What is meant by "A and B represent the same functor whence are isomorphic" in this solution? 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.

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.