3
$\begingroup$

Suppose we have an exact sequence of sheaves of abelian groups, $$ 0 \longrightarrow \mathcal{F}' \stackrel{\phi'}{\longrightarrow} \mathcal{F} \stackrel{\phi}{\longrightarrow} \mathcal{F}''. $$

I want to show that the sequence of abelian groups $$0 \longrightarrow \mathcal{F}'(U) \stackrel{\phi'_{U}}{\longrightarrow} \mathcal{F}(U) \stackrel{\phi_{U}}{\longrightarrow} \mathcal{F}''(U) $$ is exact for all open sets $U$. So far I have that exactness at $\mathcal{F}'(U)$ is immediate since injectivity for morphisms of sheaves can be checked on injectivity of morphisms of sections. However I am stuck at exactness at $\mathcal{F}(U)$. I feel as though I am very close. I have determined that exactness at $\mathcal{F}(U)$ is equivalent to the presheaf image of $\phi '$ being equal to the kernel sheaf of $\phi$.

From exactness of the original sequence, I have that $\text{im} \phi ' = \ker \phi $, but how do I then show that this is equal to the image presheaf of $\phi '$? In other words, the image presheaf of $\phi'$ was already a sheaf?

This is related to another question asked here Functor of section over U is left-exact, where user Future suggests that we can use the fact that the image sheaf can be identified with a subsheaf of the target sheaf, but I am still not sure how to do it.

I realize there are other ways to prove left exactness of the section functor, but I am very curious how to make this particular approach work.

Any help is appreciated.

$\endgroup$
  • $\begingroup$ The image presheaf of $\phi'$ is a sheaf because the restrictions are (per injectivity on section level) "the same" as in $\mathcal F'$ $\endgroup$ – Hagen von Eitzen Jul 23 '17 at 18:36
  • $\begingroup$ So I can intuitively see that this is true, although I am struggling to make it rigorous. If I have an exact sequence $$ 0 \longrightarrow \mathcal{P} \longrightarrow \mathcal{F}, $$ where $\mathcal{P}$ is a presheaf and $\mathcal{F}$ is a sheaf, then I can easily see that the sheafification $\theta: \mathcal{P} \longrightarrow \mathcal{P}^{+}$ is injective, but I am struggling to show that it is also surjective. $\endgroup$ – Joe Jul 23 '17 at 20:01
  • $\begingroup$ Another possibility : you just proved that the presheaf image is equal to the sheaf kernel... $\endgroup$ – Roland Jul 23 '17 at 20:47
  • $\begingroup$ By the way, in your last comment you are trying to prove something false. A sheaf can have sub-presheaves which are not sheaves. What Hagen Von Eitzen was saying is that the image presheaf is isomorphic to $\mathcal{F}'$ since the map $\mathcal{F}'\rightarrow\mathcal{F}$ is injective. $\endgroup$ – Roland Jul 23 '17 at 20:49
  • $\begingroup$ So I have put some more thought into this. Would I be able to verify with someone that this proof is valid for what I am trying to show? I have used some results from previous exercises in Hartshorne, so if you know those it will be easier, but otherwise I am happy for you to assume that those are true. $\endgroup$ – Joe Jul 23 '17 at 21:26
1
$\begingroup$

Okay so if I understand you right, you finally want to see why the image presheaf is already a sheaf, so I will give it a try.

(Sorry for giving the morphisms new names.)

If we can show that for $U \subseteq X$ open we have $ker(\psi_U)=im(\phi_U)$, we are done, since then the image presheaf is given by the kernel of a morphism of sheaves, which is indeed a sheaf. So let us try this.

Consider the induced sequence \begin{align*} 0 \longrightarrow \Gamma(U, \mathscr{F}') \overset{\phi_U}{\longrightarrow} \Gamma(U, \mathscr{F}) \overset{\psi_U}{\longrightarrow} \Gamma(U, \mathscr{F}''), \end{align*} and for each $x \in U$ the induced sequence \begin{align*} 0 \longrightarrow \mathscr{F}'_x \overset{\phi_x}{\longrightarrow} \mathscr{F}_x \overset{\psi_x}{\longrightarrow} \mathscr{F}''_x, \end{align*} which is exact, since the exactness of a sequence of sheaves is equivalent to the exactness of the induced sequence on stalks for all points. Now putting these together with the natural morphisms into stalks, one gets a commutative diagram (which I unfortunately was not able to type (mea culpa)).

