3

Good afternoon : I whould like to know how to define the canonical Godement resolution of a flasque sheaf. Thanks a lot.

Bryan
  • 827

1 Answers1

8

Let $\mathscr F$ be a sheaf on a topological space $(X,\tau)$. One can define the Godement sheaf of $\mathscr F$ as follows: for any $U\in\tau$, set $$ \mathcal G^0\mathscr F(U)=\prod_{x\in U}\mathscr F_x, $$ and for any inclusion $V\to U$ look at the restriction $$ \mathcal G^0\mathscr F(U)\to\mathcal G^0\mathscr F(V),\,\,\,\,\,\,\,\,s\mapsto s|_V. $$ Note that this is a restriction in the usual sense, as an element $s\in\mathcal G^0\mathscr F(U)$ is an actual function $s:U\to\coprod_{x\in U} \mathscr F_x$. Here are some important facts:

  1. The (covariant) functor $\mathcal G^0:\textrm{Sh}(X)\to \textrm{Sh}(X)$ is exact. (easy)
  2. The Godement sheaf $\mathcal G^0\mathscr F$ is flasque for every $\mathscr F\in \textrm{Ob Sh}(X)$. (easy)
  3. There is a natural monomorphism $0\longrightarrow \mathscr F\overset{\eta}{\longrightarrow} \mathcal G^0\mathscr F$.
  4. There is a flasque resolution $\mathcal G^\bullet\mathscr F$ of $\mathscr F$: $$ 0\longrightarrow \mathscr F\overset{\eta}{\longrightarrow} \mathcal G^0\mathscr F \overset{d^0}{\longrightarrow} \mathcal G^1\mathscr F \overset{d^1}{\longrightarrow} \mathcal G^2\mathscr F \longrightarrow\dots\overset{d^{q-1}}{\longrightarrow}G^q\mathscr F. \,\,\,\,\,\,\,\,\,\,\,\,\,(\star) $$

Proof of 3.

Let $U\in\tau$. Define $$ \eta_U:\mathscr F(U)\to\prod_{x\in U} \mathscr F_x,\,\,\,\,\,\,\,\,s\mapsto (s_x)_{x\in U}. $$ Very explicitly, $\eta_U(s)$ is the function $$ \eta_U(s):U\to\coprod_{x\in U} \mathscr F_x,\,\,\,\,\,\,\,\,\,x\mapsto s_x. $$ We want to show that $\eta_U$ is injective (so that $\eta$ is injective too). Let $t\in\ker\eta_U$. Then $t_x=0$ for any $x\in U$. Thus for every $x\in U$ there exists an open subset $W_x$ such that $t|_{W_x}=0$. The $W_x$'s cover $U$, so by a sheaf axiom we get $t=0$ and $\eta_U$ is injective.

Proof of 4.

Prove existence and exactness of $(\star)$ by induction on $q$, the step $q=0$ being clear by 3 and 2. Now suppose we got $(\star)$ as we want, up to $q$. Let us define:

  • $\mathcal G^{q+1}\mathscr F:=\mathcal G^0\textrm{coker}(d^{q-1})$; this definition is a good candidate because this sheaf is flasque by 2. Now the arrows:
  • Define $d^q:\mathcal G^{q}\mathscr F\to\mathcal G^{q+1}\mathscr F$ to be the composition $$ \mathcal G^{q}\mathscr F\to \textrm{coker}(d^{q-1})\to \mathcal G^0(\textrm{coker}(d^{q-1})). $$ The second arrow is injective, so we get the desired exactness of $(\star)$.
Brenin
  • 14,072
  • Thank you very much ! It's clear now. Thank you again. :) – Bryan Jun 08 '13 at 11:17
  • @Brenin, what is $q$? Why is the resolution finite in your answer. The Godement resolution can be infinite, no? Your point 4. is misleading the way it is stated. – Flavius Aetius Jul 22 '23 at 22:02