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Given a commutative square in a nice category, say, manifolds $Mfd$. Suppose all edges are submersions (I guess transverse should be OK), then the square is a pullback if and only if it locally is, i.e. for a (or all, no difference) cover $U_i$ of C, we have $P_i=A_i \times_{U_i} B_i$, where $A_i=A\times_C U_i$, similarly define $B_i$, $P_i$.

Pullback square

I can show it by hand, the question is how to show this assertion more abstractly? (It seems related to decent theory?) Thanks a lot.

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Yes, this is a question of descent. More precisely you are asking whether the base change functor $F : \mathbf{Mfd}_{/ C} \to \prod_i \mathbf{Mfd}_{/ U_i}$ reflects pullbacks. It certainly preserves pullbacks (an easy exercise in abstract nonsense), so it suffices to show that $F$ is conservative (i.e. reflects isomorphisms). But this is easy enough: given a morphism $f$ in $\mathbf{Mfd}_{/ C}$ such that $F f$ is an isomorphism in $\prod_i \mathbf{Mfd}_{/ U_i}$, we can construct $f^{-1}$ by gluing together $(F f)^{-1}$. In short, $F$ is conservative because jointly surjective families of open embeddings satisfy the descent condition in $\mathbf{Mfd}$.

By the way, I would not call $\mathbf{Mfd}$ a nice category. It has some terrible deficiencies.

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  • $\begingroup$ Thanks a lot. What I intended is to ask the statement in a general category with Grothendieck pretopology. It seems that your argument does work fine in general. $\endgroup$
    – Ma Ming
    Commented Nov 15, 2013 at 10:05

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