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:


*

*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?

*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.
 A: I don't know if it's too late since I'm new to this site.

*

*It is correct when the space is "reasonable", or when "proper" means something different, like in algebraic geometry. For general topological space, I don't know the answer but my bet is it's wrong in general since (quasi-)compact subsets would no longer need to be closed. But for reasonable spaces and in algebraic geometry where "proper" implies "universally closed", the argument is simple as shown below.

*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 $x\not\in U$.

