# An irreducible topological space $X$ admits a constant sheaf iff it is indiscrete.

I can't see where I've gone wrong in the following proof. The statement above is wrong, as any topological space can be endowed with a constant sheaf.

Proof. That any set $X$ endowed with the trivial topology admits a constant sheaf is obvious: for any set $E,$ let $\mathscr{F}(X) = E.$

Conversely, suppose that $X$ admits a constant sheaf. Then since $X$ is irreducible, for every nonempty open set $U$ of $X,$ the restriction $\mathscr{F}(X)\to \mathscr{F}(U)$ is a bijection$^1$. We recall that a sheaf is a functor, $\mathscr{F}\colon \mathbf{Ouv}_X^{\text{op}} \to \mathbf{Set},$ where $\mathbf{Ouv}_X$ is the category whose objects are open $U\subseteq X$ and whose morphisms are the inclusions. Functors preserve isomorphisms; therefore $\mathscr{F}(X) \cong_{\mathbf{Set}} \mathscr{F}(U)$ implies that $X \cong_{\mathbf{Ouv}_X} U,$ i.e., $X = U.$

1. Exercise $2.13$ in Algebraic Geometry I: Schemes, With Examples and Exercises by Ulrich Görtz and Torsten Wedhorn.
• "Functors preserve isomorphisms". Yes. But you actually use it the other way around.
– user301452
Jan 27, 2018 at 20:06
• For example, if $C$ is any category, the functor $F \colon C \to \mathrm{Set}$ sending any object of $C$ to a fixed singleton set is a perfectly good functor. Further, $F(A) \cong F(B)$ for any objects $A, B$ in $C$, but $A$ and $B$ need not be isomorphic. Jan 27, 2018 at 20:08

It is not true that $\mathscr{F}(X)\cong\mathscr{F}(U)$ implies $X\cong U$! "Functors preserve isomorphisms" goes the other direction: if $F$ is a functor and $f:A\to B$ is an isomorphism in the domain category, then $Ff:FA\to FB$ is also an isomorphism, so if $A\cong B$, then $FA\cong FB$. The converse is not necessarily true.