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.
Best Answer
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.