Why are these two stalks of sheaves equal?

Question

Let $\mathcal{F}$, be sheaf on a topological space $X$. Let $U\subset X$ an open subset and take $V\subset U$ open and $x\in V$. Then we have the claim that the stalk $\mathcal{F}|_U$ at $x$ is the same as the stalk of $\mathcal{F}|_V$ at $x$.

I somehow don't see why this should be true. If we take an element $\left[(\phi, Q)\right]\in \mathcal{F}|_{U,x}$ in the stalk, i.e. $Q\subset U$ open and $\phi\in \mathcal{F}|_U(Q)$, $x\in Q$, then I want to show that $\left[(\phi, Q)\right]\in \mathcal{F}|_{V,x}$. But I don't see why this should be true.

Can maybe someone explain this to me?

Answer 1

Let $\mathcal{F}, \mathcal{G}$, be sheaf on a topological space $X$. Let $U\subset X$ an open subset and take $V\subset U$ open and $x\in V$. Then we have the claim that the stalk $\mathcal{F}|_U$ at $x$ is the same as the stalk of $\mathcal{F}|_V$ at $x$.

If $x\in U \subseteq X$ there is a sequence of maps

$$ i:\{x\} \rightarrow U \rightarrow^j X$$

and by definition

$$F_x:=(j\circ i)^{-1}(F) \cong i^{-1}j^{-1}(F) \cong i^{-1}(F_U) \cong (F_U)_x$$

Comment: "Sorry I don't get your answer. We have defined Fx={(f,U):U⊂X,f∈F(U)}/∼ where (f,U)∼(g,V) iff there exists W⊂U∩V s.t. f|W=g|W and x∈W"

Answer: If $i:\{x\} \rightarrow X$ is the inclusion map of $x$ into $X$ you may check there is an isomorphism

$$F_x \cong i^{-1}F.$$

If $I(x)$ is the set of open substes of $X$ containing $x$, it follows this is a directed set in the following sense: We say for two elements $U,V \in I(x)$ that $U \leq V$ iff $V \subseteq U$. This set is partially ordered: For any two sets $U,V \in I(x)$ it follows $W:=U \cap V \in I(x)$ and $U \leq W, V \leq W$. We get a directed set of abelian groups $F(U)$ for $U \in I(x)$ with maps $\rho_{U,V}:F(U) \rightarrow F(V)$ for any sets $U,V \in I(x)$ with $U \leq V$. We usually define

$$F_x :=lim_{U\in I(x)} F(U)$$

as the direct limit of this direct system of abelian groups. This is by definition the following: Let $D(F):=\oplus_{U \in I(x)} F(U)$ and let $S(F)$ be the submodule generated by the following elements $s_U \in F(U), s_V\in F(V)$. The element $s_U-s_V\in S(F)$ iff there is a $U \leq W, V\leq W$ with $(s_U)_W=(t_V)_W$. Then we define

$$lim_{U\in I(x)} F(U):=D(F)/S(F).$$

This construction (when applied to any subset $j: Z \subseteq X$ instead of $x\in X$) is used to define the topological inverse image of a sheaf under $j$: We define

$$j^{-1}F:=lim_{Z \subseteq U} F(U).$$

Hence by definition $F_x:=i^{-1}F$.

Direct limits is defined in Matsumuras book on commutative ring theory, Hartshorne defines the topological inverse image of a sheaf of abelian groups using such direct limits in Chapter II.1.

With this definition using the directed system $F(U)$ for $U \in I(x)$ you will find that you may describe $F_x$ as you do in your comments: It is the set of equivalence classes of pairs $(U,s)$ with $U \in I(x), s\in F(U)$ and where $(U,s) \cong (V,t)$ iff there is an open set $x\in W \subseteq U \cap V$ and where $s_W=t_W$.

There is a map $f:F_x \rightarrow (F_U)_x$ defined by

$$f(V,s):=(V\cap U,s).$$

The map in the other direction is $g(W,s):=(W,s)$. These two maps are inverse maps in the sense that $f \circ g=g \circ f=Identity$.

Answer 2

An option is to find an isomorphism $\mathcal F_x\cong(\mathcal F|_U)_x$, for any open $U\subset X$ and $x\in U$. By definition, $\mathcal F_x:=\operatorname{colim}_{X\supset V\ni x}\mathcal F(V)$ and $(\mathcal F|_U)_x:=\operatorname{colim}_{U\supset V\ni x}\mathcal F(V)$; label the canonical maps of $\mathcal F_x$ with $\lambda_V$ and those of $(\mathcal F|_U)_x$ with $\epsilon_V$.

Obviously any open set $X\supset V\supset x$ contains some open set $U\supset V'\ni x$; the maps $\varepsilon_V:=\epsilon_{V'}\circ \mathcal F_{V',V}:V\to (\mathcal F|_U)_x$ make $(\mathcal F|_U)_x$ into a cocone under the (diagram of) open sets $X\supset V\ni x$. This construction is well defined: choosing another open set $U\supset V''\ni x$ contained in $V$, the map $\varepsilon_V$ just defined is equal to $\epsilon_{V''}\circ\mathcal F_{V'',V}$ (as they are both equal to $\epsilon_{V'\cap V''}\circ\mathcal F_{V'\cap V'',V}$).

It is also an easy check that this cocone $((\mathcal F|_U)_x,\varepsilon _V)_{X\supset V\ni x}$ has the universal property of a colimit, so that we get an isomorphism $\mathcal F_x\cong(\mathcal F|_U)_x$, the unique commuting with $\varepsilon _V$ and $\lambda_V$.