Characterization: Injective morphism of sheaves Let $X$ be an arbitrary topological space and $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism of sheaves over $X$. We call $\varphi$ injective if $\mathcal{K}:=\ker(\varphi) = 0$.
I need to show that this is equivalent to the statement that for all $p \in X$ the morphisms $\varphi_p : \mathcal{F}_p \to \mathcal{G}_p$ induced on the germs are all injective.
My main problems are the following:


*

*Embarrassingly, I do not get what $\ker(\varphi) = 0$ really means. Let us say we have sheaves over rings. Then the kernel is the zero ring. But does this also mean that for any open set $U \subset X$ the sections $\mathcal{K}(U) = \ker(\varphi_U)$ is also the zero ring?

*I still have no intuition for stalks and germs. We have defined stalks as some set equipped with some weird equivalence relation and germs as their equivalence classes. Neither do I know what injectivity means for the $\varphi_p$ nor how I get the desired conclusions with these.


Could anyone explain this to me? Thanks in advance.
 A: For your first question, yes, that's exactly what it means to say that the kernel sheaf is 0.  For your second question, have you studied complex analysis?  The identity theorem states that if two holomorphic functions agree on an open subset, then they are identical wherever they are both defined.  Thus it makes sense to identify holomorphic functions that agree on some open subset and only consider equivalence classes, aka germs.  This is exactly the equivalence relation defining the stalk of a sheaf.
The stalk of a sheaf is defined as the direct limit of the sheaf with its restriction maps:
$$
\mathcal{F}_p = \varinjlim_{U \ni p} \mathcal{F}(U)
$$
taken over all open sets $U \subseteq X$ containing $p$.  Given a morphism $\varphi$ of sheaves, the universal property of the direct limit yields an induced map on the stalks.  More concretely, we can choose a representative $(s,U)$ for a germ $s|_p$.  Since the morphism commutes with the restriction maps, then
\begin{align*}
\varphi_p(s|_p) &= \varphi_p(\operatorname{res}^U_p(s)) = \operatorname{res}^U_p(\varphi_U(s)) \, .
\end{align*}
(Note that the first restriction map is for the sheaf $\mathcal{F}$, $\operatorname{res}^U_p: \mathcal{F}(U) \to \mathcal{F}_p$, while the second is for the sheaf $\mathcal{G}$, $\operatorname{res}^U_p: \mathcal{G}(U) \to \mathcal{G}_p$.)
So to compute the action of the stalk map $\varphi_p$ on a germ, we just compute the action of the morphism on a representative for the germ, and then restrict to the stalk.
EDIT: Here's a solution to the problem:
Assume $\varphi_p$ is injective for all $p$.  Suppose $U$ is an open set and $s \in \ker(\varphi_U)$, so $\varphi_U(s) = 0$.  Given $p \in U$, then
$$
\varphi_p(s|_p) = \varphi_p(\text{res}^U_p(s)) = \text{res}^U_p(\varphi_U(s)) = \text{res}^U_p(0) = 0 \, .
$$
Since $\varphi_p$ is injective, then $s|_p = 0$, so there exists an open set $V_p \subseteq U$ with $p \in V_p$ such that $\text{res}^U_{V_p}(s) = 0$.  Note that the collection $\{V_p\}_{p \in U}$ forms an open cover of $U$.  Since $\mathcal{F}$ is a sheaf and $\text{res}^U_{V_p}(s) = 0$ for all $p \in U$, then $s = 0$ by local determination (the first sheaf axiom).  Thus $\ker(\varphi)(U) = \ker(\varphi_U) = 0$ for all $U$.
Conversely, assume $\ker(\varphi_U) = 0$ for all $U$.  Given $s|_p \in \ker(\varphi_p)$, choose a representative $(s,U)$ for $s|_p$.  Then
$$
0 = \varphi_p(s|_p) = \varphi_p(\text{res}^U_p(s)) = \text{res}^U_p(\varphi_U(s)) \, .
$$
Then there exists an open set $V \subseteq U$ containing $p$ such that
\begin{align*}
0 = \operatorname{res}^U_V(\varphi_U(s)) = \varphi_V(\operatorname{res}^U_V(s)) \, .
\end{align*}
Since $\varphi_V$ is injective, then $\operatorname{res}^U_V(s) = 0$, so
$$
s|_p = \operatorname{res}^U_p(s) = \operatorname{res}^V_p(\operatorname{res}^U_V(s)) = \operatorname{res}^V_p(0) = 0 \, .
$$
