What's definition $(\ker \varphi)_P$ the stalk of kernel presheaf

Question

If $\varphi : \mathcal{F} \to \mathcal{G}$ a morphism of sheaves then what definition of $(\ker \varphi)_P$? I digested that an element is $\langle U,r \rangle $ where $r \in \mathcal{F}(U)$ and $\varphi(U)(r) = 0$ and pairs are identified as in usual identifications.

Answer 1

If $\mathscr{F}$ is a sheaf on any topological space $X$ (which I assume is what you're interested in here), the stalk of $\mathscr{F}$ at a point $x\in X$ is $\mathscr{F}_x:=\varinjlim_{x\in U}\mathscr{F}(U)$, the direct limit over the sets $\mathscr{F}(U)$ over all open sets $U\subseteq X$ containing $x$. So, an element of this set is an equivalence class of a pair $(s,U)$ where $x\in U\subseteq X$ is open and $s\in\mathscr{F}(U)$, and this equals the class represented by $(s^\prime,V)$ if and only if there is an open $W\subseteq U\cap V$ and a section $t\in\mathscr{F}(W)$ such that $s\vert_W=t=s^\prime\vert_W$.

In particular, for a morphism $\varphi:\mathscr{F}\rightarrow\mathscr{G}$, the sheaf kernel $\ker\varphi$ is defined by $(\ker\varphi)(U)=\ker(\varphi(U))$, where $\varphi(U):\mathscr{F}(U)\rightarrow\mathscr{G}(U)$ is the map of sections over $U$. So, as you say, an element of $(\ker\varphi)_x$ is specified by an open set $x\in U\subseteq X$ and a section $s\in\ker(\varphi(U))$.

EDIT: The answer to the question posed in the comments is yes. Even though the image presheaf $U\mapsto\mathrm{im}(\varphi(U))$ is not generally a sheaf, the natural map from this presheaf to its sheafification (which is by definition the image sheaf) induces isomorphisms on stalks, so the stalk of the image sheaf is the same as the stalk of the image presheaf, which is $\varinjlim_{x\in U}\mathrm{im}(\varphi(U))$, and so an element of this stalk is represented by a pair $(t,U)$ where $t=\varphi(U)(s)$ for some $s\in\mathscr{F}(U)$.

Moreover, one has $\mathrm{im}(\varphi)_x=\mathrm{im}(\varphi_x)$ via the stalk at $x$ of the canonical map $\mathrm{im}(\varphi)\rightarrow\mathscr{G}$, and similarly $\ker(\varphi)_x=\ker(\varphi_x)$ under the stalk at $x$ of the canonical inclusion $\ker(\varphi)\rightarrow\mathscr{F}$.

Answer 2

From your title I guess you know that $U\mapsto \ker\phi(U)$ is a presheaf (it is in fact a sheaf). We call "$\ker\phi$" this presheaf. As such, it has stalks at every point: $$ (\ker\phi)_P:=\varinjlim_{U\ni P}\ker\phi(U), $$ and one can show that $(\ker\phi)_P=\ker \phi_P$, where $\phi_P$ is the map on stalks $\mathcal F_P\to\mathcal G_P$ induced by $\phi$.