{"action":"created","comment":{"url":"https://api.github.com/repos/HoTT/book/pulls/comments/35286249","id":35286249,"diff_hunk":"@@ -1441,6 +1441,245 @@ \\subsection*{Solution to \\cref{ex:acconn}}\n But also by \\cref{ex:connectivity-inductively}, each $Y(x)$ is $(-1)$-connected and all its path types are $n$-connected.\n Applying function extensionality to characterize the path types of $\\prd{x:X} Y(x)$, the claim follows from the $(-1)$-connected and $n$-connected axioms of choice.\n \n+\n+\\section*{Exercises from \\cref{cha:homotopy}}\n+\n+\\subsection*{Solution to \\cref{ex:homotopy-groups-resp-prod}}\n+Remember that the fundamental group of a pointed space is defined as the\n+$0$-truncation of the loop space of that space.\n+If we prove the equivalence without applying $\\trunc 0{\\blank}$ to both sides\n+we are done.  We use induction on the natural numbers, with base case $1$.\n+\n+Suppose we have two pointed types $(A,a)$ and $(B,b)$.\n+For the base case, we have $\\eqv{\\Omega(A \\times B)}{\\Omega(A) \\times \\Omega(B)}$\n+because it is, by definition, the equivalence of pointed types:\n+\n+\\[\\eqv{ \\Parens{(a,b)=(a,b), \\refl{(a,b)}} }{\\Parens{(a=a \\times b=b), (\\refl{a}, \\refl{b})}}.\\]\n+\n+By Univalence Axiom this is the same as proving the equality.\n+By the characterization of \\cref{thm:path-sigma} (and using Univalence) we see\n+that it suffices to give a:\n+\\[p:\\eqv{\\Parens{(a,b)=(a,b)}}{\\Parens{(a=a)\\times (b=b)}}\\]\n+\n+And to show that transporting $\\refl{(a,b)}$ along $\\ua(p)$ is equal to\n+$(\\refl{a}, \\refl{b})$.  To construct $p$ just use \\cref{thm:path-prod}. To\n+show that transport respects the equality we use\n+the propositional computation rule in \\cref{axiom:univalence} and the definition\n+of our function $p$, that is just an application of the projections and\n+functoriality (see \\cref{eq:path-prod}).\n+\n+For the inductive step we use the inductive hypothesis to get an equivalence:\n+$\\eqv{\\Omega^n(A \\times B)}{\\Omega^n(A) \\times \\Omega^n(B)}$.\n+Then we apply $\\Omega(\\blank)$ on both sides and use exactly the same\n+propositions we used in the base\n+case. This settles the inductive case, because of the inductive definition of\n+$\\Omega^{\\suc(n)}(\\blank)$.\n+\n+\n+\\subsection*{Solution to \\cref{ex:contr-infinity-sphere-colim}}\n+\n+To solve this exercise we must first define the spheres as a type family\n+$\\Sn^{\\blank} : \\nat \\to \\UU$.\n+Then we must define the inclusions that appear the in the diagram:\n+\\[ \\Sn^0 \\to \\Sn^1 \\to \\Sn^2 \\to\\cdots \\]\n+\n+Then we will be able to define the colimit as\n+a higher inductive type.\n+\n+So, by induction on the natural numbers, we define $\\Sn^{\\blank} : \\nat \\to \\UU$\n+with the base case being the two point type $\\bool$, and the inductive case\n+being the iterated application of the suspension.\n+\n+Now we define the inclusions $i_n : \\Sn^n \\to \\Sn^{n+1}$.\n+For the base case we have to give a function $i_0 : \\bool \\to \\susp \\bool$.\n+This is easy: send one point to $\\north$ and the other to $\\south$.\n+\n+For the inductive case we notice that the domain of the function $i_n$ that\n+we have to define is the suspension $\\susp \\Sn^{n-1}$. So we can use the\n+induction of $\\susp \\Sn^{n-1}$. The codomain is also a suspension, so we can\n+use any of the constructors:\n+\\begin{align*}\n+  \\north_{n+1} &: \\Sn^{n+1}\\\\\n+  \\south_{n+1} &: \\Sn^{n+1}\\\\\n+  \\merid_{n+1} &:\\Sn^n \\to \\north_{n+1} = \\south_{n+1}\n+\\end{align*}\n+to define our function.\n+We send $\\north_n \\mapsto \\north_{n+1}$ and $\\south_n \\mapsto \\south_{n+1}$.\n+Now we must give a $\\susp \\Sn^{n-1}$-indexed family of equalities between\n+$\\north_{n+1}$ and $\\south_{n+1}$. By inductive hypothesis we have\n+$i_{n-1} : \\Sn^{n-1} \\to \\Sn^n$, so we can use:\n+\\[ \\merid_{n+1} \\circ i_{n-1} \\]\n+It is also interesting to note that this construction works in a more general\n+setting: any function $f : A \\to B$ induces a function $\\susp f : \\susp A \\to \\susp B$.