Uniqueness of Exponential Objects up to Isomorphism in any Category I want to prove that for any pair of objects $a,b$ in a category $\mathcal{C}$, the exponential object $a^b$ of $a$ and $b$, if it exists, is unique up to isomorphism. It looks to be really simple, but I can't figure out the proof myself. (I need a proof that doesn't mention adjoints or Yoneda's lemma, since I'm working from a category theory textbook that hasn't defined these concepts yet)
 A: Awodey's great book (2nd ed), chapter 6, has a good introduction to exponentials, without Yoneda, ect.
The proof works out in the same way that is used for products (for example) or any other universal contruction. 
Remember: in the definition of an exponential E1 there is a unique arrow, so if you suppose that there is another exponential E2......in the end there must be a unique arrow from E1 to E2 and a unique arrow from E2 to E1, so E1 and E2 must be isomorphic. 
Study the technique used for products, equalizers or pullbacks for inspiration (always in Awodey's book)
A: Shortest answer. Right adjoints are unique up to unique isomorphism.
Short answer. The definition of the exponential object $b^a$ implies there is a natural bijection between arrows $c \to b^a$ and arrows $c \times a \to b$; the latter does not depend on $b^a$, if $d$ also has the universal property of $b^a$, then there is a natural bijection between arrows $c \to b^a$ and arrows $c \to d$, hence, $b^a$ must be isomorphic to $d$, by the Yoneda lemma.
Long answer. Let's spell out exactly where the isomorphism comes from. Suppose $d$ has the universal property of $b^a$: so there is a universal morphism $e : d \times a \to b$ such that for any $g : c \times a \to b$, there is a unique $g' : c \to d$ such that $e \circ (g' \times \textrm{id}_a) = g$. But in particular we can take $c = b^a$ and $g = \textrm{eval}$, and this gives us $g' : b^a \to d$ such that $e \circ (g' \times \textrm{id}_a) = \textrm{eval}$. But by the universal property of $b^a$ there is a unique $\lambda e : d \to b^a$ such that $\textrm{eval} \circ (\lambda e \times \textrm{id}_a) = e$, so we get
$$\textrm{eval} \circ (\lambda e \times \textrm{id}_a) \circ (g' \times \textrm{id}_a) = \textrm{eval}$$
but $(\lambda e \times \textrm{id}_a) \circ (g' \times \textrm{id}_a) = (\lambda e \circ g') \times \textrm{id}_a$, hence $\lambda e \circ g' = \textrm{id}_{b^a}$ by uniqueness. Conversely, we have
$$e \circ (g' \times \textrm{id}_a) \circ (\lambda e \times \textrm{id}_a) = e$$
so by uniqueness again we conclude $g' \circ \lambda e = \textrm{id}_d$. Thus $d \cong b^a$.
Remark. All of these answers are actually the same, just expressed in different ways. As others have said: go learn the Yoneda lemma and representable functors.
A: $b^a$ is by definition a representation (or just the representing object) of the functor $c \mapsto \hom(c \times a , b)$, i.e. we have a natural bijection $\hom(c,b^a) \cong \hom(c \times a,b)$. The Yoneda Lemma tells us that representations are unique up unique isomorphism.
PS: First learn the Yoneda Lemma, then the rest ...
