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Let $\pi:X\to Y$ be a continuous map and $\mathcal{G}$ be a sheaf on $Y$, then there is a natural isomorphism $f_p:(\pi^{-1}\mathcal{G})_p \to \mathcal{G}_{\pi(p)} $ (by using the adjunction of inverse image and pushforward or whatever means). $f_p's$ will induce a map $f: \bigsqcup_{p\in X}(\pi^{-1}\mathcal{G})_p \to \bigsqcup_{q\in Y}\mathcal{G}_{q}$.

I wonder if this map $f$ is also a continuous map with respect to the "space of section" (espace etale) topology, whose base are sets $\{(x,s_x):x\in U\}$, where $U\subset X$ is open and $s\in \mathcal G(U)$. The main difficulty for me is to find the explicit formula for $f_p$'s.

I ask this because I need this result to prove:

The pullback of the "space of sections" is the same as the "space of sections" of the pullback

No One
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Yes it is. Moreover, if we let $\mathcal E=\coprod_{q\in Y}\mathcal{G}_q$ be the etale space of $\mathcal G$ and $\mathcal{E}' =\coprod_{p\in X}\mathcal{G}_{\pi (p)}$ be the etale space of $\pi^{-1}\mathcal G$ then we have a commutative square $\require{AMScd}$ \begin{CD} \mathcal{E'} @>f>> \mathcal{E}\\ @V V V @VV V\\ X @>>\pi> Y \end{CD} which is a pullback (fibre product) in the category of topological spaces. Indeed we can define the inverse image of sheaf maps by insisting that $\mathcal{E}'=\mathcal{E}\times_Y X$ be the fibre product.

Angina Seng
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  • I am sorry but I have said that I need the continuity to prove the inverse image is a fibre product (the linked thread). In the proof, I NEED $f$ to be continuous. How to prove it? – No One Dec 27 '17 at 05:33
  • @TiWen Define a topology on $\coprod_p \mathcal{G}_{\pi(x)}$ by taking as basis pullbacks of sections of $\mathcal G$. Then it is a straightforward slog to show this is the etale space of the inverse image sheaf. A reference on the right may be relevant: https://math.stackexchange.com/questions/146996 . – Angina Seng Dec 27 '17 at 05:40