because that's precisely how this CN of A plus B was defined, that it's all n chains, which can be written as where all the simplices map either to A or to B. And I've now just divided them in such a way that I first write those map to A, and then I write those map to B, that I can certainly do. But that just means that alpha is equal to psi of what? Well, what are we supposed to do? We are supposed to write this as a difference of these two. So I could say sum i, alpha i, sigma i, comma minus sum j, beta j, tau j. Because if we take the difference between these two, this is precisely alpha. So the map is also subjective. So thus, we can apply the long. So now apply the snake lemma. And this precisely gives us the long exact homology sequence. Which one? I have not understood the question. Yeah? So I made it the wrong way around, yeah. Okay, so there was a sign around. Anyway, what matters is the principle. But I understand it's a, okay. So I want to also state without proof, but it's not difficult. It would be that the following remark, which is the same as for the exact homology sequence, it also holds in reduced homology. So, and again, this is practical because you have less case distinctions for the low homology groups. So the Meyer-Viertos sequence also reduced homology. And that's essentially an exercise to modify the proof here. Okay, now we want to give some examples. So first we have to prove the statement that I made at the beginning of the lecture that computes the homology of the n-sphere. I was using this for this invariance of dimension. So I should better prove it. So we have the n-sphere. So we have, as you know, have Sn set of all x0 to xn in n plus 1. Such that the sum of the squares of the coordinates is equal to 1. And so we want to compute the homology. So the claim was, after all, that, so I just say it for reduced homology. So hk of Sn is equal to z if k is equal to n and 0 otherwise. Okay, let's do it. So we want to use the Meyer-Viertos sequence. So we write it as a union of Sn as a union of two open subsets, of which the homology is easier to compute. So we write Sn equal to a union b, where a is equal to Sn minus, say, I could call it the north pole. So n is the point 0, 0, 1. And b is Sn minus the south pole. And I think you can imagine which one that is. And so we have this picture that, so we have here the sphere Sn. And so here we somehow have the north pole. And here we somehow have the south pole. And we have for a, we leave out this one point. For b, we leave out this one point. And then we also consider a intersected b, where we just leave out both points. So now I say that a and b are contractible. And a intersected b is homotopy equivalent to Sn minus 1. So maybe you can imagine this I will, so I make this some kind of exercise. That means I don't carry out the details. So check this. So we have, so basically we can, I want to claim. So we can retract. So we have a, so S is a neighborhood retract of a. n is a neighborhood retract of b. And there's n minus 1, which is the equator. It's 0 to xn minus 1 in n plus 1, so 0. Such that sum xi squared is equal to 1 is a neighborhood retract of a intersected b. And it's clear that e is homeomorphic to Sn minus 1. Yeah, a neighborhood retract. So S is a neighborhood retract of a. I introduced the last time what a neighborhood retract is. And OK, so I want to, it's an exercise, I say. But the exercise is just to carry out the detail. If you look at it, for instance, for a, so how does it look like, we have this picture. So here's the south pole, and we kind of throw, we take out the north pole. There's one point missing. So we can just make the kind of the homotopy by pulling it down along great circles. So each point lies on a great circle, and we just shorten it down according to n equals 0. It's the identity of this thing. At the point 1 in the homotopy, it will be here. And in between, for every point, we go half the way from the point until S along the great circle. And obviously, for b the same, and you would have to write down the formula to see that everything is continuous. And for a intersected b, it's basically the same story. We have now this thing. And we are now missing the north pole and the south pole. And again, we can just take, we can again just pull down along great circles, but only until the middle. So if we have any point anywhere here, if you are at the homotopy for t, you go so and so much along the way until you reach here. You can just write down the formula. This gives you a homotopy between the identity on this thing and the identity on the equator. So you have to write down the formula to see that it is a homotopy in a homotopy equivalence. And now, given this, we can apply the Mylvia tour sequence. I haven't done this very symmetrically to apply the Mylvia tour sequence. So what do we have? I wiped it out, but we have, at some point, we have hk of a plus hk of b maps to hk of sn, maps to hk minus 1a intersected b. And maybe one more to hk minus 1 of a plus hk minus 1 of b. So, and now we do everything with reduced homology. So in principle, we only have to put it here because the other ones will be the same. But for any k, we do it with reduced homology to make it easier. So we know that a and p are