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> [!definition] Definition. ([[(pre)sheaf kernel]])
>
Let $f:\mathcal{F} \to \mathcal{G}$ be a [[morphism of (pre)sheaves|morphism of presheaves]] valued in a [[category]] where [[categorical kernel|kernels]] make sense. The **presheaf kernel** of $f$, $\ker f$, is the [[presheaf]] specified by $(\ker f)(U):=\ker \big(f_{U}:\mathcal{F}(U) \to \mathcal{G}(U) \big)$.
>
If $f:\mathcal{F} \to \mathcal{G}$ is a [[morphism of (pre)sheaves|morphism of]] [[sheaf|sheaves]], then $\ker \mathcal{F}$ is in fact a [[sheaf]].
^definition
> [!justification]
> Quick check that $\ker \mathcal{F}$ is in fact a [[sheaf]] when $\mathcal{F}$, $\mathcal{G}$ are.
^justification
> [!basicproperties]
> Taking kernels and [[(pre)sheaf stalk|stalks]] are compatible: $(\ker f)_{p}=\ker (f_{p}: \mathcal{F}_{p} \to \mathcal{G}_{p})$ for all $p \in X$.
^properties
> [!proof]
>
> There is really not much to show.
>
> Have a map $\begin{align}
> (\ker f)_{p} &\to \ker f_{p}
> \end{align}$
> taking a germ $[U, s] \in (\ker f)_{p}$ to itself viewed as an element of $\mathcal{F}_{p}$ sent to zero by $f_{p}$. Well-defined because if $[U,s] \in (\ker f)_{p}$, then $s \in \ker f_{U}$, so $f_{p}([U,s])=[U, f_{p}(s)]=[U,0]$ and thus $[U,s] \in \ker f_{p}$. Now we want to show injective and surjective.
>
> **Injective.** If $[U,s] =0 \text{ in } \mathcal{F}_{p}$, then $s$ restricts to zero on some neighborhood $V \ni p$, hence $[U,s]=0$ in $(\ker f)_{p}$.
>
> **Surjective.** If $[U,s] \in \ker f_{p}$, then $0=f_{p}([U,s])=[U, f_{U}(s)]$ in $\mathcal{G}_{p}$. Thus there exists $p \in V \subset U$ such that $f_{U}(s) |_{V}=0$. So $s |_{V} \in (\ker f)(V)$. Thus $(V, s |_{V}) \in (\ker f)_{p}$.
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####
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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
> ```