Extension of an open embedding by zero (a.k.a lower-shriek functor)

Motivating problem. We would like to show the following statement. Let $\mathscr{I}$ be an injective sheaf on a topological space $X$ valued in $\textbf{Ab}.$ Then it is flasque, meaning that for every open $U \subset X,$ the restriction map $\Gamma(X, \mathscr{I}) \rightarrow \Gamma(U, \mathscr{I})$ is surjective.

Denote by $i : U \hookrightarrow X$ the corresponding open embedding. We define the functor $$i_{!} : \mathrm{Mod}_{\mathscr{O}_{U}} \rightarrow \mathrm{Mod}_{\mathscr{O}_{X}}$$ as follows: given an $\mathscr{O}_{U}$-module $\mathscr{F},$ we define $O(X)^{\mathrm{op}} \rightarrow \textbf{Ab}$ by $V \mapsto \Gamma(V, \mathscr{F})$ if $V \subset U$ and $V \mapsto 0$ if $V \not\subset U.$ This is a presheaf on $X$ because the restriction maps are either given by restriction maps of $\mathscr{F}$ if the relevant open subsets are within $V$ or zero maps if any of the relevant open subsets is not contained in $V.$ Functoriality is easy to check because if any bigger open subset is zero, then we would have to get zero maps.

We define $i_{!}\mathscr{F}$ the sheafification of this presheaf. Since the presheaf has $\mathscr{O}_{X}$ action on it, this gives an $\mathscr{O}_{X}$-module structure on $i_{!}\mathscr{F}.$ Given any $\mathscr{O}_{U}$-linear map $\mathscr{F}_{1} \rightarrow \mathscr{F}_{2},$ we can easily get the map between corresponding presheaves on $X,$ which gives us a sheaf map $i_{!}\mathscr{F}_{1} \rightarrow i_{!}\mathscr{F}_{2}.$ Functoriality for a composition $\mathscr{F}_{1} \rightarrow \mathscr{F}_{2} \rightarrow \mathscr{F}_{3}$ is also easy to check, which tells us that $i_{!}$ is indeed a functor.

What's important is the following.

Adjunction. We have $$\mathrm{Hom}_{\mathscr{O}_{X}}(i_{!}\mathscr{F}, \mathscr{G}) \simeq \mathrm{Hom}_{\mathscr{O}_{U}}(\mathscr{F}, i^{-1}\mathscr{G}),$$ functorial in $\mathscr{F}$ and $\mathscr{G},$ where the isomorphism is taken in $\textbf{Set},$ and it is important to remember how this map is given.

If we start with an $\mathscr{O}_{X}$-linear map $i_{!}\mathscr{F} \rightarrow \mathscr{G},$ for any open $V \subset X,$ we have $$\Gamma(U \cap V, \mathscr{F}) \rightarrow \Gamma(U \cap V, i_{!}\mathscr{F}) \rightarrow \Gamma(U \cap V, \mathscr{G}),$$ and this is compatible with inclusions in $X$ as we consider other open subsets. Hence, this gives rise to $\mathscr{F} \rightarrow \mathscr{G}|_{U} = i^{-1}\mathscr{G}.$

Conversely, if we are given $\mathscr{F} \rightarrow i^{-1}\mathscr{G},$ then for any open $V \subset X,$ if $V \subset U,$ consider $\Gamma(V, \mathscr{F}) \rightarrow \Gamma(V, \mathscr{G})$ and if $V \not\subset U,$ consider $0 \rightarrow \Gamma(V, \mathscr{G}).$ As we vary $V,$ with respect to inclusion, we get a map from the presheaf we discussed above whose sheafification is $i_{!}\mathscr{F}$ to $\mathscr{G}.$

It is not difficult to check that these two procedures are inverses to each other (partially because $\mathscr{G}$ is a sheaf, which lets us use the universal property of the sheafification).

