Is the exponential object $B^A$ related to $hom_C(A,B)$? In category $Set$, do $B^A$ and $hom_{Set}(A,B)$ mean the same: the set of all the functions from $A$ to $B$?
In a category $C$, is the exponential object $B^A$ of objects $A$ and $B$ related to $hom_C(A,B)$? 
Thanks.
 A: For a cartesian closed category, at least, we have $$\mathsf{Hom}(A,B)\cong\mathsf{Hom}(1\times A,B)\cong\mathsf{Hom}(1,B^A)$$ where $1$ is the terminal object. This would be described as the arrows $A\to B$ correspond to the global elements (or global points) of $B^A$ (which just means the arrows $1\to B^A$).
Crucially, however, in most categories (with $\mathbf{Set}$ being the main exception) an object isn't determined by its global elements. For example, in the category of $M$-sets for a monoid $M$, an arrow $1\to (X,\alpha)$ is an element of $X$ such that $\alpha(m,x)=x$ for all $m\in M$. There are plenty of non-trivial $M$-sets where this is never true meaning there are no global elements. For example, consider the action of $\mathbb N$ on itself via addition.
The upshot is that there can be a lot more going on in $B^A$ than the global elements (i.e. the morphism $A\to B$) suggest. (It's also possible to define $B^A$ in a category that doesn't have a terminal object.)
A: Yes. Intuitively, the exponential object $B^A$ is somehow the "object version" of $Hom(A,B)$ - or at least can be interpreted as such after we add to the picture a morphism satisfying an appropriate universal property.
Specifically, remember that (fixing $A,B$) the (up to iso) exponential $B^A$ is an object $X$ together with an evaluation morphism $$ev: X\times A\rightarrow B$$ (we're working in a category with finite products, so this in fact makes sense) such that for any object $U$ and any morphism $m:U\times A\rightarrow B$ there is a unique map $h: U\rightarrow X$ such that the appropriate diagram commutes. 
Roughly speaking, in this definition we're viewing $m$ as a "$U$-indexed family" of maps from $A$ to $B$ (after de-Currying, $m$ takes an element of $U$ and outputs a map from $A$ to $B$); saying that $X$ represents the collection of maps from $A$ to $B$ says that we can "interpret" $U$ in $X$ appropriately (this is our $h$).

EDIT: That said, don't push this sort of set-theoretic analogy too far - basic intuitions about how sets work can break down in natural categories, per Derek Elkins' answer. 
