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Suppose that $A, B$ are objects of a category with all finite products and exponentials, $\mathbb{C}$. Show that there is a bijection between hom$(A, B)$ and hom$(1, B^A)$ where $1$ is a terminal in $\mathbb{C}$.

My attempt: Consider the isomorphism given by transposition, hom$(1 \times A, B) \cong$ hom$(1, B^A)$. It suffices to show that hom$(A, B) \cong$ hom$(1 \times A, B)$. Since $1$ is terminal, there is unique morphism from any object $Z$ to $1$, $k_Z$. Consider $g \in $ hom$(A, B)$. We can define an isomorphism by $g \mapsto k_A \times g$. This must be an isomorphism since $k_A$ is unique.

This seems correct to me but I'm a little doubtful.

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Your idea is right, but I don't think your proposed isomorphism $\newcommand\Hom{\operatorname{Hom}}\Hom(A,B)\simeq \Hom(1\times A,B)$ makes any sense at all.

I.e., you propose sending $g : A\to B$ to $k_A\times g : 1\times A \to A\times B$, which is not an element of $\Hom(1\times A,B)$.

Instead, it suffices to show that $A\simeq 1\times A$, since then $\Hom(A,B)\simeq \Hom(1\times A,B)$.

For this, we use the Yoneda lemma. Since we have $$\Hom(Z,1\times A)\simeq \Hom(Z,1)\times \Hom(Z,A) \simeq \{*\}\times \Hom(Z,A)\simeq \Hom(Z,A),$$ the Yoneda lemma tells us that $1\times A\simeq A$.

Edit

If you're not familiar with the Yoneda lemma, Matematleta points out that you can also see $A\simeq 1\times A$ by noticing that $(A,(k_A,\textrm{id}_A))$ is also a product for $1$ and $A$. (If $f:Z\to 1$ and $g:Z\to A$, then $g$ is the unique map from $Z$ to $A$ such that $k_A\circ g = f = k_Z$, and $\textrm{id}_A\circ g = g$).

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  • $\begingroup$ Isn't this overkill? $A\simeq 1\times A$ simply because $(A,\langle id_A,!\rangle)$ is also a product. $\endgroup$ – Matematleta Mar 20 '19 at 2:28
  • $\begingroup$ @Matematleta Sure, that also works, I just personally find this more immediately clear. But that's also quite clear and nice. I think I just generally like to think of products as representing functors in the first place. $\endgroup$ – jgon Mar 20 '19 at 2:33
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    $\begingroup$ I only pointed this out because from the OP's attempt, it seemed to me that he/she hasn't seen Yoneda yet. No problem with your answer (which I was happy to upvote!), I think of products that way too, --- $\endgroup$ – Matematleta Mar 20 '19 at 2:37
  • $\begingroup$ @Matematleta Hm, good point, I'll edit that in as a second viewpoint. $\endgroup$ – jgon Mar 20 '19 at 2:39
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    $\begingroup$ @Matematleta Thanks! I was trying to read up on Yoneda lemma, but the isomorphism you provided is cleaner. $\endgroup$ – real_father Mar 20 '19 at 2:58

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