The Stacks project

Proof. This immediately follows from the fact that if $V = \bigcup V_ j$ is an open covering in $Y$, then $f^{-1}(V) = \bigcup f^{-1}(V_ j)$ is an open covering in $X$. $\square$

Proof. This is because $(g \circ f)_*\mathcal{F}(W) = \mathcal{F}((g \circ f)^{-1}W)$ and $(g_* \circ f_*)\mathcal{F}(W) = \mathcal{F}(f^{-1} g^{-1} W)$ and $(g \circ f)^{-1}W = f^{-1} g^{-1} W$. $\square$

Proof. The colimit is over the partially ordered set consisting of open subsets $V \subset Y$ which contain $f(U)$ with ordering by reverse inclusion. This is a directed partially ordered set, since if $V, V'$ are in it then so is $V \cap V'$. Furthermore, if $U_1 \subset U_2$, then every open neighbourhood of $f(U_2)$ is an open neighbourhood of $f(U_1)$. Hence the system defining $f_ p\mathcal{G}(U_2)$ is a subsystem of the one defining $f_ p\mathcal{G}(U_1)$ and we obtain a restriction map (for example by applying the generalities in Categories, Lemma 4.14.8).

Note that the construction of the colimit is clearly functorial in $\mathcal{G}$, and similarly for the restriction mappings. Hence we have defined $f_ p$ as a functor.

A small useful remark is that there exists a canonical map $\mathcal{G}(U) \to f_ p\mathcal{G}(f^{-1}(U))$, because the system of open neighbourhoods of $f(f^{-1}(U))$ contains the element $U$. This is compatible with restriction mappings. In other words, there is a canonical map $i_\mathcal {G} : \mathcal{G} \to f_* f_ p \mathcal{G}$.

Let $\mathcal{F}$ be a presheaf of sets on $X$. Suppose that $\psi : f_ p\mathcal{G} \to \mathcal{F}$ is a map of presheaves of sets. The corresponding map $\mathcal{G} \to f_*\mathcal{F}$ is the map $f_*\psi \circ i_\mathcal {G} : \mathcal{G} \to f_* f_ p \mathcal{G} \to f_* \mathcal{F}$.

Another small useful remark is that there exists a canonical map $c_\mathcal {F} : f_ p f_* \mathcal{F} \to \mathcal{F}$. Namely, let $U \subset X$ open. For every open neighbourhood $V \supset f(U)$ in $Y$ there exists a map $f_*\mathcal{F}(V) = \mathcal{F}(f^{-1}(V))\to \mathcal{F}(U)$, namely the restriction map on $\mathcal{F}$. And this is compatible with the restriction mappings between values of $\mathcal{F}$ on $f^{-1}$ of varying opens containing $f(U)$. Thus we obtain a canonical map $f_ p f_* \mathcal{F}(U) \to \mathcal{F}(U)$. Another trivial verification shows that these maps are compatible with restriction maps and define a map $c_\mathcal {F}$ of presheaves of sets.

Suppose that $\varphi : \mathcal{G} \to f_*\mathcal{F}$ is a map of presheaves of sets. Consider $f_ p\varphi : f_ p \mathcal{G} \to f_ p f_* \mathcal{F}$. Postcomposing with $c_\mathcal {F}$ gives the desired map $c_\mathcal {F} \circ f_ p\varphi : f_ p\mathcal{G} \to \mathcal{F}$. We omit the verification that this construction is inverse to the construction in the other direction given above. $\square$

Proof. This you can see as follows

Here we have used Categories, Lemma 4.14.10, and the fact that any $V$ open in $Y$ containing $f(x)$ occurs in the third description above. Details omitted. $\square$

Proof. This is a combination of Lemmas 6.17.2 and 6.21.4. $\square$

Proof. To see this use that adjoint functors are unique up to unique isomorphism, and Lemma 6.21.2. $\square$

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

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. $\square$

Proof. Immediate from Definition 6.21.9 and the definition of the map on stalks above. $\square$