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 $f^{-1}(\mathcal{F}) = X\times_Y \mathcal{F}$?
 A: Without change the names, for any $x\in X$ one defines
\begin{equation}
(f^{-1}\mathcal{F})_x=\lim_{\overrightarrow{x\in U\,\text{open}}}(f^{-1}\mathcal{F})(U)
\end{equation}
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:
\begin{equation}
\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)};
\end{equation}
by this (canonical) isomorphism, one can state that the following diagram
\begin{equation}
\require{AMScd}
\begin{CD}
f^{-1}\mathcal{F} @>>> \mathcal{F}\\
@VVV & @VVV\\
X @>>f> Y
\end{CD}
\end{equation}
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 $X\times_{Y,f}\mathcal{F}$.
