1

Goal : Let $f: (X,\mathcal{O}_X ) \to (Y, \mathcal{O}_Y)$ be a morphism of ringed spaces. Let $\mathcal{F}$ be a (quasi-coherent) $\mathcal{O}_X$-module, which is generated by a family $(\gamma_i)_{i\in I} \subseteq \Gamma(X,\mathcal{F})$. Then the counit map ( canonical morphism ) $c: f^{*}(f_{*}\mathcal{F}) \to \mathcal{F}$ is surjective ?

EDIT : Here the quasi-coherence of $\mathcal{F}$ over a ringed space $(X,\mathcal{O}_X$) means an $\mathcal{O}_X$-module $\mathcal{F}$ such that for each $x\in X$ there exists an open neighborhood $U$ of $x$ and an exact sequence of $\mathcal{O}_X|_U$-modules of the form $\mathcal{O}_X^{(J)}|_U \to \mathcal{O}_X^{(I)}|_U \to \mathcal{F}|_U \to 0,$ where $I$ and $J$ are arbitrary index sets ( depending on $x$ ).

I am following Will Sawin's answer to this question: his argument is as follows.
Fix an open subset $U\subseteq X$. We want to show that $\Gamma(U, f^{*}f_{*}\mathcal{F}) \to \Gamma(U,\mathcal{F})$ is surjective. So let $s \in \Gamma(U,\mathcal{F})$. Then $s$ is of the form $s= \sum_{i=1}^{n}a_i \gamma_i|_U$, where $a_i \in \Gamma(U,\mathcal{O}_X)$. Note that $\Gamma(X,\mathcal{F}) = \Gamma(Y,f_{*}\mathcal{F})$. Let's take the pullback $f^{*}\gamma_1 , \dots, f^{*}\gamma_n \in \Gamma(X, f^{*}f_{*}\mathcal{F})$ under the natural map $\Gamma(Y,f_{*}\mathcal{F}) \to \Gamma(X, f^{*}f_{*}\mathcal{F})$.

Q.1. What is the explicit form of $f^{*}\gamma_i$ 's ?

( Continuing argument ) Then the restricted counit map $c_U$ to $U$ sends $$ c_U : \Gamma(U,f^{*}f_{*}\mathcal{F}) \to \Gamma(U,\mathcal{F}) $$ $$ \sum_{i=1}^n a_i((f^{*}\gamma_i)|_U) \mapsto s= \sum_{i=1}^{n}a_i \gamma_i|_U $$ .

Q.2. This statement is true? If so, why?

At first glance, this correspondence perhaps seems to work, but the definition of the counit map seems so abstract that I can't do explicit computation more. I want to enhance understanding about the counit morphism. Can anyone help? Thank you.

Shaun
  • 47,747
Plantation
  • 3,710
  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Shaun May 28 '25 at 09:41
  • Please ask only one question per post. You ask two questions in this post; they should be posted separately. Please don't use "EDIT". Instead, revise the question to read well for someone who encounters it for the first time. See https://cs.meta.stackexchange.com/q/657/755. – D.W. May 28 '25 at 17:52

2 Answers2

2

Let's place the definitions in the post to start. If $f:X\to Y$ is a morphism of ringed spaces and $\mathcal{G}$ is a sheaf of $\mathcal{O}_Y$-modules, then $f^*\mathcal{G}$ is defined as the tensor product $f^{-1}\mathcal{G} \otimes_{f^{-1}\mathcal{O}_Y} \mathcal{O}_X$, where $f^{-1}\mathcal{G}$ is the sheafification of the presheaf $f^{-1,pre}\mathcal{G}$ defined by sending $U\subset X$ to $\lim_{V\supset f(U)} \mathcal{G}(V)$. We therefore have a sequence of morphisms $$ f^{-1,pre}\mathcal{G} \to f^{-1}\mathcal{G} \to f^{-1}\mathcal{G}\otimes_{f^{-1}\mathcal{O}_Y} \mathcal{O}_X \cong f^*\mathcal{G}$$ where the first map is sheafification and the second map is given by $s\mapsto s\otimes 1$ for all sections over all opens. For a section $s\in \Gamma(U,\mathcal{G})$, we define the pullback section $f^*s \in \Gamma(f^{-1}(U),f^{*}\mathcal{G})$ to be the image of $s\in \Gamma(f^{-1}(U),f^{-1,pre}\mathcal{G})$ under this chain of morphisms. (A pleasing feature of this is that if you view $s\in\Gamma(U,\mathcal{G})$ as the image of a morphism $s:\mathcal{O}_U\to\mathcal{G}|_U$ given by $1\mapsto s$, then $f^*s$ is exactly the image of $1\in\mathcal{O}_{f^{-1}(U)}$ under the pullback map $f^*s:f^*\mathcal{O}_U\to f^*\mathcal{G}$ where we make the canonical identification $\mathcal{O}_{f^{-1}(U)}\cong f^*\mathcal{O}_U$. Details left to the reader.)

