11
$\begingroup$

The lower shriek functor is defined by

$$f_{!}F(U)=\{s\in\Gamma(f^{-1}(U),F)\;:\; f|_{\mathrm{supp}(s)}:\mathrm{supp}(s)\rightarrow U\text{ is proper}\}$$

On the other hand, if $j:V\subset X$ is the inclusion of an open set, the extension by zero functor is defined by $$ j_{!}F(U)=\begin{cases} F(jU)=F(U) & U\subset V\\ 0 & \text{otherwise} \end{cases}$$

How can I prove these definitions coincide for inclusions of open sets? I don't know anything about base-change theorems, so I'd like to avoid them.

$\endgroup$
5
  • $\begingroup$ What is $f$ and $F$? $\endgroup$
    – Babai
    Commented Dec 19, 2015 at 19:20
  • $\begingroup$ @Babai $f$ is a continuous map $X\rightarrow Y$ of topological spaces. $F$ is some sheaf on $X$. $\endgroup$
    – user153312
    Commented Dec 19, 2015 at 22:00
  • 1
    $\begingroup$ I think there is a typo; it should state that $j_!F(U) = F(U\cap V)$ $\endgroup$
    – 54321user
    Commented Jul 25, 2016 at 1:33
  • 10
    $\begingroup$ Your second description of $j_!$ is not correct. In general it only describes a presheaf, not a sheaf. To get a sheaf, you need to sheafify. Equivalently, you can define $(j_! F)(U)$ as the set of those sections $s \in F(U \cap V)$ such that locally $s = 0$ or $U \subseteq V$ (by which I mean that there exists an open covering $U = \bigcup_i U_i$ such that for each $i$, $s|_{U_i \cap V} = 0$ or $U_i \subseteq V$). $\endgroup$ Commented Nov 8, 2016 at 12:17
  • $\begingroup$ Very well said, @Ingo. $\endgroup$ Commented Nov 11, 2016 at 9:45

0

You must log in to answer this question.