Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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. – Zhen Lin Nov 1 '12 at 20:47

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 $\mathcal{G}$ that 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 equality, 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 space and continuous map; that is, the éspace étalé of $f^{-1}\mathcal{F}$ is (canonically homeomorphic) to the topological space $X\times_{Y,f}\mathcal{F}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.