contractable. So their reduced homology is all 0. So this is 0 plus 0. And this is 0 plus 0. So we find that hk reduced of sn is equal to hk minus 1 reduced of, well, a intersected b, which we know is, which has the same homology groups as sn minus 1. And now we know, so for instance, we have that s0 is two points. The set consisting of the points 1 and minus 1. And you can easily compute that the reduced homology of a set of two points is z. So hk0, if k is different from 0 because it's just points. We know that for a point, all the higher homology vanishes have a union of path components. It's the whole homology is just the sum. And then we take the reduced homology, so we have one sum and less, and z of k is equal to 0. And the statement says that, therefore, by induction, by this formula, the result follows. Thus, by induction, hk of sn is equal to 0. Different from n, z, k is equal to n. I mean, one could have done the whole argument also without reduced homology, but with the usual homology. But then it's always a bit complicated because you do not have the just things are 0, but there's some z's and so on. And you will have some maps of which you have to show the injective or something. So it always is a shortcut to use the reduced homology. Because if something is path connected, the reduced homology, the reduced homology is 0. And that somehow makes things easier. So this was this example. I want to make a couple of small applications of this, which I think you already had done for n equal 1 using the fundamental group of the circle. We can now do it for all. It was the only thing for all spheres. So we have the following corollary. sn minus 1. So we write dn to be the unit sphere. So the unit disk in dimension n. So this is the set of all x0. Well, maybe I write now x1 to xn. Well, maybe I write that. Whatever. It doesn't matter. In rn, the sum of the xi squared is smaller equal to 1. So it's just the unit disk. And sn is the, now with sn minus 1, we take, we also changing the numberation. This is OK. It's the boundary. And the claim is that sn minus 1 is not a retract of dn. And the proof is, I think, something you already had with homo is very similar to what you had for homology groups. So assume we have such a thing. So let f from dn to sn minus 1 be a retraction. So that means it's continuous. And f restricted to sn minus 1 is the identity on sn minus 1. And we also look at the inclusion to dn be the inclusion. Then you actually had an exercise. I mean, you had an exercise which you have already solved, which says that i lower star, say, from hk of sn minus 1 to hk of dn is injective for all k.  Otherwise, I assume that n is bigger equal to 2. It's an exercise. I mean, in the following, it's an exercise to conclude if n is equal to 1. So in this case, we see that, for instance, if we do this for the n minus 1st homology, we have that i lower star is a map from hn minus 1 of sn minus 1 to hn minus 1 of dn. But this one is equal to z. And this one is obviously equal to 0, because dn is contractable. I mean, everybody can write a homotopy which contracts this thing to a point. And so this is a contradiction. Obviously, it can be no injective map from z to 0. And so we deduce that there can be no such retraction. And from this, one gets in the standard way the Brouwer fixed point theorem that I think you also have seen already in the case n equal to 2. So every continuous map, f from dn to dn has a fixed point. So by this, I mean that is an x in dn, such that f of x is equal to x. And so how do we do that? Well, that's very simple. I first make a drawing. So I will anyway also just say it in words. You have to write down the formula. So let's assume we have such an f without fixed point. We want to use it to make a retraction of dn to sn minus 1. We do this as follows. We define dn to sn minus 1. And now I don't write the formula as follows by letting r of x to be the intersection point of the line from x to f of x with sn minus 1. So there will be two intersection points, no? But I take the intersection point on the side of x. So the picture is as follows. So we have here our sphere, the disk. And here's our point x. Somewhere is the point f of x. We just, you know, we are in rn. We can connect them by line. And the line will intersect the sphere in two points. And we take the one which lies nearer to x. We call it r of x, OK? And obviously, one has to check exercise r is continuous. By definition, r is a retraction. Because if we already are on sn minus 1, then the intersection point, if x is already in sn minus 1, then the intersection point with this line with this n minus 1 will be x. By definition, r is a retraction. And to prove that r is continuous, you have to write down a formula. And you know, if you write down some rational expression in continuous function, where the denominator is not 0, this will be continuous. So this proves r is fixed point 0. So let me see. I have half an hour. OK, I want to see that one more example. Just to show that we are now able to at least compute some homology groups. So say I call it g2, with a union of two circles which touch in a point. So this thing looks like this. And in fact, these two circles are called c1. So c1 and c2 are the two circles. And here we have