We now come to our claim that $ker(\psi_U)=im(\phi_U)$:

$[\supseteq]:$ Let $s \in \Gamma(U, \mathscr{F}')$ and consider for each $x \in U$ the germ of $\psi_U(\phi_U(s))$ in the stalk $\mathscr{F}''_x$: \begin{align*}(\psi_U(\phi_U(s)))_x = \psi_x(\phi_x(s_x)). \end{align*} But by exactness, $\psi_x(\phi_x(s_x))=0$ for all $x \in U$. Hence $\psi_U(\phi_U(s))=0$, so $im(\phi_U) \subseteq ker(\psi_U)$.

$[\subseteq]:$ Let $t \in ker(\psi_U)$, so $\psi_U(t)=0$. Then for all $x \in U$ we have that $\psi_x(t_x)=(\psi_U(t))_x = 0$, so the germ of $t$ at $x$ is an element in $ker(\psi_x)=im(\phi_x)$ by exactness again. Hence for every $x \in U$ there is a $s'_x \in \mathscr{F}'_x$, say of the form $[(s'_{(x)},V_{(x)})]$ for some open neighborhood $V_{(x)} \subseteq U$ of $x$ and $s'_{(x)} \in \Gamma(V_{(x)}, \mathscr{F'})$, such that $\phi_x(s'_x)=t_x$. Then we have that for $x,y \in U$ \begin{align*} \phi_{V_{(x)}\cap V_{(y)}}(s'_{(x)} |_{V_{(x)}\cap V_{(y)}}) = t | _{V_{(x)}\cap V_{(y)}} = \phi_{V_{(x)}\cap V_{(y)}}(s'_{(y)} |_{V_{(x)}\cap V_{(y)}}), \end{align*} so that by the injectivity of $\phi_{V_{(x)}\cap V_{(y)}}$ (which you already proved), we get the required condition \begin{align*} s'_{(x)} |_{V_{(x)}\cap V_{(y)}} = s'_{(y)} |_{V_{(x)}\cap V_{(y)}} \end{align*} for the gluing of the $s'_{(x)}$ for $x \in U$. Therefore we have a section $s \in \Gamma(U, \mathscr{F})$ with the property that for all $x \in U$ \begin{align*} s | _{V_{(x)}} = s'_{(x)}. \end{align*} Now we can conclude that for every $x \in U$ \begin{align*} (\phi_U(s))_x = \phi_x(s_x) = \phi_x(s'_x) = t_x, \end{align*} since $s_x=s'_x$, which gives $\phi_U(s)=t$ as desired.

$\endgroup$
1
$\begingroup$

If I understand you correctly, you say that you already proved that the presheaf image of $\phi'$ is equal to the sheaf kernel of $\phi$. In that case, the presheaf image of $\phi'$ is already a sheaf, hence equal to its sheafification.

But as you said, there are other ways to prove this. Here is a more general approach for example:

We have a global section functor

\begin{align} \Gamma\colon Sh(X) &\to Ab \\ \mathcal{F} &\mapsto \mathcal{F}(X) \end{align}

and for every abelian group $G$ in $Ab$ we have the functor taking the group to the sheaf associated to the constant presheaf $G$

\begin{align} a \colon Ab &\to Sh(X) \\ G &\mapsto \underline{G} \end{align}

We have an adjunction $ a \colon Ab \leftrightarrow Sh(X) \colon \Gamma $, which means that

$$ \mathrm{Hom}_{Sh(X)}(\underline{G},\mathcal{F}) \cong \mathrm{Hom}_{Ab}(G,\mathcal{F}(X))$$

Indeed, given any group homomorphism $f\colon G\to \mathcal{F}(X)$ we have a unique presheaf morphism (hence by universal property of associated sheaf a unique sheaf morphism) whose component at any open $U$ is $f$ followed by the restriction to $U$.

Now you can use the nice behaviour of hom-set functors with exact sequences to show that being right adjoint (resp. left adjoint) implies being left exact (resp. right exact).

So the global section functor is left exact. And then apply this for any open subset $U$ and the restriction sheaf.

$\endgroup$
  • $\begingroup$ Thank you Pedro for your answer. I do quite like that more abstract approach. However I just wanted to pick up on the first part of your answer. What I have already proved is that the image sheaf, $\text{im} \phi '$, can be identified with some subsheaf of $\mathcal{F}$. Perhaps l am missing something trivial, but it is not immediately obvious to me that this implies that the presheaf image of $\phi'$ is necessarily equal to the kernel sheaf, $\ker \phi$. Why could it not be that the image presheaf is not at all a sheaf in $\mathcal{F}$, and only equals $\ker \phi$ after sheafification? $\endgroup$ – Joe Jul 24 '17 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.