The Stacks project

Proof. There are two conditions to check.

First, suppose we are given three objects $\phi : V \to u(U)$, $\phi ' : V \to u(U')$, and $\phi '' : V \to u(U'')$ and morphisms $a : U' \to U$, $b : U'' \to U$ such that $u(a) \circ \phi ' = \phi$ and $u(b) \circ \phi '' = \phi$. We have to show there exists another object $\phi ''' : V \to u(U''')$ and morphisms $c : U''' \to U'$ and $d : U''' \to U''$ such that $u(c) \circ \phi ''' = \phi '$, $u(d) \circ \phi ''' = \phi ''$ and $a \circ c = b \circ d$. We take $U''' = U' \times _ U U''$ with $c$ and $d$ the projection morphisms. This works as $u$ commutes with fibre products; we omit the verification.

Second, suppose we are given two objects $\phi : V \to u(U)$ and $\phi ' : V \to u(U')$ and morphisms $a, b : (U, \phi ) \to (U', \phi ')$. We have to find a morphism $c : (U'', \phi '') \to (U, \phi )$ which equalizes $a$ and $b$. Let $c : U'' \to U$ be the equalizer of $a$ and $b$ in the category $\mathcal{C}$. As $u$ commutes with equalizers and since $u(a) \circ \phi = u(b) \circ \phi = \phi '$ we obtain a morphism $\phi '' : V \to u(U'')$. $\square$

Proof. The assumptions imply that the assumptions of Lemma 7.5.1 are satisfied (see the discussion in Categories, Section 4.18). By Categories, Lemma 4.19.8 we see that $\mathcal{I}_ V$ is a (possibly empty) disjoint union of directed categories. Hence it suffices to show that $\mathcal{I}_ V$ is connected.

First, we show that $\mathcal{I}_ V$ is nonempty. Namely, let $X$ be the final object of $\mathcal{C}$, which exists by assumption. Let $V \to u(X)$ be the morphism coming from the fact that $u(X)$ is final in $\mathcal{D}$ by assumption. This gives an object of $\mathcal{I}_ V$.

Second, we show that $\mathcal{I}_ V$ is connected. Let $\phi _1 : V \to u(U_1)$ and $\phi _2 : V \to u(U_2)$ be in $\mathop{\mathrm{Ob}}\nolimits (\mathcal{I}_ V)$. By assumption $U_1\times U_2$ exists and $u(U_1\times U_2) = u(U_1)\times u(U_2)$. Consider the morphism $\phi : V \to u(U_1\times U_2)$ corresponding to $(\phi _1, \phi _2)$ by the universal property of products. Clearly the object $\phi : V \to u(U_1\times U_2)$ maps to both $\phi _1 : V \to u(U_1)$ and $\phi _2 : V \to u(U_2)$. $\square$

Proof. This is just the map $c(\text{id}_{u(U)})$ introduced above. $\square$

Proof. Let $\mathcal{G}$ be a presheaf on $\mathcal{D}$ and let $\mathcal{F}$ be a presheaf on $\mathcal{C}$. We will show that the displayed formula holds by constructing maps either way. We will leave it to the reader to verify they are each others inverse.

Given a map $\alpha : u_ p \mathcal{F} \to \mathcal{G}$ we get $u^ p\alpha : u^ p u_ p \mathcal{F} \to u^ p \mathcal{G}$. Lemma 7.5.3 says that there is a map $\mathcal{F} \to u^ p u_ p \mathcal{F}$. The composition of the two gives the desired map. (The good thing about this construction is that it is clearly functorial in everything in sight.)

Conversely, given a map $\beta : \mathcal{F} \to u^ p\mathcal{G}$ we get a map $u_ p\beta : u_ p\mathcal{F} \to u_ p u^ p\mathcal{G}$. Let $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. We claim that the functor $u^ p\mathcal{G}_ V$ on $\mathcal{I}_ V$ has a canonical map to the constant functor with value $\mathcal{G}(V)$. Namely, for every object $(U, \phi )$ of $\mathcal{I}_ V$, the value of $u^ p\mathcal{G}_ V$ on this object is $\mathcal{G}(u(U))$ which maps to $\mathcal{G}(V)$ by $\mathcal{G}(\phi ) = \phi ^*$. This is a transformation of functors because $\mathcal{G}$ is a functor itself. This leads to a map $u_ p u^ p \mathcal{G}(V) \to \mathcal{G}(V)$. Another trivial verification shows that this is functorial in $V$ leading to a map of presheaves $u_ p u^ p \mathcal{G} \to \mathcal{G}$. The composition $u_ p\mathcal{F} \to u_ p u^ p\mathcal{G} \to \mathcal{G}$ is the desired map. $\square$

Proof. By adjointness of $u_ p$ and $u^ p$ we have

and hence by Yoneda's lemma we see that $u_ ph_ U = h_{u(U)}$ as presheaves. $\square$