3
$\begingroup$

Let $X$ be an irreducible space, and let $\{U_{i}\}_{i\in I}$ be an open covering of $X$. Let $\mathcal{F}$ be a sheaf on $X$ such that the restriction of $\mathcal{F}$ to each open $U_{i}$ is constant. I want to show that $\mathcal{F}$ has to be constant.

Note that the constant sheaf is the sheafification of the constant presheaf with value $A$ which assigns to every open the value $A$.

Notice that since $X$ is irreducible every pair $U_{i}$ and $U_{j}$ from the open covering have non-empty intersection. Define for $i\in I$ the sheaf $\mathcal{F}_{i}:=\mathcal{F}|_{U_{i}}$, I tried to show that we have isomorphisms $\mathcal{F}_{i}|_{U_{i}\cap U_{j}}$ to $\mathcal{F}_{j}|_{U_{i}\cap U_{j}}$ which satisfies the desired properties such that we can glue them uniquely to the constant sheaf. But I failed doing this.

Any help would be appreciated! I think my main struggles is how to deal with the sheafification here. Do I have a explicit description of the sets $\mathcal{F}_{i}(V)$ for $V\subset U_{i}$?

$\endgroup$

1 Answer 1

3
$\begingroup$

One way of doing this is to work like you have a generic point, a point with closure the whole space, contained in every open set. That is, define the set $A$ to be the colimit of $F(U)$ over all nonempty open sets, ordered by inclusion. Then, we have a canonical map $F(U)\rightarrow A$, so we can view $A$ as the constant sheaf, and we have our proposed isomorphism to a constant sheaf, since the constant presheaf is a sheaf on an irreducible space.

To verify that this map is an isomorphism, we can work locally, noting that this colimit can be taken with respect to the (cofinal) system of subsets contained in $U_i$, for any $i$. The reason we can do this is that our original colimit was a directed set, which is equivalent to our space being irreducible, after unravelling the definitions.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.