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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 $colim_{f(U) \subset V} \mathcal{F}(V)$ 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 espace etale(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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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
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