\n+\n+Now is a good time to make a drawing to convince oneself that this is the\n+right way to define the inclusions between consecutive spheres, and that this\n+is the diagram intended in he exercise.\n+\n+Lets define the type $\\Sn^{\\infty}$ as the colimit of the diagram we just\n+constructed. The constructors are:\n+\\begin{align*}\n+  j_{\\blank} &: \\prd{n : \\nat} \\Sn^n \\to \\Sn^{\\infty}\\\\\n+  \\glue_{\\blank} &: \\prd{n : \\nat} \\prd{x: \\Sn^n} j_n(x) = j_{n+1}\\circ i_n (x)\n+\\end{align*}\n+And the induction:\n+\\[\n+  \\ind{\\Sn^{\\infty}} : \\prd{C:\\Sn^\\infty \\to \\UU} \\prd{J_{\\blank} : \\prd{n : \\nat} \\prd{x : \\Sn^n} C(j_n(x))}\n+  \\Parens{\\prd{n:\\nat} \\prd{x:\\Sn^n} \\dpath C {\\glue_n(x)} {J_n(x)} {J_{n+1}\\circ i_n (x)}}\n+  \\to \\prd{e:\\Sn^{\\infty}} C(e)\n+\\]\n+\n+We also have the usual computational rules.\n+Now, we can interpret the proof that we are going to give as follows.\n+Each sphere is the ``equator'' of the next one. We say ``equator'' because it passes\n+trough $\\north$ and $\\south$ so technically you have to rotate the bigger sphere for\n+the inclusion to be exactly in the equator.\n+The intuitive idea is that we can retract to a point a circle that is the equator of a sphere.\n+Just as we can retract to a point two points that are the equator of a circle.\n+This idea extends to all $n$-spheres. If we have all the spheres at the same time\n+we can retract them all. But we have to be careful: the homotopies\n+must be coherent for the total homotopy to be well defined. This can be done in\n+classical homotopy theory using the fact that a CW complex is the colimit of\n+the inclusions between its skeletons. In our case we constructed the colimit.\n+So, if we define a homotopy for each sphere, and prove that this functions\n+coincide in the glued parts (respect the equalities given by $\\glue$), by\n+induction on the colimit we get a homotopy defined in the colimit.\n+\n+The outline of the argument is as follows. It suffices to show that every point in\n+$\\Sn^{\\infty}$ is equal to $j_0(\\north_0)$, we prove it in two steps.\n+The first step is to prove it only for the points of the form $j_n(\\north_n)$.\n+For the second step we will take an arbitrary $x:\\Sn^{\\infty}$\n+and, by induction, we will assume it comes from a $y : \\Sn^n$ for some $n$,\n+that is $x \\defeq j_n(y)$.\n+Then we note that using the commutative diagram given by $\\glue_n(x)$:\n+\\begin{center}\n+\\begin{tikzpicture}\n+\\node (N0) at (0,2) {$\\Sn^{n}$};\n+\\node (N1) at (2,2) {$\\Sn^{n+1}$};\n+\\node (N2) at (2,0) {$\\Sn^{\\infty}$};\n+\\node (N3) at (1.1,1.1) {};\n+\\node (N4) at (1.5,1.5) {};\n+\\draw[->] (N0) -- node[above]{\\footnotesize $i_n$} (N1);\n+\\draw[->] (N0) -- node[left]{\\footnotesize $j_n$} (N2);\n+\\draw[->] (N1) -- node[right]{\\footnotesize $j_{n+1}$} (N2);\n+\\draw[double, double equal sign distance] (N3) -- node[left,above]{\\footnotesize $glue_n$} (N4);\n+\\end{tikzpicture}\n+\\end{center}\n+we get a path $x \\defeq j_n(y) = j_{n+1} \\circ i_n(y)$.\n+But, as we will show, the image of $\\Sn^n$ under $i_n$ is contractible,\n+every point in the image is equal to $\\north_{n+1}$. Thus, we are able to show that\n+$j_{n+1}\\circ i_n(y) = j_{n+1}(\\north_{n+1})$. Composing the proof of the first step\n+with the proof of the second step we conclude the exercise.\n+\n+To construct a $D_{\\blank} : \\prd{n:\\nat} j_n(\\north_n) = j_0(\\north_0)$ we\n+proceed by induction on $n$. For the base case we can \n+use $\\refl{j_0(\\north_0)}$. For the inductive case, we have by inductive\n+hypothesis $j_n(\\north_n) = j_0(\\north_0)$. By our definition of $i_{\\blank}$,\n+we have that $\\north_{n+1} \\equiv i_n(\\north_n)$. So $j_{n+1} (\\north_{n+1})$\n+equals $j_{n+1}\\circ i_n(\\north_n)$. By concatenation with $\\glue_n(\\north_n)$,\n+we reduce our goal to the inductive hypothesis.