Personal remark. Note that $$\begin{align*} i^{*}\mathscr{G} &= i^{-1}\mathscr{G} \otimes_{i^{-1}\mathscr{O}_{X}}\mathscr{O}_{U} \\ &= i^{-1}\mathscr{G} \otimes_{\mathscr{O}_{U}}\mathscr{O}_{U} \\ &= i^{-1}\mathscr{G} \end{align*}$$ and both of them are $\mathscr{G}|_{U}.$ Note that this is a special situation where $i$ is an open embedding.

Understanding local sections. For our main application, we take $\mathscr{F} = \mathscr{O}_{U}$ in the adunction formula so that we have $$\mathrm{Hom}_{\mathscr{O}_{X}}(i_{!}\mathscr{O}_{U}, \mathscr{G}) \simeq \mathrm{Hom}_{\mathscr{O}_{U}}(\mathscr{O}_{U}, \mathscr{G}|_{U}).$$ Now, note that $$\mathrm{Hom}_{\mathscr{O}_{U}}(\mathscr{O}_{U}, \mathscr{G}|_{U}) \simeq \Gamma(U, \mathscr{G}),$$ again where we consider the isomorphism only as a bijection given by $\phi \mapsto \phi_{U}(1).$

We are now ready to tackle our motivating problem. We are given a topological space $X$ and an injective sheaf $\mathscr{I}$ valued in $\textbf{Ab}.$ We can view $X$ as a ringed space $(X, \mathscr{O}_{X}),$ where $\mathscr{O}_{X} = \underline{\mathbb{Z}}.$ Sheaves on $X$ valued in $\textbf{Ab}$ are precisely $\underline{\mathbb{Z}}$-modules. Hence, we will be done if we show the following.

Theorem. Let $(X, \mathscr{O}_{X})$ be a ringed space. Any injective $\mathscr{O}_{X}$-module is flasque.

Proof. Consider any injective $\mathscr{O}_{X}$-module $\mathscr{I}$ and an open subset $U \subset X.$ Let $s \in \Gamma(U, \mathscr{I}).$ Our goal is to find $t \in \Gamma(X, \mathscr{I})$ such that $t|_{U} = s.$

We know $s$ corresponds to an $\mathscr{O}_{U}$-linear map $\mathscr{O}_{U} \rightarrow \mathscr{I}|_{U}$ whose local section $$\Gamma(U \cap V, \mathscr{O}_{U}) \rightarrow \Gamma(U \cap V, \mathscr{I})$$ is given by $f \mapsto f s|_{U \cap V}.$ By adjunction, this corresponds to an $\mathscr{O}_{X}$-linear map $i_{!}\mathscr{O}_{U} \rightarrow \mathscr{I}.$ Note that the sequence of $\mathscr{O}_{X}$-linear maps $$0 \rightarrow i_{!}\mathscr{O}_{U} \rightarrow \mathscr{O}_{X}$$ the second of which is corresponding to $\mathrm{id} : \mathscr{O}_{U} \rightarrow \mathscr{O}_{X}|_{U}$ is exact as we can check its stalks. Since $\mathscr{I}$ is an injective $\mathscr{O}_{X}$-module, the map $i_{!}\mathscr{O}_{U} \rightarrow \mathscr{I}$ extends to an $\mathscr{O}_{X}$-linear map $\psi : \mathscr{O}_{X} \rightarrow \mathscr{I}.$ Let $t = \psi_{X}(1) \in \Gamma(X, \mathscr{I})$ be the corresponding section. Then we have $t|_{U} = s$ by sending $1$ according to the following identical compositions:

• $\Gamma(U, \mathscr{O}_{U}) = \Gamma(U, \mathscr{O}_{X}) \rightarrow \Gamma(U, \mathscr{I}),$
• $\Gamma(U, \mathscr{O}_{U}) \rightarrow \Gamma(U, i_{!}\mathscr{O}_{U}) \rightarrow \Gamma(U, \mathscr{I}).$

This finishes the proof $\Box$