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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 $\alpha_x$.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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