We can also identify the sections of $f^{-1,pre}f_*\mathcal{F}$ over an open $U\subset X$: by the definition of $f^{-1,pre}$, $\Gamma(U,f^{-1,pre}f_*\mathcal{F}) = \lim_{V\supset f(U)} (f_*\mathcal{F})(V)$, and then by the definition of $f_*$, we have that $(f_*\mathcal{F})(V) = \mathcal{F}(f^{-1}(V))$, so $\Gamma(U,f^{-1,pre}f_*\mathcal{F}) = \lim_{V\supset f(U)} \mathcal{F}(f^{-1}(V))$, where $V$ runs over open sets of $Y$ containing $f(U)$.

With all this background, let's construct the counit map. The counit map $f^*f_*\mathcal{F}\to\mathcal{F}$ is induced from the map $f^{-1,pre}f_*\mathcal{F}\to\mathcal{F}$ given over an open $U\subset X$ by sending an element of our colimit represented by $s_V\in\mathcal{F}(f^{-1}(V))$ to $s_V|_U$ - this works because $V\supset f(U)$ implies $f^{-1}(V)\supset U$, and it plays nice with restrictions because it is itself a restriction map. The way we use this map $f^{-1,pre}f_*\mathcal{F}\to\mathcal{F}$ to induce the counit map $f^*f_*\mathcal{F}\to\mathcal{F}$ is by noting first that since $\mathcal{F}$ is a sheaf, the map $f^{-1,pre}f_*\mathcal{F}\to\mathcal{F}$ factors uniquely through the sheafification $f^{-1}f_*\mathcal{F}$, then as the target is an $\mathcal{O}_X$-module, the map $f^{-1}f_*\mathcal{F}\to\mathcal{F}$ factors through $f^{-1}f_*\mathcal{F}\to f^{-1}f_*\mathcal{F}\otimes_{f^{-1}\mathcal{O}_Y} \mathcal{O}_X\to\mathcal{F}$ where the first map is $s\mapsto s\otimes 1$ for all sections $s$ over all opens. As $f^{-1}f_*\mathcal{F}\otimes_{f^{-1}\mathcal{O}_Y} \mathcal{O}_X\cong f^*f_*\mathcal{F}$, we have our desired counit map. (The proof that this is the counit map is somewhat involved and covered elsewhere, so let me skip that in this post.)

Tracing through this definition, we see that if $s\in \mathcal{F}(X)$ is a global section, then under the map $f^{-1,pre}f_*\mathcal{F}\to\mathcal{F}$, $s$ is again sent to $s$: for any $V\supset f(X)$, we have $f^{-1}(V)=X$, so the colimit is just $\mathcal{F}(X)$, and the restriction map is just the identity! From our background on the definition of $f^*s$, we see that this implies $f^*s$ maps to $s$ under the counit map, providing the answer to your question.

