This is a follow up question to something I asked earlier: What does it mean for a sequence of sheaves to be exact

Let $F, G, H$ be sheaves on a topological space $X$, and let $$F \xrightarrow{\alpha} G \xrightarrow{\beta} H$$ be morphisms of sheaves. Let $\mathcal O = \textrm{Im}^{\textrm{pre}}(\alpha)$ be the presheaf $U \mapsto \textrm{Im }( \alpha(U))$, and let $\textrm{Im } \alpha$ be a sheafification of this presheaf. There is a canonical choice of $\textrm{Im } \alpha$ which is actually a subsheaf of $G$, which is directly obtained by using the universal property on any sheafification of $\mathcal O$, and does not depend on the specific choice of sheafification. Assuming $(\textrm{Im } \alpha, \theta)$ (where $\theta: \mathcal O \rightarrow \textrm{Im } \alpha$ is the universal map) is this canonical sheafification, we say that the sequence is exact if $\textrm{Im } \alpha$ is equal to the sheaf $\textrm{Ker } \beta$.

I'm having trouble understanding the proof of the result that the sequence is exact if and only if the corresponding sequence on the stalks $F_x \xrightarrow{\alpha_x} G_x \xrightarrow{\beta_x} H_x$ is exact for all $x \in X$.

For example, let me suppose that $\textrm{Im } \alpha = \textrm{Ker } \beta$. Let $i, i^+$ be the respective inclusion morphisms of $\mathcal O, \textrm{Im } \alpha$ into $G$. Since $i^+ \circ \theta = i$, and $i^+$ maps $\textrm{Im } \alpha$ onto the kernel of $\beta$, we have that also $i$ maps $\mathcal O$ into $\textrm{Ker } \beta$, i.e. $\textrm{Im } (\alpha(U)) \subseteq \textrm{Ker } \beta(U)$ for all $U$. Thus $\beta \circ \alpha$ is the zero morphism, which implies $\beta_x \circ \alpha_x = 0$ for all $x$. Now I'm having trouble seeing that the kernel of $\beta_x$ is contained in the image of $\alpha_x$.

  • 1
    $\begingroup$ Almost by definition $\beta_x\circ \alpha_x =0$ implies that the image of $\alpha_x$ is contained in the kernel of $\beta_x$, right? Do you mean to ask how to do the other part, that the exactness of the sequence of sheaves implies that the kernel of $\beta_x$ is contained in the image of $\alpha_x$? $\endgroup$ – Potato Dec 6 '15 at 1:29
  • 3
    $\begingroup$ Hint: sheafification does not alter the stalks. $\endgroup$ – Remy Dec 6 '15 at 1:39
  • $\begingroup$ Sorry, that comment might be confusing, but in any case I think you've mixed up $\alpha$ and $\beta$ in your last sentence. $\endgroup$ – Potato Dec 6 '15 at 1:40
  • $\begingroup$ I sure did mix them up $\endgroup$ – D_S Dec 6 '15 at 2:33

You are left to show $Ker(\beta _x)\subset Im(\alpha_x)$

We have the sequence of abelian groups $F_x\xrightarrow{\alpha_x}G_x\xrightarrow{\beta_x}H_x$

Let us choose an element $g_x\in Ker(\beta_x)$, we need to show $g_x\in Im(\alpha_x)$ i.e., we need to find an element such that $\alpha_x$ maps that element to $g_x$

There exists open set $(x\in )U\subset X$ and $g\in G(U)$ corresponding to $g_x\in G_x$

Look at the commutative diagram

$\require{AMScd} \begin{CD} G(U) @>\beta(U)>> H(U)\\ @VVV @VVV\\ G_x @>\beta_{x}>> H_x \end{CD}$

$\beta_x(g_x)=0$ implies $(\alpha(U)(g))_x=0$ Therefore, there exists some $x\in W\subset U$ such that $(\alpha(U)(g))|_W=0$

Let us restrict the previous commutative diagram

$\require{AMScd} \begin{CD} G(W) @>\beta(W)>> H(W)\\ @VVV @VVV\\ G_x @>\beta_{x}>> H_x \end{CD}$

Here, $\beta(U)(g|_W)=0$ i.e., $g|_W \in ker (\beta(W))$ which is equal to $Im(\alpha (W))$

Now consider the sequence of sheaves $F\rightarrow O \rightarrow Im (\alpha) \rightarrow (G)$

Consider the composition $F\rightarrow Im(\alpha)$

Use the fact $Im(\alpha)_x=im(\alpha_x)$ (Hartshorne Chapter-II Ex 1.2)

Now, $g|W\in Im(\alpha)(W) $ i.e., $g_x\in Im(\alpha)_x$ but $Im(\alpha)_x=im(\alpha_x) \implies g_x\in Im(\alpha_x)$ Hence Proved.

  • $\begingroup$ Thank you for answering. For the converse, I can just argue that $\beta \circ \alpha = 0$ (if it's zero on the stalks, it must be the zero morphism), and then the inclusion morphism $Im(\alpha) \rightarrow \Ker(\beta)$ is both injective and surjective on the stalks, so it must be an isomorphism of sheaves, and in particular surjective on the global parts. $\endgroup$ – D_S Dec 7 '15 at 0:44
  • $\begingroup$ @D_S I didn't understand the hint you gave for the converse. Have you written down the proof of the converse? $\endgroup$ – Babai Dec 15 '15 at 8:46

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.