Issues with defining the stalk $(\ker\varphi)_p$

Question

I was solving the following problem from Hartshorne: Given a morphism of sheaves on a topological space $\varphi:\mathscr{F}\to\mathscr{G}$, show that $\ker(\varphi_p)=(\ker\varphi)_p$ for any point $p$.

My question is not about the actual problem, which I think I can solve, but rather a confusion over how $(\ker\varphi)_p$ is defined. From my understanding of stalks, $(\ker\varphi)_p$ is a subgroup of $\mathscr{F}_p$, and can be viewed as the set of $(U,s)$ where $U$ is open and $s\in(\ker\varphi)(U)=\ker(\varphi_U)$ (i.e. $s\in \mathscr{F}(U)$ and $\varphi_U(s)=0$), along with the relation $(U,s)=(V,t)$ if there is open $W$ with $p\in W\subseteq U\cap V$ so $s|_W=t|_W$.

My confusion is the following: Suppose we have an element of $\mathscr{F}_p$ and we want to check if it is in the subgroup $(\ker\varphi)_p$. To do so, I would pick a representative $(U,s)$ of the element and check if it meets the above condition. My thinking was if $(\ker\varphi)_p$ makes sense, the outcome of this check should be independent of the representative chosen, so if $(U,s)=(V,t)$ and $(U,s)$ satisfies the condition to be in $(\ker\varphi)_p$, $(V,t)$ should too. However, I haven't been able to show this. Here is my attempt:

If $(U,s)\in (\ker\varphi)_p$ by our definition above $s\in\mathscr{F}(U)$ and $\varphi_U(s)=0$. Since $(U,s)=(V,t)$, pick $W$ so $p\in W\subseteq U\cap V$ and $s|_W=t|_W$. Then we have $\varphi_V(t)|_W=\varphi_W(t|_W)=\varphi_W(s|_W)=\varphi_U(s)|_W=0$. However, I don't see any way to conclude $\varphi_V(t)=0$, which seems like what we want to conclude $(V,t)\in(\ker\varphi)_p$. Of course, we have $(V,t)=(W,t|_W)$ with the latter satisfying the property for $(\ker\varphi)_p$, but this doesn't address the concern.

So, is there some way to show $\varphi_V(t)=0$, or am I misunderstanding something about how $(\ker\varphi)_p$ is defined? Thank you!

Answer 1

You are misunderstanding something. The correct characterization is that $(U,s)\in(\ker \varphi)_p$ iff there exists a nonopen empty $U'\subset U$ such that $\varphi(U')(s|_{U'})=0$. Once you make this adjustment, you're good: if $(U,s)$ and $(V,t)$ are two representatives of the same element and $(U,s)\in(\ker\varphi)_p$, then $(U'\cap V,t|_{U'\cap V})$ is sent to zero by $\varphi(U'\cap V)$ and so $(V,t)\in(\ker\varphi)_p$ as well.

Answer 2

Your definition of $(\ker \phi)_p$ is correct, but it is not strictly speaking a subgroup of $\mathcal F_p$. Both use "the same" equivalence relation, but they are quotients of different sets. On one hand, $\mathcal F_p$ is a quotient of $$\{(U,f) \mid U\subseteq X, f\in \mathcal F(U)\}$$ while $(\ker \phi)_p$ is a quotient of $$\{(U,f) \mid U\subseteq X, f\in \ker \phi_U\}.$$ In this way, equivalence classes of $(\ker \phi)_p$, as sets, might be "smaller". I will give an example where the equivalence class of $(U,f)$ in $\ker \phi_U$ is smaller than the equivalence class of $(U,f)$ in $\mathcal F(U)$ below.

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Let $C^1$ and $C$ be the sheaves of continuously differentiable and continuous sheaves on $\mathbb R$ respectively. We have a morphism of sheaves $$C^1 \xrightarrow{d} C$$ sending a differentiable function $f$ to its derivative $f'$. Now, we can define $b: \mathbb R \to \mathbb R$ by $$b(x) = \begin{cases} 0 & x\in [-1,1]\\ (x+1)^2 & x\in (-\infty, -1)\\ (x-1)^2 & x\in (1,\infty) \end{cases}$$ so that $b\in C^1(\mathbb R)$. Define $U=(-1,1)$. Now, notice that $b|_{U} \in \ker d_U$, but $b \not \in \ker d_{\mathbb R}$. Therefore, by construction we have $$(\mathbb R, b) \sim (U, b|_U)$$ as elements of the stalk $(C^1)_0$, but $(U, b|_U)$ is literally an element of the equivalence class of $(U, b|_U)$ in $(\ker d)_p$, but $(\mathbb R, b)$ is NOT.

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Instead, the way we should think about this is that there is a natural inclusion where you send $$[(U,f)] \in (\ker \phi)_p \longmapsto [(U,f)] \in \mathcal F_p,$$ where you include an equivalence class into the bigger equivalence class. The content of the Hartshorne exercise is to prove that this is an injection with image being $\ker(\phi_p)$ (which is literally a subset of $\mathcal F_p$).

In terms of the example, although $(\mathbb R, b) \not\in [(U,b)] \in (\ker d)_0$, we do that that $(\mathbb R, b) \in [(U,b)] \in (C^1)_0$, and that $[(\mathbb R, b)]$ is sent to $0$ by $d_0$.