In Moerdijk, Classifying spaces and classifying topoi, on page 22, we have a functor from the topos $\mathcal BG$ of right $G$-sets ($G$ is a group) to the topos $Sh(X)$, the sheaves (étale spaces) over a topological space $X$. We want to prove that it preserves colimits and finite limits. The Author says that "it suffices to check this for the stalk at each point $x\in X$". He then computes the stalk $F(S)_x$ for every $S\in \mathcal BG$ (call $F$ the functor) and finds that $F(S)_x\cong S$ for every $x$, so it is clear that $F$ preserves colimits and finite limits.

My question is: why does it suffice to study stalks? My idea is that, as always, the sheaf structure allows one to "recover" the global properties later, and focus on what happens on stalks. Can one help me formalizing this in relationship to the specific context of limits and colimits?

Thank you in advance.


1 Answer 1


For any functor between any categories, there are canonical maps $\varphi:F(\operatorname{lim}S_i)\rightarrow\operatorname{\lim} F(S_i)$ and $\psi:\operatorname{colim}F(S_i)\rightarrow F(\operatorname{colim}S_i)$. To say that $F$ preserve limits/colimits means that these maps are isomorphisms.

Now, if the target category is a category of sheaves over some topological space, there is a useful criterion for a map to be an isomorphism : it is enough to check it on stalks.

You get by composition $(F(\lim S_i))_x\overset{\varphi}\rightarrow (\lim F(S_i))_x\rightarrow \lim F(S_i)_x$. If this is a finite limit, the last map is an isomorphism (stalks commute with finite limits), hence $\varphi:F(\operatorname{lim}S_i)\rightarrow\operatorname{\lim} F(S_i)$ for any map $(F(\operatorname{lim}S_i))_x\rightarrow\operatorname\lim F(S_i)_x$ is an isomorphism.

The same works for colimits, but in that case, you don't need to assume that the colimit is finite because stalks commute with arbitrary colimits.

  • $\begingroup$ @DerekElkins See, I don't like very much this argument because stalks aren't exactly colimits of sheaves, aren't they ? More precisely, stalks are obtained by taking a colimit of $\Gamma(U_i,.)$ of $\mathcal{F}$, and because colimits in functor categories are computed termwise, stalks are in fact $(\operatorname{colim}\Gamma(U_i,.))(\mathcal{F})$ but we are taking colimit of left exact functors... From this, this is not obvious at all that stalks commute with arbitrary colimits... $\endgroup$
    – Roland
    Commented Jan 15, 2018 at 16:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .