----- > [!proposition] Proposition. ([[characterizing short exact sequences of sheaves]]) > Let $\mathcal{F}'$ be a [[subsheaf]] of a [[sheaf]] $\mathcal{F}$. The natural map $\mathcal{F} \to \frac{\mathcal{F}}{\mathcal{F}'}$ is [[surjective sheaf morphism|surjective]] and has [[(pre)sheaf kernel|kernel]] $\mathcal{F}, so that there is an [[exact sequence]] $0 \to \mathcal{F}' \to \mathcal{F} \to \frac{\mathcal{F}}{\mathcal{F}'} \to 0.$ > Conversely, if $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}''\to 0$ is an [[exact sequence]] of [[sheaf|sheaves]], then $\mathcal{F}'$ is [[isomorphism|isomorphic]] to a [[subsheaf]] of $\mathcal{F}$ and $\mathcal{F}''$ is [[isomorphism|isomorphic]] to the [[quotient sheaf|quotient]] of $\mathcal{F}$ by this [[subsheaf]]. > [!proof]- Proof. ([[characterizing short exact sequences of sheaves]]) > Let $U$ be an open subset of $X$. > > The 'natural map' in question is $\theta \circ \pi$, where the [[natural transformation]] $\pi$ has $U$-component $\begin{align} > \pi_{U} : \mathcal{F}(U) &\to \frac{\mathcal{F}(U)}{\mathcal{F}'(U)} \\ > s & \mapsto s + \mathcal{F}'(U) > \end{align}$ > and $\theta$ is the [[sheafification]] [[morphism of (pre)sheaves|morphism]]. > > **Surjectivity.** Let $p \in U$ be arbitrary. By [[sheaf morphism injectivity and surjectivity can be tested on stalks]], it is enough to check that $(\theta \circ \pi)_{p}:\mathcal{F}_{p} \to (\frac{\mathcal{F}}{\mathcal{F}'})_{p}$ is a [[surjection]]. Note that passing to the stalk map is [[covariant functor|functorial]], i.e., $(\theta \circ \pi)_{p}=\theta_{p} \circ \pi_{p}$. We know $\theta_{p}$ is an [[isomorphism]] ("$\theta$ preserves [[(pre)sheaf stalk|stalks]]"), and $\pi_{p}$ is the usual natural projection [[group homomorphism]], hence $\theta_{p} \circ \pi_{p}$ is [[surjection|surjective]] as a composition of [[surjection|surjections]]. > > **Kernel.** Let $s \in \ker(\theta \circ \pi)(U)$, i.e., $s \in \ker \theta_{U} \circ \pi_{U}$. We know how $\theta_{U}$ is defined; it does not kill anything nontrivial. So $\pi_{U}(s)=0$, hence $s \in \mathcal{F}'(U)$. Thus, we have an inclusion $\ker (\theta \circ \pi) \subset \mathcal{F}'$. Now we can appeal to [[equality of subsheaf and sheaf can be tested on stalks]]: $(\ker(\theta \circ \pi))_{p}=\ker ((\theta\circ \pi)_{p})$ because [[(pre)sheaf kernel|taking kernels and stalks commute]], and then by functoriality of stalk maps this equals $\ker(\theta_{p} \circ \pi_{p})$ which in turn just equals $\ker \pi_{p}$ because $\theta_{p}$ [[sheafification|is an isomorphism of stalks]]. WTS $\ker \pi_{p}=\mathcal{F}'_{p}$. Indeed, $[U,s] \in \ker \pi_{p}$ iff $[U, \pi_{U}(s)]=0$ in $\left( \frac{\mathcal{F}}{\mathcal{F}} \right)_{p}$, iff (shrinking $U$ if necessary) $\pi_{U}(s)=0$ in $\frac{\mathcal{F}(U)}{\mathcal{F}'(U)}$ iff $s \in \mathcal{F}'(U)$ so that $[U,s] \in \mathcal{F}'_{p}$. > > > **Conversely,** suppose $0 \to \mathcal{F}' \xrightarrow{f} \mathcal{F} \xrightarrow{g}\mathcal{F}'' \to 0$ is an [[exact sequence]] of [[sheaf|sheaves]]. Certainly $\ker f=0$, so $f$ is an [[injective sheaf morphism]]. Not hard to convince oneself that $f$ is an isomorphism onto its image. So $\mathcal{F}'$ may indeed be viewed as a [[subsheaf]] of $\mathcal{F}$. We are given that $g$ is [[surjective sheaf morphism|surjective]] with [[(pre)sheaf kernel|kernel]] equal to $\im f$; the goal now is to show $\frac{\mathcal{F}}{f(\mathcal{F}')} \cong \mathcal{F}''$. > > **Claim:** The map $\overline{g}: \frac{\mathcal{F}}{\mathcal{F}'} \to \mathcal{F}''$ specified via $U$-component $\overline{g}_{U}(s + \mathcal{F}'(U)):= g(s)$ is [[well-defined]] and is an [[isomorphism]]. > > *Well-definition.* If $s_{1}+\mathcal{F}'(U)=s_{2}+\mathcal{F}'(U)$, then > > > *Isomorphism.* [[sheaf isomorphism iff isomorphism on stalks|It is enough to]] check that the [[(pre)sheaf stalk|stalk map]] $\overline{g}_{p}:\left( \frac{\mathcal{F}}{\mathcal{F}'} \right)_{p} \to \mathcal{F}''_{p}$ is an [[isomorphism]] for all $p$. By [[sheaf morphism injectivity and surjectivity can be tested on stalks]], [[surjective sheaf morphism|surjectivity]] of $g$ implies surjectivity of each ${g}_{p}$. Moreover, the map $\pi:\mathcal{F} \to \frac{\mathcal{F}}{\mathcal{F}'}$ induces on stalks the natural projection $\pi_{p}:\mathcal{F}_{p} \to \frac{\mathcal{F}_{p}}{\mathcal{F}_{p}'}\cong (\frac{\mathcal{F}_{}}{\mathcal{F}'})_{p}$[^1]; by the [[first isomorphism theorem]], $\overline{g}_{p}$ fits into a diagram > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAB12BbOnACwDGjYADEAvgH00IMaXSZc+QigCM5KrUYs2nHvyENRktAHITMuSAzY8BImRUb6zVog7deg4eID0uzwZGUuZiGjBQAObwRKAAZgBOEFxIZCA4EEgATNTO2m6caFhSINQMdABGMAwACgq2yiDxWBF8OBZxicmIqelIapoubBHF1HwwdFBsOADuEGMTCLIdSVnUvYj9OHRYDGx8EBAA1iUg85NuM3PjUAilWGCuIFB0cGOTOVqPnDAAHlhwODgAAIAIRAzgQGgweIMe4wYARYwnMqVGp1JRsJotNqhMRAA > \begin{tikzcd} > \mathcal{F}_p \arrow[d, "\pi_p"'] \arrow[r, "g_p", two heads] & \mathcal{F}_p'' \\ > \mathcal{F}/\mathcal{F}_p' \arrow[ru, "\exists ! \overline{g}_p"', two heads, dashed, hook] & > \end{tikzcd} > \end{document} > ``` > and is hence an isomorphism. > > (or something like this, there are small details left to check after class) > [^1]: Here we've used the easy fact that taking quotients and taking stalks commute; see [[quotient sheaf|quotient sheaf]]. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```