Hartshorne's question II.1.2 (a) is pretty straightforward:
For any morphism of sheaves $\varphi: \mathcal{F} \rightarrow \mathcal{G}$, show that for each point $P$, $(\ker \varphi)_P = \ker (\varphi_P)$ and $(\text{im } \! \varphi)_P = \text{im } \! (\varphi_P)$.
A solution can be found here, c/o the University of Arizona, the first part reading:
Observe $(\ker \varphi)_P = \lim_{\rightarrow U \ni P} (\ker \varphi)(U) = \lim_{\rightarrow U \ni P} \ker \varphi_U$ is a subgroup of $\mathcal{F}_P$, as is $\ker \varphi_P$, so we show equality inside $\mathcal{F}_P$. For $x∈(\ker \varphi)_P$ pick $(U,y)$ representing $x$, with $y \in \ker \varphi_U$. Then the image of $y \in \mathcal{F}_P$, i.e. $x$, is mapped to zero by $\varphi_P$. Conversely, if $x \in \ker \varphi_P$ there exist $(U,y)$ with $y \in \mathcal{F}(U)$ and $\varphi_U (y) = 0$ so $x \in (\ker \varphi)_P$.
I don't have a problem with this solution at all, right up until the fact that I realize I cannot understand why it doesn't hold true for presheaves. Now, of course, if it were to hold true for presheaves, then it would imply that any morphism of presheaves $\varphi: \mathcal{F} \rightarrow \mathcal{G}$ is injective if and only if the induced morphism $\varphi_P: \mathcal{F}_P \rightarrow \mathcal{G}_P$ is injective, which we know is not necessarily the case.
Can someone point out to me wherein in the Uni of Arizona answer they implicitly make use of the fact that we are dealing with sheaves and not presheaves?
Thanks in advance, and looking forward to your answers!
The proof is actually fine, and establishes that $\ker{\phi}_P = (\ker{\phi})_P$. The only issue is the conclusion that you’re drawing from it. This result implies that a map $\phi$ of presheaves is injective on stalks if and only if the stalks of $\ker{\phi}$ are all zero. However, if $\mathscr{F}$ is not a sheaf, then $\ker{\phi}$ will generally not be a sheaf, and so the fact that all of the stalks of $\ker{\phi}$ are zero does not imply that the presheaf $\ker{\phi}$ is zero (i.e. that $\phi$ is injective).
Of course, if you assume that $\mathscr{F}$ is a sheaf, the same will be true of $\ker{\phi}$, and it will follow that $\phi$ is injective if and only if it is injective on stalks.