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?


1 Answer 1


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


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


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

$$lim_n G_n \cong G.$$

  • $\begingroup$ 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? $\endgroup$ Jun 2, 2021 at 13:02

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