# Inverse image sheaf and éspace étalé

Let $f\colon X \rightarrow Y$ be a continuous map of topological spaces. Let $\mathcal{F}$ be a sheaf of abelian groups on $Y$. The inverse image sheaf $f^{-1}(\mathcal{F})$ is the sheaf associated to the presheaf which assigns $\operatorname{colim}_{f(U) \subset V} \mathcal{F}(V)$ for every open subset $U$ of $X$, where $V$ runs through every open subset $V$ of $Y$ containing $f(U)$. We identify $\mathcal{F}$ with its éspace étalé (e.g. Hartshorne's algebraic geometry, Ch. II). Let $X\times_Y \mathcal{F}$ be the fiber product of topological spaces. Then how do we prove $f^{-1}(\mathcal{F}) = X\times_Y \mathcal{F}$?

• Painfully. A proof that $f^{-1}$ is left adjoint to $f_*$ is given in Mac Lane and Moerdijk's Sheaves in geometry and logic, Chapter II §9, and then one appeals to the fact that left adjoints are unique up to unique isomorphism. Commented Nov 1, 2012 at 20:47

Without change the names, for any $x\in X$ one defines $$(f^{-1}\mathcal{F})_x=\lim_{\overrightarrow{x\in U\,\text{open}}}(f^{-1}\mathcal{F})(U)$$ where $(f^{-1}\mathcal{F})$ is the sheafification of the presheaf $\mathcal{G}$, which assigns to any open subset $U$ of $X$ the Abelian group $\displaystyle\lim_{\overrightarrow{f(U)\subseteq V\,\text{open}}}\mathcal{F}(V)$; therefore $f^{-1}\mathcal{F}$ and $\mathcal{G}$ have the same stalks.
In other words: $$\forall x\in X,\,(f^{-1}\mathcal{F})_x=\mathcal{G}_x=\lim_{\overrightarrow{x\in U\,\text{open}}}\left(\lim_{\overrightarrow{f(U)\subseteq V\,\text{open}}}\mathcal{F}(V)\right)\cong\lim_{\overrightarrow{f(x)\in V\,\text{open}}}\mathcal{F}(V)=\mathcal{F}_{f(x)};$$ by this (canonical) isomorphism, one can state that the following diagram $$\require{AMScd} \begin{CD} f^{-1}\mathcal{F} @>>> \mathcal{F}\\ @VVV & @VVV\\ X @>>f> Y \end{CD}$$ is Cartesian in the category $\mathbf{Top}$ of topological spaces and continuous maps; that is, the éspace étalé of $f^{-1}\mathcal{F}$ is (canonically homeomorphic) to the topological space $X\times_{Y,f}\mathcal{F}$.