Like what the title said, I want to use Yoneda's lemma to "guess" the definition of exponential object in $SET$. So basically I want to say that given any two sets $A, B$ in $SET$, the exponential object "$A^B$" exists in the category of sets. So let's pretend for a moment, that a person has no notions of set theory, how would one give an good guess on how we should define "$A^B$". I have seen similar things being done using Yoneda's lemma on the category of presheafs, but I want to take a step back even further, just to see how to use Yoneda's lemma in this manner.



This depends on how you define the exponential object at all. Without any knowledge of set theory, how do you motivate the isomorphism $\mathcal{C}(A, B^C) \simeq \mathcal{C}(A \times C, B)$?

If you accept that, though, it becomes fairly clear. $A^B \simeq \mathcal{Set}(1, A^B) \simeq \mathcal{Set}(1 \times B, A) \simeq \mathcal{Set}(B, A)$. So $A^B$ is the set of functions from $B$ to $A$ (if it exists at all: you'll still need to check that $\mathcal{Set}(A, B^C) \simeq \mathcal{Set}(A \times C, B)$ in general).

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  • $\begingroup$ ah thanks. So I don't really need yoneda to for the "guess", I just need to accept (check) that the natural isomorphism you stated. $\endgroup$ – user16319 Nov 20 '19 at 21:38
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    $\begingroup$ @user16319 Yes, though in a sense the Yoneda lemma is needed to show that $\mathcal{C}(A, B^C) \simeq \mathcal{C}(A \times C, B)$ uniquely defines the exponential object. That is, if $\mathcal{C}(A, E) \simeq \mathcal{C}(A \times C, B) \simeq \mathcal{C}(A, E')$, then $E \simeq E'$. $\endgroup$ – SCappella Nov 21 '19 at 4:24

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