Let $A$ be an abelian group, and $X$ a topological space. Define the constant sheaf $\mathcal{A}$ on $X$ determined by $A$ as follows: for any open set $U \subset X$,

$\mathcal{A}(U)=$the group of continuous functions from $U$ to $A$, where $A$ is endowed with the discrete topology; with the usual restrictions we obtain a sheaf.

Moreover, the constant presheaf $\mathcal{F}$ of $A$ is defined by $\mathcal{F}(U)=A$ when $U \neq \phi$, again considering the usual restrictions we obtain a presheaf.

I am asked to show that the sheafification of $\mathcal{F}$ is $\mathcal{A}$.For this, I wanted to use the universal property of sheafification, and considered the application $\theta :\mathcal{F} \to \mathcal{A}$, $\theta(U)(a):=s_a$, where $s_a$ is the constant function on $U$, equal to $a$. I managed to show it is a morphism of sheaves. but given a sheaf $\mathcal{G}$ and a morphism $\varphi: \mathcal{F} \to \mathcal{G}$, I'm having trouble finding the unique morphism $\psi :\mathcal{A} \to \mathcal{G}$ s.t. $\psi \theta =\varphi$

  • $\begingroup$ Are you familiar with the 'local homeomorphism' view of sheaves? Then $\psi$ becomes a continuous function between the total spaces, where $\mathcal A$ has all stalks isomorphic to $A$. $\endgroup$ – Berci Dec 3 '12 at 13:56

Let $\mathcal{F}'$ be the sheaf associated to $\mathcal{F}$ and let $\theta : \mathcal{F} \to \mathcal{F}'$ be the canonical morphism. Recall that $\theta$ induces an isomorphism between the stalks at every point (Liu, 2.2.14); hence the stalks of $\mathcal{F}'$ are $A$ everywhere. Now consider the morphism $\phi : \mathcal{F} \to \mathcal {A}$ which sends every element of $A$ to the correaponding constant map. The universal property of the sheaf associated to $\mathcal{F}$ gives a unique morphism $\psi : \mathcal{F}' \to \mathcal{A}$ through which $\mathcal{F}$ factors. Prove that this is an isomorphism by showing the induced maps on the stalks are isomorphisms (Liu, 2.2.12). For surjectivity, consider for any germ $f_x \in \mathcal{A}_x$ the germ of the section $f(x)$, and use the fact that continuous maps to a discrete space are locally constant.


I'm late to the party, but here is the answer for reference's sake to the question of how to define the morphism $\psi$ in the verification of the universal property, as asked in the original question.

Let $f \in \mathcal{A}(U)$ for $U \subseteq X$ open. The key observation is that $\{f^{-1}(a)\}_{a \in A}$ forms a pairwise disjoint open cover of $U$. So it is enough to specify the value of $\psi(U)(f)|_{f^{-1}(a)}$ for each $a \in A$. The condition $\psi \circ \theta = \varphi$ forces us to define $\psi(U)(f)|_{f^{-1}(a)} = \varphi(U)(a)$. Define $\psi(U)(f)$ to be the section in $\mathcal{F}(U)$ obtained by gluing these together.


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.