Lemma 6.21.8. Let $f : X \to Y$ be a continuous map. There are bijections between the following four sets

the set of maps $\mathcal{G} \to f_*\mathcal{F}$,

the set of maps $f^{-1}\mathcal{G} \to \mathcal{F}$,

the set of $f$-maps $\xi : \mathcal{G} \to \mathcal{F}$, and

the set of all collections of maps $\xi _{U, V} : \mathcal{G}(V) \to \mathcal{F}(U)$ for all $U \subset X$ and $V \subset Y$ open such that $f(U) \subset V$ compatible with all restriction maps,

functorially in $\mathcal{F} \in \mathop{\mathit{Sh}}\nolimits (X)$ and $\mathcal{G} \in \mathop{\mathit{Sh}}\nolimits (Y)$.

**Proof.**
A map of sheaves $a : \mathcal{G} \to f_*\mathcal{F}$ is by definition a rule which to each open $V$ of $Y$ assigns a map $a_ V : \mathcal{G}(V) \to f_*\mathcal{F}(V)$ and we have $f_*\mathcal{F}(V) = \mathcal{F}(f^{-1}(V))$. Thus at least the "data" corresponds exactly to what you need for an $f$-map $\xi $ from $\mathcal{G}$ to $\mathcal{F}$. To show that the sets (1) and (3) are in bijection we observe that $a$ is a map of sheaves if and only if corresponding family of maps $a_ V$ satisfy the condition in Definition 6.21.7.

Recall that $f^{-1}\mathcal{G}$ is the sheafification of $f_ p\mathcal{G}$. By the universal property of sheafification a map of sheaves $b : f^{-1}\mathcal{G} \to \mathcal{F}$ is the same thing as a map of presheaves $b_ p : f_ p\mathcal{G} \to \mathcal{F}$ where $f_ p$ is the functor defined earlier in the section. To give such a map $b_ p$ you need to specify for each open $U$ of $X$ a map

\[ b_{p, U} : \mathop{\mathrm{colim}}\nolimits _{f(U) \subset V} \mathcal{G}(V) \longrightarrow \mathcal{F}(U) \]

compatible with restriction mappings. We may and do view $b_{p, U}$ as a collection of maps $b_{p, U, V} : \mathcal{G}(V) \to \mathcal{F}(U)$ for all $V$ open in $Y$ with $f(U) \subset V$. These maps have to be compatible with all possible restriction mappings you can think of. In other words, we see that $b_ p$ corresponds to a collection of maps as in (4). Of course, conversely such a collection defines a map $b_ p$ and in turn a map $b : f^{-1}\mathcal{G} \to \mathcal{F}$.

To finish the proof of the lemma you have to show that by "forgetting structure" the rule that to a collection $\xi _{U, V}$ as in (4) associates the $f$-map $\xi $ with $\xi _ V = \xi _{f^{-1}(V), V}$ is bijective. To do this, if $\xi $ is a usual $f$-map then we just define $\tilde\xi _{U, V}$ to be the composition of $\xi _ V : \mathcal{G}(V) \to \mathcal{F}(f^{-1}(V))$ by the restruction map $\mathcal{F}(f^{-1}(V)) \to \mathcal{F}(U)$ which makes sense exactly because $f(U) \subset V$, i.e., $U \subset f^{-1}(V)$. This finishes the proof.
$\square$

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