Let $(X,\mathcal{T}_X)$ be a topological space, where $X=\{x\}$ is a point space, and hence $\mathcal{T}_X=\{\emptyset,\{x\}\}$.

Is it true that any sheaf $\mathcal{F}$ on $(X,\mathcal{T}_X)$ is a skyscraper sheaf? It seems that it is so by definition.

I define a skyscraper sheaf $S_x$ on the point $x\in X$ to be a sheaf $S_x:\mathcal{O}(X)^{op}\to\mathcal{C}$ where $\mathcal{C}$ is any category with a terminal object $T$, such that where $S\in\text{Ob}(\mathcal{C})$:

$$S_x(U)=\begin{cases}S,& x\in U,\\ T,&\text{ otherwise }\end{cases}$$ and for $f:V\hookrightarrow U$ I have $S_x(f)=id$ when $x\in V$ and is the unique terminal morphism otherwise.

It seems then that what I suggest is immediately true, just wanted a sanity check.

  • $\begingroup$ A skyscraper sheaf is one that's been pushed forward from a sheaf on a point. Every sheaf on a point is trivially pushed forward through the identity map $\endgroup$ – leibnewtz Sep 12 '18 at 14:37
  • $\begingroup$ @leibnewtz It's tautological in your definition, but you can see that is not my definition. Although I can take that as confirmation of the validity of above :-) $\endgroup$ – user591482 Sep 12 '18 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.