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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)

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It's the same as the uniqueness proof for any universal object: define a category such that the object you want is an initial or terminal object in that category. –  Qiaochu Yuan May 25 '12 at 19:23
Please define what you mean by exponential object, because the one that immediately comes to mind uses adjoints. –  Zhen Lin May 25 '12 at 19:39
$b^a$ is an exponential object of two objects $a,b$ in a category $\mathcal{C}$ if there is an associated evaluation function $eval:b^a \times a \rightarrow b$ such that for any object $c$ in $\mathcal{C}$ there is a unique arrow $\lambda g: c \times a \rightarrow b$ such that $eval \circ (\lambda g \times 1_a) = g.$ –  Anas May 25 '12 at 20:07

3 Answers 3

up vote 3 down vote accepted

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.

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Thanks for the worked out solution. Yeah I think I need to work with another text alongside Goldblatt's Topoi, one that introduces functors and adjoints much earlier. –  Anas May 26 '12 at 20:58

Awodey's great book, 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)

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Thanks for the recommendation and link. –  Anas May 26 '12 at 21:02
I am glad it helped :-) –  magma May 27 '12 at 9:12

$b^a$ 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 ...

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