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While browsing some old questions I came across the following: tensor product of sheaves commutes with inverse image

It seemed like something interesting was going on in the answer, but I don't quite understand it. I understand the calculation on Hom sets, but I don't understand what is meant by:

"So $f^*\mathcal{M} \otimes_{\mathcal{O}_X} f^*\mathcal{N}$ and $f^*(\mathcal{M} \otimes_{\mathcal{O}_Y} \mathcal{N})$ represent the same functor, whence they are canonically isomorphic."

Can someone give a general statement of the fact being used here? I think this might have something to do with the Yoneda lemma, but unfortunately, I have never really been able to understand the Yoneda lemma very well. At some point I verified that the Yoneda lemma was indeed true, but my understanding was poor so it didn't really stick.

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Seth
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2 Answers2

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Let $\mathcal{C}$ be a (locally) small category and let $F \colon \mathcal{C} \to \bf{Set}\hspace{1mm}$ be a functor, then we say that $F$ is representable if there is a natural isomorphism between $F$ and $\textrm{Hom}_{\mathcal{C}}(A, -)$ for some object $A \in \mathcal{C}$. In this case, we say that the object $A$ ''represents'' the functor $F$.

A consequence of the Yoneda Lemma is that this object $A$ that represents a given functor $F$ is unique up to isomorphism (to prove this, either just appeal to the Yoneda embedding or see the comments for a direct approach). This is precisely what was used in the link you referenced: the only way the functors $\textrm{Hom}(f^* \mathcal{M} \otimes_{\mathcal{O}_X} f^* \mathcal{N}, -)$ and $\textrm{Hom}(f^*(\mathcal{M} \otimes_{\mathcal{O}_Y} \mathcal{N}), -) \hspace{1mm}$ can be naturally isomorphic is if the first arguments themselves are isomorphic.

msteve
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  • Thank you, that answers the question about what "fact" is being used. Now I just want to understand why this follows from the Yoneda lemma. (probably this is supposed to be obvious, but I'm still confused by the Yoneda lemma) – Seth Mar 04 '15 at 19:57
  • The wikipedia article on representable functors that I linked explains this pretty well (you're looking for 4.1 Uniqueness). If that isn't clear, I could re-explain it here. – msteve Mar 04 '15 at 19:59
  • Ok, I guess we use the fact that $h^-$ is a fully faithful functor from $C^\text{op}$ to $\text{Set}^C$, which follows from Yoneda embedding. – Seth Mar 04 '15 at 20:10
  • Oh you're right, I suppose we can also just appeal to the statement of Yoneda embedding. More directly (though this is saying the exact same thing), if $A_1,A_2$ represent the same functor, then there is a natural isomorphism $\textrm{Hom}(A_1,-) \to \textrm{Hom}(A_2,-)$. The Yoneda lemma says that such natural isomorphisms are in 1-1 correspondence with isomorphisms $A_1 \to A_2$. – msteve Mar 04 '15 at 20:20
  • I see, I guess I'm still struggling to understand how all this follows from the Yoneda lemma to be honest (I'll keep trying). I didn't realize that isomorphisms were preserved in the correspondence you mentioned. I guess this has something to do with naturality of the correspondence? (I really struggle to unpack all these definitions sometimes) – Seth Mar 04 '15 at 20:38
  • For the isomorphisms, you're correct that Yoneda gives an 1-1 correspondence between the natural transformations and $Hom(A_1,A_2)$, and the invertible elements in each will be mapped to one another under this correspondence (I may be using that the category $\mathcal{C}$ is enriched, but this is certainly true for the category of sheaves). (And I understand your pain - all the category theory didn't really sink in for me until I had a critical mass of examples to work with.) – msteve Mar 04 '15 at 21:29
  • This is just naturality, why would it have anything to do with enrichments? –  Mar 04 '15 at 21:48
  • Ah true, I was thinking something along the lines of invertible objects being sent to invertible ones under an isomorphism if these were rings, but that's unnecessary. – msteve Mar 04 '15 at 21:58
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Think of representing a functor as satisfying a universal property. Two objects satisfying the same universal property are isomorphic up to a unique isomorphism compatible with property, which means that they are canonically isomorphic.

See my answer here :

What is a universal property?

for details.

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