In response to a question in the comments about why this is the counit map, here's the essential logic:

  1. For a map of topological spaces $f:X\to Y$ there is an adjunction between the functors $f^{-1,pre}: Psh(Y) \to Psh(X)$ and $f_*: Psh(X)\to Psh(Y)$. The counit $\epsilon:f^{-1,pre}f_*\mathcal{F} \to \mathcal{F}$ is defined on an open set $U\subset X$ by sending a section $s_V\in \lim_{V\supset f(U)} \mathcal{F}(f^{-1}(V))$ to $s_V|_U$, which works since $V\supset f(U)$ implies $f^{-1}(V)\supset U$. The unit $\eta: \mathcal{G}\to f_*f^{-1,pre}\mathcal{G}$ is defined on an open $V\subset Y$ by sending $s\in\mathcal{G}(V)$ to $s_V \in \lim_{W\supset f(f^{-1}(V))} \mathcal{G}(W)$, which works because $V\supset f(f^{-1}(V))$. The map $$\operatorname{Hom}_{Psh(X)}(f^{-1,pre}\mathcal{G},\mathcal{F}) \to \operatorname{Hom}_{Psh(Y)}(\mathcal{G},f_*\mathcal{F})$$ is defined by $$(f^{-1,pre}\mathcal{G}\stackrel{\varphi}{\to} \mathcal{F}) \mapsto (\mathcal{G}\stackrel{\eta}{\to} f_*f^{-1,pre}\mathcal{G}\stackrel{f_*\varphi}{\to} f_*\mathcal{F})$$ while the map $$\operatorname{Hom}_{Psh(Y)}(\mathcal{G},f_*\mathcal{F}) \to \operatorname{Hom}_{Psh(X)}(f^{-1,pre}\mathcal{G},\mathcal{F})$$ is defined by $$(\mathcal{G}\stackrel{\psi}{\to} f_*\mathcal{F}) \mapsto (f^{-1,pre}\mathcal{G}\stackrel{f^{-1,pre}\psi}{\to} f^{-1,pre}f_*\mathcal{F} \stackrel{\epsilon}{\to} \mathcal{F}).$$ Checking these are adjunctions boils down to checking these two maps are mutually inverse. Everyone needs to do this computation for themselves once in their lives, and I am not going to deprive you of that opportunity. (It's a definition chase with a bit of bookkeeping. I would recommend a blackboard or a whiteboard. If you need to use paper, don't be shy about multiple sheets or a good amount of eraser work.)

  2. If $\mathcal{F}$ and $\mathcal{G}$ are sheaves, we may upgrade the adjunction $$\operatorname{Hom}_{Psh(X)}(f^{-1,pre}\mathcal{G},\mathcal{F}) \cong \operatorname{Hom}_{Psh(Y)}(\mathcal{G},f_*\mathcal{F})$$ to an adjunction $$\operatorname{Hom}_{Sh(X)}(f^{-1}\mathcal{G},\mathcal{F}) \cong \operatorname{Hom}_{Sh(Y)}(\mathcal{G},f_*\mathcal{F})$$ via the universal property of sheafification: given a presheaf map $f^{-1,pre}\mathcal{G}\to \mathcal{F}$, as $\mathcal{F}$ is a sheaf the map factors uniquely through the sheafification $f^{-1}\mathcal{G}$, and so the left-hand sides of those two lines are canonically in bijection. On the other hand, since sheaves are a full subcategory of presheaves, the right hand sides are canoncially in bijection.