intersection point p. And we want to compute the homology of this wonderful space. And so I again compute the reduced homology. Then you can compute from it the non-reduced one. So we do this by using, again, the Maya-Vietou sequence. So we said, so we have to write g2. So the thing which would be the easiest thing to do, which doesn't quite work because the assumptions don't apply, would be to take u equal to c1 and v equal to c2. And then apply the Maya-Vietou to this. This doesn't quite work because we have assumed that u and v must be open. And they are not open here. I mean, there is a sharper form of the Maya-Vietou sequence which actually would apply. And this would be OK. But we now have to use what we have. So we take u, so let epsilon be some small positive number. So I mean, really small. So maybe if the radius of this is 1, it's to be smaller than 1. And we take and let, so e epsilon p be a ball of radius epsilon around p. So we assume we are in R2 here. And we just take a very small ball. So we have here our small ball.  So that is, we have this thing. And we have also this little thing here. And now it's easy to see. So first thing is note, if you take u intersected v, this is just this. And it's not difficult to say to see. I mean, you can do this as an exercise that this is contractable. In fact, p is a neighborhood retract of it. And we just have to write down some coordinates how to pull this together. And in the same way, we have that c1 is homotopy equivalent to u. So you can make a homotopy which pulls this extra part in. And c2 is homotopy equivalent to v. So now we are in the situation as if we had taken just u equal to c1 and v equal to c2. Because now we can again apply the Myovir-Tour sequence. Is it? So we have hn.  Now that's the usual thing. You have the intersection, the direct sum of the open subsets goes to the union. And then we have the boundary map. We need it, the delta to hn minus 1 of u intersected v. And now u intersected v is contractable. So the reduced homology vanishes. This is 0. This is 0. This u is homotopy equivalent to this circle, c1. So this is equal to hn of s1 plus hn of s1. So we find this. So it's reduced. So which means that these two are isomorphic. We know that for s1, the reduced homology is z if n is equal to 1 and 0, otherwise. So we find hn of g2 is equal to 0 and different from 1, z, n is equal to 1. And in the usual way, we can know as g2 is path connected non-empty for the non-reduced homology, we find that hn of g2 is z if n is equal to 0 or 1. So nobody protested. I've just proven something else than what I said.  So you should complain. Anyway, and then for the non-reduced one, you get z also in case n is equal to 0, because it's path connected. OK, so we can see what I can do. One more example. So I do not think I will. So I want to do one more way of one more application, which is about attaching of cells. So what happens to the homology if you make your space more complicated by adding some piece, which looks like a disk, but connecting it wrong, the boundary. So let's see. So let dn, well, we have the dn we had now 1,000 times, but I mean, dn is the same as it always was. d0 to an minus 1 in n, so it's the sum in i squared is equal to 1, and it contains sn minus 1 in the usual way, the unit sphere. And we now want to say what we mean by attaching cells. So definition. So assume, so let y be a topological space. I mean, this is not, and let f from sn minus 1 to y be a continuous map. We want to somehow attach dn to y using this map, so that it's kind of only fixed along the sn minus 1 given by this map f. So we define y union over f dn to be, we take first the disjoint union of these two, model some equivalence relation. And it's with a quotient topology in the usual way. The quotient space where, so obviously, so the equivalence relation is induced by identifying a point in sn minus 1 with its image point in y. And everything else is not identified. So is the equivalence relation induced by p in sn minus 1 is equivalent to f of p in y. So if you want to do it explicitly, what it means is if you have a point in y, which does not lie in the image of sn minus 1, it's only, the equivalence class consists only of the point y itself. If you have a point in dn, which does not lie in sn minus 1, its equivalence class only consists of that point itself. And if you have a point in the image, then it's identified with any point which maps to it. No? And so this gives you, if you then give this thing the quotient topology from the disjoint union model of this thing, you get some nice topological space. And you can somehow think you have y here. And somehow this sn minus 1 gets mapped into y. And then you have to somehow, now this is not going to work, then you somehow have to add this dn just attaching it as this as a boundary. So we cannot really make a picture, but you have this thing here, which kind of sticks out. So OK. And now the statement is that again we get some exact sequence, which so lemma. So assume we are in this situation. So let f from sn minus 1 to y continues. And we have, we write now just yf for this space here. Then we have a long exact sequence. And I again will do it in reduced topology, because it is what works best. So say hk of sn minus 