This is paraphrased from exercise 2.2.5 in Qing Liu's book:
Let $X$ be a topological space. Fix a sheaf $\mathcal{G}$ on $X$ and a
closed point $x_{0}\in X$. We define a new sheaf $\mathcal{F}$ as follow:
$\mathcal{F}(U)=\mathcal{G}(U)$ if $x_{0}\notin U$ and
$\mathcal{F}(U)=\left\{s\in \mathcal{G}(U):s_{x_{0}}=0\right\}$
otherwise. Prove that $\text{Supp}$ $\mathcal{F}= \text{Supp}$
$\mathcal{G}\setminus \left\{x_{0}\right\}$. Here $\text{Supp}$
$\mathcal{F}=\left\{x\in X:\mathcal{F}_{x}\neq 0 \right\}$.
It is easy to see that $\text{Supp}$ $\mathcal{F}\subseteq \text{Supp}$ $\mathcal{G}\setminus \left\{x_{0}\right\}$, but I don't know how to prove the reverse that if $x\in \text{Supp}$ $\mathcal{G}\setminus \left\{x_{0}\right\}$ then $x\in \text{Supp}$ $\mathcal{F}$. Can anyone help? I'm sorry if my question is silly.
Best Answer
Assume that $x\in \text{Supp}$ $\mathcal{G}\setminus \left\{x_{0}\right\}$. Because $\mathcal{G}_x$ is non zero, we may find an open neighbourhood $U$ of $x$ together with a section $s\in \mathcal{G}(U)$ such that $s_x \not = 0$ in $\mathcal{G}_x$. Intersecting $U$ with $X\setminus \{x_0\}$ (which is an open subset of $X$), we may assume that $x_0\not \in U$. Now, because $\mathcal{F}(U)=\mathcal{G}(U)$, the section $s$ defines a non-zero element of the stalk of $\mathcal{F}$ at $x$, showing that $\mathcal{F}_x$ is not zero.
NB: Actually, if $x\not = x_0$, then $\mathcal{F}_x=\mathcal{G}_x$. This can be seen by writing the stalk as an inductive limit on open neighbourhoods of $x$, and intersecting these neighbourhoods with $X\setminus \{x_0\}$, which does not change the actual limit. That is: $$\begin{align} \mathcal{G}_x&= \varinjlim_{x\in U\subset X}\mathcal{G}(U)\\&= \varinjlim_{x\in U\subset X}\mathcal{G}(U\cap(X\setminus \{x_0\}))\\&= \varinjlim_{x\in U\subset X}\mathcal{F}(U\cap(X\setminus \{x_0\}))\\&= \varinjlim_{x\in U\subset X}\mathcal{F}(U)\\&= \mathcal{F}_x \end{align}$$