\n+\n+Lets now show that the inclusion of $\\Sn^n$ in $\\Sn^{n+1}$ can be\n+continuously retracted to $\\north_{n+1}$.\n+That is, lets construct a homotopy:\n+\\[ H_{\\blank} : \\prd{n:\\nat}\\prd{x:\\Sn^n} i_n(x) = \\north_{n+1} \\]\n+For the case $n\\equiv 0$ we know that $i_0(\\btrue) \\equiv \\north_1$ and\n+$i_0(\\bfalse) \\equiv \\south_1$.  This is because we constructed the inclusion\n+that way. So we can prove the equalities\n+using $\\refl{\\north_1}$ and $\\merid_1(\\btrue) : \\north_1 = \\south_1$.\n+\n+For the inductive case we defined, previously, $i_n(\\north_n) \\equiv \\north_{n+1}$\n+and $i_n(\\south_{n}) \\equiv \\south_{n+1}$. So we can prove the equalities using\n+$\\refl{\\north_{n+1}}$ and $(\\merid_{n+1}(\\north_n))^{-1}$.\n+Then we have to prove that the function respects $\\merid_n$:\n+\\[ \\prd{x:\\Sn^{n-1}} \\dpath {x\\mapsto (i_n(x) = \\north_{n+1})} {\\merid_n(x)} {\\refl{\\north_{n+1}}} {(\\merid_{n+1}(\\north_n))^{-1}} \\]\n+By \\cref{thm:transport-path} (and some straightforward computation) this reduces to:\n+\\[ i_n(\\merid_n(x)) = \\merid_{n+1}(\\north_{n}) \\]\n+But, by our definition of $i_{\\blank}$, and the computation rule of the\n+suspension induction $i_n(\\merid_n(x))$ equals $\\merid_{n+1} \\circ i_{n-1}(x)$.\n+And, by inductive hypothesis, $i_{n-1}(x) = \\north_n$, which gives us the\n+desired result.\n+\n+Composing the two proofs we just gave we get a function:\n+\\[ J_{\\blank} \\defeq \\glue_n(x)\\ct \\apfunc{j_{n+1}}{H_n(x)}\\ct D_{n+1}","path":"exercise_solutions.tex","position":164,"original_position":164,"commit_id":"019f39ae63cd39e01a4383268d100404140712ad","original_commit_id":"019f39ae63cd39e01a4383268d100404140712ad","user":{"login":"mikeshulman","id":646600,"avatar_url":"https://avatars.githubusercontent.com/u/646600?v=3","gravatar_id":"","url":"https://api.github.com/users/mikeshulman","html_url":"https://github.com/mikeshulman","followers_url":"https://api.github.com/users/mikeshulman/followers","following_url":"https://api.github.com/users/mikeshulman/following{/other_user}","gists_url":"https://api.github.com/users/mikeshulman/gists{/gist_id}","starred_url":"https://api.github.com/users/mikeshulman/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/mikeshulman/subscriptions","organizations_url":"https://api.github.com/users/mikeshulman/orgs","repos_url":"https://api.github.com/users/mikeshulman/repos","events_url":"https://api.github.com/users/mikeshulman/events{/privacy}","received_events_url":"https://api.github.com/users/mikeshulman/received_events","type":"User","site_admin":false},"body":"I think if J is to have the type given that is a double dependent product, then it needs to be abstracted over n and x, i.e. to start with `\\lam{n}{x}`.  In that case I would also write just `J \\defeq ` at the beginning.  Or you could instead write `J_n(x) \\defeq` and give the term you've written, but then the type should omit the two `\\prd`s.","created_at":"2015-07-23T02:49:39Z","updated_at":"2015-07-23T02:49:39Z","html_url":"https://github.com/HoTT/book/pull/840#discussion_r35286249","pull_request_url":"https://api.github.com/repos/HoTT/book/pulls/840","_links":{"self":{"href":"https://api.github.com/repos/HoTT/book/pulls/comments/35286249"},"html":{"href":"https://github.com/HoTT/book/pull/840#discussion_r35286249"},"pull_request":{"href":"https://api.github.com/repos/HoTT/book/pulls/840"}}},"pull_request":{"url":"https://api.github.com/repos/HoTT/book/pulls/840","id":40592324,"html_url":"https://github.com/HoTT/book/pull/840","diff_url":"https://github.com/HoTT/book/pull/840.diff","patch_url":"https://github.com/HoTT/book/pull/840.patch","issue_url":"https://api.github.com/repos/HoTT/book/issues/840","number":840,"state":"open","locked":false,"title":"solutions to exercises: 8.1 8.3 8.4 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