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Let $X$ be a topological space, $F$ an abelian sheaf on $X$ and $Z$ be a locally closed subset. Then, we can choose an open subset $V$ such that $Z \subset V$ and $Z$ is a closed subset in $V$. Define $\Gamma_Z(X,F)$ the subgroup of $F(V)$ consisting of all those sections of $F$ whose support is contained in $Z$. I reffered "Hartshorne's book Local cohomology". In book, $\Gamma_Z(X,F)$ is independent of $V$ chosen above. But, I can't understand this sentence. The meaning of this sentence is that for open sets $U,V$ containg $Z$ above defining subsets are isomorphic??? is it poosible to prove?? Help me....

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up vote 4 down vote accepted

Yes, for open subsets $U,V$ containing $Z$ the above defined groups are isomorphic. Here is a proof. Clearly it is enough to prove the isomorphism of groups you want using the whole space $X=U$ and another open subset $V$ containing $Z$. I.e. we claim an isomorphism of groups:

$$\Gamma_Z(X,\mathcal{F}) \simeq \Gamma_{Z \cap V}(V, \mathcal{F}|_V)= \Gamma_Z(V,\mathcal{F}|_V).$$

The map $\varphi:\Gamma_Z(X,\mathcal{F}) \rightarrow\Gamma_Z(V,\mathcal{F}|_V)$ given by restriction of sections is a morphism of abelian groups, indeed it is a well defined restriction of the canonical morphism of abelian groups $\mathcal{F}(X) \rightarrow \mathcal{F}(V)$. We need to prove it is a bijection.

The injectivity is given by the fact that if $\varphi(s_1)=\varphi(s_2)$, then $(s_1-s_2)|_V=0$ and $(s_1-s_2)|_{X\setminus Z}=0-0=0$. Since $V$ and $X\setminus Z$ are an open covering of $X$ we deduce $s_1-s_2=0$ because $\mathcal{F}$ is a sheaf (and not only a presheaf).

For the surjectivity let $s\in \Gamma_Z(V,\mathcal{F}|_V)$, and consider the zero section $0\in\mathcal{F}(X\setminus Z)$. Since the support of $s$ is contained in $Z$ we have that $s|_{V\cap(X\setminus Z)}=0|_{V\cap(X\setminus Z)}$. Using again the fact that $\mathcal{F}$ is a sheaf (again being only a presheaf would not suffice), we deduce that there exists $t\in\mathcal{F}(X)$ such that $t|_{V}=s$, and it is zero otherwise. This implies $t\in\Gamma_Z(x,\mathcal{F})$ and $\varphi(t)=s$.

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thanks Giovanni De Gaetano – Sang Cheol Lee Feb 6 '13 at 10:50
but, is it possible that we assume $U=X$? $Z$ be locally closed in $X$. – Sang Cheol Lee Feb 6 '13 at 11:23
The assumption $X=U$ is harmless. Indeed if $U,V$ are two open subsets containing $Z$ then $\Gamma_{Z\cap V}(V,\mathcal{F}|_V)\simeq\Gamma_Z(X,\mathcal{F})\simeq \Gamma_{U\cap Z}(U,\mathcal{F}|_U)$. Regarding the fact that $Z$ is only locally closed in $X$ I have been indeed a bit sloppy. As it is stated my proof works assuming $Z$ closed, it works also for $Z$ locally closed but you have to be a bit more careful when applying arguments involving open coverings of $X$. When I'll have some time I'll write down details about it. Thank you for pointing it out! – Giovanni De Gaetano Feb 6 '13 at 11:30
thank s your kinds – Sang Cheol Lee Feb 6 '13 at 17:06

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