Exponentials in $\mathbf{Set}$ 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!
 A: 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.