  3. If $X$ and $Y$ are ringed spaces and $f$ is a morphism of ringed spaces, we may upgrade the adjunction $$\operatorname{Hom}_{Sh(X)}(f^{-1}\mathcal{G},\mathcal{F}) \cong \operatorname{Hom}_{Sh(Y)}(\mathcal{G},f_*\mathcal{F})$$ to an adjunction $$\operatorname{Hom}_{\mathcal{O}_X}(f^*\mathcal{G},\mathcal{F}) \cong \operatorname{Hom}_{\mathcal{O}_Y}(\mathcal{G},f_*\mathcal{F}).$$ The first step is that $\mathcal{G}\to f_*\mathcal{F}$ is a morphism of $\mathcal{O}_Y$-modules iff corresponding morphism $f^{-1}\mathcal{G}\to\mathcal{F}$ is a morphism of $f^{-1}\mathcal{O}_Y$-modules, so that our maps give an isomorphism $$\operatorname{Hom}_{f^{-1}\mathcal{O}_Y}(f^{-1}\mathcal{G},\mathcal{F}) \cong \operatorname{Hom}_{\mathcal{O}_Y}(\mathcal{G},f_*\mathcal{F}).$$ Next, use the tensor-hom adjunction and the fact that $\mathcal{H}om_{\mathcal{O}_X}(\mathcal{O}_X,\mathcal{F})\cong\mathcal{F}$ to establish that $$\operatorname{Hom}_{f^{-1}\mathcal{O}_Y}(f^{-1}\mathcal{G},\mathcal{F}) \cong \operatorname{Hom}_{f^{-1}\mathcal{O}_Y}(f^{-1}\mathcal{G},\mathcal{H}om_{\mathcal{O}_X}(\mathcal{O}_X,\mathcal{F})) \cong \\ \cong \operatorname{Hom}_{\mathcal{O}_X}(f^{-1}\mathcal{G}\otimes_{f^{-1}\mathcal{O}_Y} \mathcal{O}_X,\mathcal{F}) \cong \operatorname{Hom}_{\mathcal{O}_X}(f^*\mathcal{G},\mathcal{F}).$$ Tracing through the counit map for each of these adjunctions, one sees that the counit for the adjunction between $f^*$ and $f_*$ is as specified above.