1 goes to, so by the map given f to hk of y goes to hk of yf goes to hk minus 1 of sn minus 1 and so on. And as I said, I wanted to do it with reduced topology. It also holds non-reduced, but with reduced it's better to use. So we have such a long exact sequence. So well, we just have to see that this is in a suitable sense an instance of the Myriviator sequence. But obviously, if it was just the Myriviator sequence, it would be with two trivial, you have to see why. So proof. And again, I will have to cheat a little bit, so to leave some of the more geometric sides as an exercise. So we have to write yf equal to u intersected v. So for instance, if we write, maybe I should write it here, we certainly have a map, or maybe write here, j is the map from dn to y union over f dn, which just sends a point here to its equivalence class. So this is a completely fine map. We see that if we, for instance, now take u. So we write it as union like that with u equal to, say, the image under j of the ball with radius 1 half. So this I just mean, you know, so what do I want? It's 1 to xn in dn, such that the sum of the xi squared is equal to 1 half, smaller equal to 1 half. So this is just some subset, and I think I have to take the open ball, so it's smaller than 1 half, because it's supposed to be open subsets. And v, I just take the whole thing minus the image of the center of the disk. So we have the 0.0 in the disk. We just take away this one point, make a little hole into this thing. Now clearly, yf is the union of these two sets, and they are open. So we can apply the Myriator's sequence, but there are a few things we can see. So j, if we restrict j to this disk, it's a homeomorphism, because there's nothing we identify once we are inside the disk. The identification is only on the boundary. So we have that this is homeomorphic to just the ball with this, and so it is contractable. So we have u is contractable. And what we did with more difficult to see is that v is a homotopy equivalent to y itself. Because if you think of it, once you have taken one point, you can again kind of, after all, I mean, one would have to write down this in formulas, and this would be an exercise. But you have this disk, vn, you have taken away one point. So the origin, you can certainly kind of retract this to the boundary by just moving out along the radii. And you can make a homotopy equivalence of this thing with y by taking the identity on y in this map on the n, and then taking the identification. And I claim you still get a nice homotopy, because you would have to work this out. So exercise proof this. So but once we have this, the statement is clear. Because again, this is a retraction. So nothing happens on the boundary. And the boundary is the only place where we have identified things. So it's in this picture, which I just wiped out, where we have this kind of something like a bubble on this thing. We just make a hole here, and we can contract it. And then we are left only with the thing we had before. So anyway, once we believe this, it's always simple. We apply the My-Vietto sequence, which says that if we take hk of u intersected v, this maps to hk of u plus hk of v, maps to hk of this wonderful space to hk minus 1 of u intersected v. And again, in the same situation, this space is contractable. No, no, no. So it's correct. But so u intersected r. So I haven't said what is u intersected v. Now look at the intersection of these two. We take this ball, and we take out this point. So this just looks precisely, again, like a smaller disk taking out one point. So that's homotopy equivalent to Sn minus 1. Sn minus 1. So this is just the k-thomology of Sn minus 1. u and v are so u is contractable. So it means all the reduced homology is 0. This is 0 plus. And we have seen that v is homotopy equivalent to y. And then we have hk of yf. And then we have hk minus 1 of Sn minus 1. So this is an example. So it somehow tells us that we have a way, if we, for instance, built our space by successively adding such pieces, we have a way to compute the homology in an inductive way. What? Ah, yeah, yeah, certainly. So we write yf as a union of two open subsets. Yes. We want to apply the Myviator sequence, and that is by doing that. And it was a misprint. But u intersected v. Here it is u intersected v is homotopy equivalent to a sphere. So now there is no further time. So one application of this result, which obviously will not be able to give you, because I first also would have to define this thing, is that one can now prove, for instance, if I'm in particular bad mood, I will make this an exercise, but we will have to see. So we can look at p and c. So this is the set of all, say, so this is cn plus 1 without the origin, divided by some equivalence relation, where we have that if we have some n tuple a0 to an is equivalent to p0 bn. So n plus 1 tuple of complex numbers is equivalent to another one, if and only if this vector is just a multiple of the other by a non-zero constant. So there exists a lambda in c without 0, such that, say, a0 to an is equal to lambda v0 until lambda bn. So we have this space. And one thing, there exists a lambda, lambda lambda, in a non-zero complex number, lambda, such that if I multiply this vector by lambda, I