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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 $f^{-1}(\mathcal{F}) = X\times_Y \mathcal{F}$?

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    $\begingroup$ 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. $\endgroup$
    – Zhen Lin
    Commented Nov 1, 2012 at 20:47

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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 $X\times_{Y,f}\mathcal{F}$.

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