# Bijection between hom$(A, B)$ and hom$(1, B^A)$

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.

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 $$\textrm{id}_A\circ g = g$$).

• Isn't this overkill? $A\simeq 1\times A$ simply because $(A,\langle id_A,!\rangle)$ is also a product. – Matematleta Mar 20 '19 at 2:28
• @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. – jgon Mar 20 '19 at 2:33
• 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, --- – Matematleta Mar 20 '19 at 2:37
• @Matematleta Hm, good point, I'll edit that in as a second viewpoint. – jgon Mar 20 '19 at 2:39
• @Matematleta Thanks! I was trying to read up on Yoneda lemma, but the isomorphism you provided is cleaner. – real_father Mar 20 '19 at 2:58