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So adjoint functors are one thing that has caused me endless pain since trying to learn algebraic geometry. They still haven't really "clicked" like they should have. I've been told that it makes it easier to work with the counit and unit of adjunction, but I tend to find these even harder.

I was trying to do what I assume is a common exercise (exercise II 5.5.3 of Hartshorne): Let $X = \text{Spec }A$ be an affine scheme and show that for any $A$-module $M$ and any sheaf of $\mathcal{O}_{X}$-modules $\mathcal{F}$, there is a natural isomorphism $$ \text{Hom}_{A} \left( M , \Gamma(X, \mathcal{F}) \right) \simeq \text{Hom}_{\mathcal{O}_{X}} \left( \widetilde{M}, \mathcal{F} \right). $$ In other words, show that the global section functor and the twidlification functor are an adjoint pair. The bijection seems easy enough. It's the naturality that is making it hard. I really wanted to use Yoneda's lemma here, so I tried that but ended up coming to a conclusion that (I think?) is absurd. This was my attempt:

Fix a sheaf of $\mathcal{O}_{X}$-modules $\mathcal{F}$. Define the functors \begin{align} h_{\Gamma(X, \mathcal{F})} &:= \text{Hom}_{A} \left( -, \Gamma(X, \mathcal{F}) \right) \\ G(-) &:= \text{Hom}_{\mathcal{O}_{X}} \left( \widetilde{-} , \mathcal{F} \right) \end{align} Then Yoneda's lemma tells us that there is a natural isomorphism (as functors into the category of sets) $$ \text{Nat} \left( h_{\Gamma(X, \mathcal{F})} , G \right) \simeq \text{Hom}_{\mathcal{O}_{X}} \left( \Gamma(X, \mathcal{F})\tilde{\,} , \mathcal{F} \right). $$ This is where I became confused. The above line seems to be suggesting that there can only exist a natural isomorphism between the functors $h_{\Gamma(X, \mathcal{F})}$ and $G $ if $\mathcal{F}$ is quasi-coherent. But then the exercise asks to show that this adjunction holds in the entire ambient category of $\mathcal{O}_{X}$-modules.

My guess is that I have misunderstood how to apply Yoneda's lemma here. In fact, I used it here specifically because I wanted some practice with applying it. Can Yoneda's lemma be used for this exercise at all? If so, where have I gone wrong?

As an aside: Does anyone have some general advice for learning and understanding adjoint functors and Yoneda's lemma well, particularly being able to show that two functors are adjoint? This is one concept that I keep struggling with despite coming to grips with other content in algebraic geometry. I've added a tag for soft question for this last remark/question.

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  • $\begingroup$ It's a good question. You might get a better response if you replaced "twidlification" with "sheafification", at least in the title. $\endgroup$ – Zach Teitler Oct 23 '17 at 7:55
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    $\begingroup$ @ZachTeitler I was tempted to call it sheafification, but I figured that word was already loaded... $\endgroup$ – Luke Oct 23 '17 at 7:58
  • $\begingroup$ My general advice is to find a whole bunch of examples of adjoint functors in other parts of mathematics that you like. As with understanding anything, it's useful to have lots of examples, but what's special about having examples of concepts in category theory is that you typically get more examples by learning more other branches of math, rather than by learning more category theory. Or rather, category theory has this property to an unusual extent, more than other branches of math. $\endgroup$ – Qiaochu Yuan Oct 23 '17 at 18:11
  • $\begingroup$ Also, once you can write these sorts of bijections down, naturality is an automatic consequence, so it is in fact not the hard part. That is, prove the following: a functor $G$ has a left adjoint iff $\text{Hom}(x, G(-))$ is always a representable functor (the representing object being $F(x)$ for $F$ the left adjoint), and similarly for right adjoints. In other words, adjoints exist iff they exist "pointwise." $\endgroup$ – Qiaochu Yuan Oct 23 '17 at 18:28
  • $\begingroup$ @QiaochuYuan, I have finally gotten back to this question after thinking about a few other related things in the meantime. This is potentially a stupid question since I am still new to this stuff, but given your comment we would still need to show that the left adjoint of $G$ is actually $F$, since all we can conclude so far is that it has a left adjoint, right? My understanding is that the fact that $F$ is the desired left adjoint should be immediate from the Yoneda lemma. Is that correct? $\endgroup$ – Luke Nov 2 '17 at 2:57
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To start with, proving that the bijection is natural in this case is simple. Recall that the isomorphism $\mathrm{Hom}(\tilde{M}, F)\to \mathrm{Hom}(M,\Gamma(F))$ just takes a morphism $f:\tilde{M}\to F$ to the morphism on global sections, since $\Gamma(\tilde{M})=M$. This is obviously natural: it doesn't matter if I compose first, then take global sections, or vice-versa. (Try convincing yourself of this just by righting the diagrams.)

What is interesting is to sort out your confusion regarding Yoneda. Note that in $\mathrm{Hom}(\tilde{F_X}, F)$ (I'm writing $F_X:=F(X)=\Gamma(X,F)$ to simplify notation) there is a distinguished morphism, namely the co-unit. An indeed this can only be an isomorphism if $F$ is quasi-coherent. This is because the co-unit being an isomorphism implies that the adjunction is an equivalence, and indeed we get an equivalence only when we restrict to quasi-coherent sheaves.

To round things off, note that when you formulated your Yoneda argument, you were indeed asking when the adjunction is an equivalence. The sheafification is full and faithful (this is easy to see), so to know when the adjunction is an equivalence, you need to know what is the essential image of the functor. In other words, you want to know when $F=\tilde{M}$ for some $M$. But since Yoned is an embedding, this is the same as asking when $$\mathrm{Hom}(\tilde{\cdot},F)\simeq\mathrm{Hom}(\tilde{\cdot},\tilde{M})\simeq\mathrm{Hom}(\cdot, M)$$ where the last equality is again because sheafification is full and faithful. But using the adjunction, you know that $\mathrm{Hom}(\tilde{\cdot},F)\simeq \mathrm{Hom}(\cdot,\Gamma (F))$, and again by the fact that Yoneda is an embedding, we must have $M\simeq\Gamma(F)$. Therefore, to check the essential image of the sheafification, we need only chech when $h_{\Gamma(F)}\simeq \mathrm{Hom}(\tilde{\cdot},F)$.

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