Plantation
  • 3,710
Hank Scorpio
  • 2,951
  • O.K. Thanks for kind explanation. Question : Q.1. In the first paragraph in your answer, you wrote " and therefore the value of the colimit on $f^{-1}(U)$ is just $\Gamma(U,\mathcal{G})$." You mean, just $\Gamma(U,\mathcal{G}) = \Gamma(f^{-1}(U),f^{-1,pre}\mathcal{G})$ ? Q.2. In the thrid paragraph in your anwer, I still don't understand the factorizability argument in showing the induction of the counit map $f^{}f_{}\mathcal{F} \to \mathcal{F}$ from $f^{-1,pre}f_{*}\mathcal{F}\to \mathcal{F}$. I still can't figure out explicit form of the counit map. – Plantation May 26 '25 at 23:34
  • Q.3. I still don't understand that $f^{*}s$ maps to $s$. – Plantation May 26 '25 at 23:35
  • Yes, that was a typo. 2. There's (IMO) too much stuff going on to really get a complete description of the map $f^f_\mathcal{F} \to \mathcal{F}$ - you have a sheafification of the inverse image presheaf (which has an annoying colimit to calculate a general section) then you have a tensor product (itself involving another sheafification). But we do know that these more complicated objects and morphisms sit in the middle of a morphism that we know a bit more about! That's the entire point of this answer - even if it's hard to get a full description, you can still get a result here.
    • – Hank Scorpio May 27 '25 at 04:50
  • 2
  • If $s\in f^{-1,pre}f_\mathcal{F}(X)$ maps to $s\in \mathcal{F}(X)$ and $s\mapsto f^s$ from $f^{-1,pre}f_\mathcal{F}$ to $f^f_\mathcal{F}$, and the map $f^{-1,pre}f_\mathcal{F}\to \mathcal{F}$ factors through $f^{-1,pre}f_\mathcal{F}(X) \to f^f_\mathcal{F}$, then $f^s$ maps to $s$. Maybe less notation will help you: if $a\in A$ maps to $c\in C$ and $A\to C$ factors as $A\to B\to C$, then the image of $a$ in $B$ maps to $c$.
    • – Hank Scorpio May 27 '25 at 04:54
  • O.K. Than you. Q. In your third paragraph, how the factorization of $f^{-1}f_{}\mathcal{F} \to \mathcal{F}$ through $f^{-1}f_{}\mathcal{F} \otimes_{f^{-1}\mathcal{O}Y} \mathcal{O}_X $ possible? Can we exhibit explicit form of $f^{-1}f{*}\mathcal{F} \otimes_{f^{-1}\mathcal{O}_Y} \mathcal{O}_X \to \mathcal{F}$? – Plantation May 28 '25 at 00:09
  • And can we really show that that map is the co unit (for example by showing that this corresponds to the identity map $\operatorname{id}{f^{*}\mathcal{F}}$ under the adjoint situation $\operatorname{Hom}{\mathcal{O}X}(f^{*}f{}\mathcal{F},\mathcal{F}) \cong \operatorname{Hom}{\mathcal{O}_Y}(f{}\mathcal{F},f_{*}\mathcal{F})$ ) ? – Plantation May 28 '25 at 00:09
  • Q.4. More question, In the first paragraph in your answer, why $\Gamma(U, \mathcal{G})= \Gamma(f^{-1}(U) , f^{-1,pre}\mathcal{G}) $, from that the $U$ is smallest member ( true? ) in the index set of the colimit $\lim_{V \supset f(f^{-1}(U))}$ ? – Plantation May 28 '25 at 00:16
  • Remember, $f^{-1}f_\mathcal{F}$ is not an $\mathcal{O}X$-module, it's only an $f^{-1}\mathcal{O}_Y$-module. Since $\mathcal{O}_X$ is also an $f^{-1}\mathcal{O}_Y$-module, and the map from the product $$f^{-1}f\mathcal{F}\times f^{-1}\mathcal{O}Y \to f^*f*\mathcal{F}$$ which is given over each open by the product of sections (i.e. $(s,t)\mapsto s\cdot t$) is a $f^{-1}\mathcal{O}_Y$-balanced map, that factors through the tensor product by universal property. – Hank Scorpio May 28 '25 at 03:17
  • Next, for showing that this map is the counit, yes, absolutely, but it's not so quick and it is a little annoying and you should have done it before. The shortest way is probably to show that this is the adjunction on the level of presheaves, then use the factorizations like we've done here to upgrade that claim a level at a time - first from $f^{-1,pre}$ to $f^{-1}$, then to $f^*$. For Q4, nevermind, that's not always correct nor important and I've removed it. – Hank Scorpio May 28 '25 at 03:19
  • Some confusion : For the first comment. Why you considered $f^{-1}f_{}\mathcal{F} \times f^{-1}\mathcal{O}_Y \to f^{}f_{*}\mathcal{F}$? C.f. Given a ring $R$ and modules $A_R, {_R}B$, then their tensor product is an abelian gorup $A\otimes_R B$ and an $R$-biadditive function $h : A \times B \to A \otimes_R B$ such that. for every abelian group $G$ and every $R$-biadditive $f: A \times B \to G$, there exists a unique $\mathbb{Z}$-homomorphism $\tilde{f} : A \otimes_R B \to G$ such that $f = \tilde{f} \circ h$. – Plantation May 28 '25 at 04:24
  • We want to show that $f^{-1}f_{}\mathcal{F} \to \mathcal{F}$ factorizes through $f^{-1}f_{}\mathcal{F} \otimes_{f^{-1}\mathcal{O}_Y } \mathcal{O}_X$. Here, in your consideration, what objects correspond to $A,B$, and $G$?

    For the second comment. You said, " this is the adjunction on the level of presheaves". What does it exactly means?

     – Plantation May 28 '25 at 04:24
  • Shoot, sorry, typo - the second factor of the product in that centered equation should be $\mathcal{O}X$. The point is that we can take the composite map $A \stackrel{a\mapsto (a,1)}{\to} A\times B \to G$ and then factor that through the tensor product to get a composition $A\to A\times B \to A\otimes B \to G$, and then use the map $A\to A\otimes B$. "Adjunction on the level of presheaves" means what it says - the functors $f^{-1,pre}$ and $f*$ are adjoint as functors between the categories of presheaves on $X$ and $Y$, and the counit map for this adjunction is what we begin with here. – Hank Scorpio May 28 '25 at 07:24
  • Through inspection of your comment, for the centered equation you exhibit, I think that $\mathcal{F}$, instead of $f^{}f_{}\mathcal{F}$, is more correct object. Do you agree?
    • – Plantation May 28 '25 at 08:26