# Computation of the stalks of the skyscraper sheaf

It is well-known that the stalks of the skyscraper sheaf $$\mathcal{G}_P$$ (where $$\mathcal{G}_P(U)=G$$ if $$P \in U$$ and $$\mathcal{G}_P=(0)$$ otherwise). Indeed, stalks are of the form

$$(\mathcal{G}_P)_q=G$$ if $$q \in \overline{\{p\}}$$ and $$(\mathcal{G}_P)_q=(0)$$ otherwise. Let then $$q \in \overline{\{p\}}$$. Then any open neighborhood containing $$q$$, must also contain $$p$$, whence

$$(\mathcal{G}_P)_q=\lim\limits_{\substack{\rightarrow \\ q \in V}} \mathcal{G}_P(V)=\lim\limits_{\substack{\rightarrow \\ q \in V}}=G$$

(the direct limit of copies of the same abelian group agrees with the group itself?)

Now, consider $$q \notin \overline{\{p\}}$$. Then $$V=X \setminus \overline{\{p\}}$$ is an open neighborhood of $$q$$ not containing $$p$$ and, hence, $$\mathcal{G}_P(V)=\emptyset$$. Thus, note that if we take any element of the stalk at $$q$$, i.e. $$[(s,W)]_\simeq$$, then $$W \cap V$$ does not contain $$p$$ and, thus, the section is the zero section. Is it my argument correct?

Question: "Now, how can I conclude that the stalks agree with the trivial group?"

Answer: If $$Y:=\{x\}$$ is the one point topological space with $$x$$ as only point and topology given by $$\{ \{x\},\emptyset\}$$, there is a sheaf $$\underline{G}$$ on $$Y$$ defined by

$$\underline{G}(\{x\}):=G, \underline{G}(\emptyset):=(e)$$

with $$e\in G$$ the identity element. The group $$(e)$$ is the trivial group. There is a canonical map $$\rho: G \rightarrow (e)$$ of groups, and hence $$\underline{G}$$ is a sheaf of abelian groups on $$Y$$. You define

$$G_x:=i_*(\underline{G})$$

and verify that for $$x \notin V$$ it follows

$$G_x(V):=\underline{G}(i^{-1}(V))=\underline{G}(\emptyset)=(e)$$

is the trivial group by definition.

If $$G_n:=G$$ for all $$n$$ and $$\lambda_n: G_n \rightarrow G_{n-1}$$ is the identity map and $$\{G_n ,\lambda_n\}$$ is a direct system, it should follow that

$$lim_n G_n \cong G.$$

• I corrected my question. Does my argument work? Moreover, how can I show in general that the direct limit of the same group agrees with the group itself? Jun 2 at 13:02