# Proper direct image and extension by zero

I was reading about the proper direct image functor, which can be defined in a general setting as follows.

Let $X$ and $Y$ be topological spaces and let $f:X\rightarrow Y$ be a continuous map. Let $\mathcal{F}$ be a sheaf of abelian groups on $X$. For a section $\sigma$ of $\mathcal{F}$ the support of $\sigma$ is defined to be the closure of $\{x\mid \sigma_x\neq 0\}$. The proper direct image $f_!\mathcal{F}$ is then defined to be the sheaf on $Y$ with $$f_!\mathcal{F}(V):=\left\{\sigma\in \mathcal{F}\left(f^{-1}(V)\right) \ \middle| \ \text{f|_{\mathrm{supp}(\sigma)}: \mathrm{supp}(\sigma)\rightarrow V is proper} \right\}.$$

Now consider the case where the map $f$ is an open embedding $U\rightarrow X$ and $\mathcal{F}$ is a sheaf of abelian groups on $U$. I have seen in many different texts stating the fact that in this case $f_!$ coincide with what is called "extension by zero", which is equivalent to saying that $$\left(f_!\mathcal{F}\right)_x=\left\{\begin{array}{ll} \mathcal{F}_x & \text{if x\in U},\\ 0 & \text{otherwise}. \end{array}\right.$$ I haven't been able to find any proof of such statement. While the first case ($x\in U$) is pretty obvious, I have not been able to prove the second case ($x\notin U$).

Just for the reference, while I was doing a search on the internet, I also came across this post on mathstackexchange from 2 years ago on the exact same topic, which has not been answered:

Prove extension by zero is a special case of lower shriek?

Here are my questions:

1. Is the statement correctly stated? Did I miss any topological conditions (such as locally compact or Hausdorff) on the spaces $X$ that would otherwise make the statement correct?

2. How to prove this statement? I feel like if the statement is correct, then one should be able to prove it just using point-set topology since we are stating all definitions in topological terms.

2. Only for $$x\not\in U$$: let $$V$$ be an open of $$X$$ containing $$x$$, and suppose $$s\in f_!\mathcal{F}(V)$$ be a non-zero section, then by definition of proper direct image, the restriction of $$f$$ to $$Z:=\mathrm{supp}(s)$$ is proper. Note $$Z\subset U\cap V\subset V$$. Since $$f|_Z:Z\to V$$ is universally closed, in particular it is closed, so $$Z$$ is closed in $$V$$. As $$x\not\in Z$$, we can take the open neighborhood $$V\setminus Z$$ of $$x$$, to which the restriction of $$s$$ is zero, hence $$s=0$$ in $$(f_!\mathcal{F})_x$$. Thus $$(f_!\mathcal{F})_x=0$$ for all $$x\not\in U$$.