# The “pullback presheaf” and the proper base change theorem in topology

Let $f:X\rightarrow Y$ be a continuous map of topological spaces and let $\mathcal{F}$ be a sheaf (say of abelian groups to fix the idea) on $Y$. Define the following rule on open sets of $X$: $$V\mapsto f'\mathcal{F}(V):=\varinjlim_{U\supseteq f(V)}\mathcal{F}(U)$$ One may check that $f'\mathcal{F}$ gives a presheaf on $X$. Under some strong assumptions one may show that $f'\mathcal{F}$ is a sheaf on $X$. Let me give two examples

1. If $Y$ is Hausdorff, locally compact and paracompact and $f$ is a closed embedding then $f'\mathcal{F}$ is actually a sheaf (this is not a trivial exercise, it took me a while to figure that out).

2. A variation of 1. is: if $Y$ is Hausdorff and locally compact, and $f$ is the embedding of a compact subset $Y\subseteq X$.

I got interested in this question since it is directly related to the proper base change theorem in topology which says the following:

Proper base change theorem: Let $f:X\rightarrow Y$ be a proper map with $Y$ Hausdorff and locally compact and $X$ paracompact. Then for any sheaf $\mathcal{F}$ on $Y$ and $y\in Y$ one has that \begin{align}\label{eqn} R^qf_*(\mathcal{F})_y\simeq H^q(X_y, f|_{X_y}^{-1}\mathcal{F}) \hspace{2cm} (\star) \end{align} where $X_y=f^{-1}(y)$ is the fiber above $y$.

So here are 3 questions:

Q 1 Is there a common generalization of 1. and 2. in the topological setting ?

Q 2 I would like to have a couple of (non-artificial ) examples where the presheaf $f'\mathcal{F}$ fails to be a sheaf in order to have a feeling for the possible geometrical (and/or topological) obstructions. (Note that this is closely related to examples of maps where the isomorphism $(\star)$ above fail).

Q 3 To what extend is it possible to generalize the proper base change theorem in the topological setting? (so here I have in mind of relaxing the assumptions on $f$ and may be adding additional restrictions on $Y$)

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Take $Y$ to be a point, $X$ a non-empty Hausdorff space that is not a point, and $\mathcal F$ a non-zero sheaf: then $f'\mathcal F$ is not sheaf. This would also seem to give counterexamples to your statement, unless I misunderstand something. – Angelo Aug 18 '11 at 18:30
Greg Muller provides an exmaple in mathoverflow.net/questions/45212/… – Peter McNamara Aug 18 '11 at 18:52
Hi @Angelo, thanks for pointing my mistake, I know what is wrong! – Hugo Chapdelaine Aug 18 '11 at 18:59
I needed my map $f$ to be a closed embedding and not just proper. – Hugo Chapdelaine Aug 18 '11 at 19:04
In my opinion, the pullback sheaf is easiest to understand using the etale space formulation of sheaves rather than the functorial description. If $Z\rightarrow Y$ is etale, then the pullback $X\times_Y Z\rightarrow X$ is etale over X. – Benjamin Steinberg Aug 18 '11 at 22:08