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.


1 Answer 1


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

  1. 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.
  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 $x\not\in U$.

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