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I am confused about the skyscraper sheaf.

Say we have set $S$, and the topological space $X$. For any $x\in X$, the skyscraper sheaf $F_x$ assigns any open set $U$ containing $x$ the set $S$, and to open sets that do not contain $x$, it assigns $\{e\}$.

What are the sections in this sheaf? Are these the elements of the set $S$? How do the restriction maps work?

freebird
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1 Answers1

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A "section" of a sheaf $F$ over a set $U$ is by definition just an element of the set $F(U)$. So a section of $F_x$ over an open set containing $x$ is an element of $S$, and the only section over an open set not containing $x$ is $e$.

The restriction maps are defined in the only "obvious" way there is to define them. If $U\subseteq V$ and $x\not\in U$, then $F_x(U)=\{e\}$, and the restriction map $F_x(V)\to F_x(U)$ is the unique map from $F_x(V)$ to $F_x(U)$ (the map that sends everything to $e$). If $U\subseteq V$ and $x\in U$, then $x\in V$ as well, so $F_x(U)=F_x(V)=S$, and the restriction map is the identity map $S\to S$.

Eric Wofsey
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  • Thanks a lot Eric! Can you also explain how this goes on to show that the only non-trivial stalk is over $x$, and that the rest of the points will have 0 stalks over them? – freebird Sep 30 '16 at 23:04
  • Well, that's not necessarily true... – Eric Wofsey Sep 30 '16 at 23:06
  • In particular, what is true is that the stalk over any point in the closure of ${x}$ is $S$, and the stalk over any other point is ${e}$. You can find some more details at http://math.stackexchange.com/questions/606773/stalks-of-skyscraper-sheaf. – Eric Wofsey Sep 30 '16 at 23:10
  • Thanks. That's exactly what I was thinking. I'd read the above erroneous assertion somewhere on the internet – freebird Sep 30 '16 at 23:11