get the other vector. This is just the space of complex lines in cn plus 1. And what one can prove with this theorem, or also in other ways, but one can prove, so this can show. And this would be, but it is a bit complicated. I might make it an exercise if I give you enough hints, but I'm not yet sure. Namely, that hn of cpn, also hk of cpn is equal to 0 if k is odd. It's also 0 if k is bigger than 2n. And it's z if k is even and 0k. So here one would have a space where it is a little bit more complicated. So one case where you can check it is for instance, if you take cp, just to see that whether this might be so true, cp0 is just a point. And that's OK. You get just z for k equals 0, otherwise you get 0. And cp1, it's not so obvious, but one can prove that this is isomorphic to s2, not s1. s1 is over, if you have r, we are here over the complex numbers. And then again, you will find that this is true. But in general, you get this. So we have now a space where there are also a few more homology groups which are non-zero. But anyway, a much harder exercise would be to do the real projective space when the answer is also much more complicated. But that, I think, is almost impossible anyway. OK, so that's, I mean, I don't know. So that's maybe all I want to say now. So I have tried to introduce some of the main results and to do some examples. And we had very little times we didn't get very far. But anyway, I hope it is enough to at least get some impression and to be able to compute some examples and some applications. If one, I mean, algebraic topology is a big field. So if one really wants to study it, then one, you know, if you look, for instance, at the book of Hedger, which I partially use, I think I covered about maybe 25 out of 500 pages. So, but anyway, that's what is just the first impression. But I mean, it is, you know, you have seen some of the methods. OK, maybe I stop then, and then we will see. No, no, you? Well, I mean, that's one thing. I don't know, this may not. I mean, algebraic topology is not only about that. So I mean, it's one thing. One can study whether two spaces are homomorphic. Or you can also see, after all, we have proven that homology is invariant under homotopy. So it's not really about homomorphic. But even if they're homotopy equivalent, you get the same homology. So you can decide that. But you know, sometimes you can prove that certain things are impossible, like with this Rauer-Fick's point theorem and so on, because you can do certain things. It's not just about things being homomorphic. But in some sense, you can always think like that. If you want to decide anything in topology or in any other subject, you can also always reformulate this as a statement about maps. So about continuous maps or something like that. And so the homology, if you have a continuous map, it will give you a map on homology. And sometimes it's impossible that such a map exists because the homology groups do not fit. And so you can see that certain things are impossible because you cannot do that. And sometimes you also can find the converse. And then one thing which I did not say is that homology actually has something to do with the fact what the space looks like. I mean, it's not clear here, but I had said in the very beginning something that if you have the zeroth homology tells you how many pieces there are, so you can really see it. The first homology tells you something about certain kind of holes. I mean, you can kind of imagine that if you can make a simplex around the hole in your space, then it is a boundary if you can somehow fill it up in the space. And so you get non-zero homology if there will be a hole. And there's some way how you can make this kind of precise. And there are also other things. There is co-homology, which is dual to homology. It's not much more difficult than homology, but it has some extra features. And one of the things you have with homology, and this then is, for instance, also related to dirham homology. It somehow turns out to be the same as dirham homology, something with differential forms. And dirham homology, for one thing you could say, it's just another way to compute homology, which is most as the same as homology. But in fact, dirham homology is about solving differential equations. So it means homology or homology tells you something about solving differential equations. You can, by computing, doing certain computations in algebraic topology, you can decide whether certain differential equations can be solved or not. Certain partial differential equations are many forms. And there are other things. So often when you have some space, you can make some construction on top of the space. And this construction will give you a certain homology class. And so then maybe there are some reasons why this homology class cannot exist or cannot do what you see. And so you see that this construction that you would like to make cannot be there. So there are many cases. It's not just about two things being homomorphic. That's just the kind of causes thing. But it's kind of a more universal tool to, you always want to associate some groups to your question and some homomorphism between the groups to the question. And then you have translated your problem into a different one.