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I am trying to understand the idea of an exponential in category theory. If we start with $\mathbf{Set}$, how do you prove that the function set $\text{Hom}(X,Z)$ is an exponential for the sets $X$ and $Z$? Does it suffice to prove the bijection $$\text{Hom}(Y\times X,Z)\cong \text{Hom}(Y,\text{Hom}(X,Z))$$ Also, something that confuses me...What is the property that $\text{Hom}(X,Z)$ has, exactly? It is the set, up to isomorphism, such that for every function ... there exists a unique ... such that what happens? I know that I am almost rephrasing the universal property, but I am asking if we can say something without reference to the evaluation map, just by mentioning the set $\text{Hom}(X,Z)$, to characterize it in some sense. I mean if, given the sets $X$ and $Z$, one would like to identify the set $\text{Hom}(X,Z)$ as the set $S$ that has property $\mathcal{A}$, what property would $\mathcal{A}$ be? Thanks!

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It is a set for which there exists a bijection of the form you mention that is natural in $Y$. – Zhen Lin Mar 11 '14 at 11:45
So, if you want to identify a set $S$ as an exponential, without mentioning the evaluation map, you have to actually mention the bijection and the naturality condition? I was thinking of a property that does not require any category theory notions...Like the very definition of an exponential object as a universal arrow... – trecer21 Mar 11 '14 at 11:54
How can you hope to express a notion from category theory without using category theory notions? – Zhen Lin Mar 11 '14 at 12:08
I am sorry, I think I did not make clear what I meant. I will make one last attempt. Is it possible to give some information about the sets $X$, $Y$ and $S$ (of course $X\times Y$ is ok as well), and functions between them, in elementary set-theory words, that will allow me to identify $S$ as the set of functions $X\to Y$? Something like : Given sets $X$ and $Y$, the set $S$ is the unique set such that..." This is what I am asking. – trecer21 Mar 11 '14 at 13:25
Given sets $X$ and $Z$, the set $S=\text{Hom}(X,Z)$ is the exponential $Z^X$, which means that $S$ is the unique set such, that for each $Y$ there is an isomorphism [blah blah]. If you want to express this condition in set-theoretical and not categorical context, (I suggested in my answer that) you do the translation from arrows of the category to functions with the set theoretical definition and then work on that. – frabala Mar 11 '14 at 13:40

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up vote 4 down vote accepted

The definition of the exponential is:

Let the category $\mathcal{C}$ have binary products. An exponential of objects $B$ and $C$ consists of an object $C^B$ and an arrow $\varepsilon_{C,B}: C^B \times B \to C$ such that, for any object $A$ and arrow $f :A\times B \to C$ there is a unique arrow $\tilde{f} : A \to C^B$ such that $\varepsilon_{C,B}\circ (\tilde{f} \times \mathbf{1}_B ) = f$.

In locally small categories (like Set), $S=\text{Hom}(X,Z)$ is the set of all morphisms with domain $X$ and codomain $Z$. So, the elements of $\text{Hom}(X,Z)$ are of the form $f:X\to Z$.

So, regarding the category Set, you have to show that the isomorphism you have written in your post holds for any object $Y$ in Set. This is equivalent to showing that for any set $Y$, there is a bijection $$\bar{(-)}_Y:\text{Hom}(Y,\text{Hom}(X,Z))\to\text{Hom}(Y\times X,Z)$$ Turns out that the bijection is $$\bar{(-)}_Y=\varepsilon_{Z,X}\circ (- \times \mathbf{1}_X )$$ where $\mathbf{1}_X(x):X\to X$, with $\mathbf{1}_X(x)=x$, and $\varepsilon_{Z,X}:\text{Hom}(X,Z)\times X\to Z$, with $\varepsilon_{Z,X}(h,x)=h(x)$. So, the evaluation function cannot be detoured.

If you show that $\bar{(-)}_Y$ as defined above is a bijection for each $Y$, then you can conclude that $S=\text{Hom}(X,Z)=Z^X$ (in other words, $S$ is the exponential of $Z$ and $X$), because you can show that $S$ satisfies all properties of the exponential's definition, $Z^X$. This actually proves that all Hom-sets over Set are exponential objects of Set.

If you want to go out of the categorical context, you need to translate each function $f:A\to B$ as the set $f=\{(a,b): a\in A, b\in B\}$ such that $\forall (a,b_1),(a,b_2)\in f, b_1=b_2$, and then show the desired bijection within the set theoretical context. But this could be quite messy, because you'll have to manipulate not only functions-as-sets, but also sets of functions-as-sets.

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In the other post, I asked how to prove the bijection of sets. Here, my question was, what is needed to be proved in order to prove that $\text{Hom}(X,Z)$ is an exponential. Thanks for your answer. Please correct me if I am wrong: If by $\text{Hom}$ I mean the set of functions from a set to an other set (not the arrows, so that there is no reference to category theory), the requirement that the sets $\text{Hom}(Y,S)$ and $\text{Hom}(Y\times X,Z)$ are isomorphic for all $Y$, where $Y\times X$ is the cartesian product of the sets $Y$ and $X$, implies that $S=\text{Hom}(X,Z)$, up to isomorphism. – trecer21 Mar 11 '14 at 15:07
@trecer21 I don't think so... You can see this by playing with an example. For a very simple one, put $X=\{a,b\}$ and $Z=\{A,B\}$. Then $f$ and $g$ defined as $$f(a)=A\text{ and }f(b)=B\\g(a)=B\text{ and } g(b)=A$$ are different but isomorphic. Check what's happening for $Y=\{c\}$. You will produce two different functions in $\text{Hom}(Y\times X,Z)$ and therefore two different functions in its isomorphic Hom-Set. So no, $\text{Hom}(X,Z)$ is up to isomorphism only if you have already defined Hom-sets to be up to isomorphism, which you haven't. – frabala Mar 11 '14 at 15:45
I am very confused now...Please, one last question, I hope this clarifies things. If the set of functions from a set $Y$ to a set $S$ is in bijection with the set of functions from the cartesian product $Y\times X$ to $Z$, what can we say about $S$? I mean the elementary (if you want high school) definitions of the words function and set and cartesian product, as we use them in, say, real analysis... – trecer21 Mar 11 '14 at 17:29
@trecer21 If the set of functions from a set $Y$ to a set $S$ is in bijection with the set of functions from the cartesian product $Y\times X$ to $Z$, then what you can say about $S$ is that it is the Hom-set Hom$(X,Z)$ (and Hom-sets are not defined up to isomorphism). That is all. A sketch for the proof of this is in my post. And if you replace all arrows with functions and all objects with sets, then you get a set theoretical proof for this. – frabala Mar 11 '14 at 17:48
Yes, but what about the fact that $\text{Hom}(X,Z)$ is an exponential? Aren't exponentials unique up to isomorphism? Thank you, anyway, for all your help! – trecer21 Mar 11 '14 at 20:59

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