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> [!proposition] Proposition. ([[the section functor is left-exact]])
> Let $X$ be a [[topological space]]. For any open set $U \subset X$, the [[section functor|section]] [[covariant functor|functor]] $\Gamma(U, \cdot)$ is [[exact functor|left-exact]], i.e., given an [[exact sequence]] of [[sheaf|sheaves]] $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}''$
we obtain an [[exact sequence]] $0 \to \Gamma(U, \mathcal{F}') \to \Gamma(U, \mathcal{F} ) \to \Gamma(U, \mathcal{F}'')$
of (say) [[abelian group|abelian groups]].
^proposition
> [!proof]- Proof. ([[the section functor is left-exact]])
> Label the maps involved as
>
> $0 \to \mathcal{F}' \xrightarrow{f} \mathcal{F} \xrightarrow{g} \mathcal{F}''$
> and $0 \to \mathcal{F}'(U) \xrightarrow{f_{U}}\mathcal{F}(U) \xrightarrow{g_{U}} \mathcal{F}''(U).$
> We need to show:
> 1. $\ker f_{U}=\{ 0 \}$
> 2. $\im f_{U}=\ker g_{U}$
>
> $(1)$ is immediate: since $f$ is an [[injective sheaf morphism]], $\ker f_{U}=\{ 0 \}$.
>
> Initially, $(2)$ does not appear immediate: while we know that sections in the [[sheafification]] of the [[presheaf image]] of $f$ at $U$ agree exactly with sections in $\ker g_{U}$, this is *a priori* not the same as having $\im f_{U}=\ker g_{U}$ because of the involved [[sheafification]]. However, since $f$ is [[injective sheaf morphism|injective]], the [[sheafification]] does nothing — the presheaf image is already a sheaf, and so actually $(2)$ *is* immediate. (See [[sheaf image#^properties]].)
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#### 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
> ```