7.5 Functoriality of categories of presheaves

Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories. In this case we denote

$u^ p : \textit{PSh}(\mathcal{D}) \longrightarrow \textit{PSh}(\mathcal{C})$

the functor that associates to $\mathcal{G}$ on $\mathcal{D}$ the presheaf $u^ p\mathcal{G} = \mathcal{G} \circ u$. Note that by the previous section this functor commutes with all limits.

For $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ let $\mathcal{I}^ u_ V$ denote the category with

7.5.0.1
$$\label{sites-equation-colim-category} \begin{matrix} \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}^ u_ V) & = & \{ (U, \phi ) \mid U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), \phi : V \to u(U) \} \\ \mathop{Mor}\nolimits _{\mathcal{I}^ u_ V}((U, \phi ), (U', \phi ')) & = & \{ f : U \to U' \text{ in }\mathcal{C} \mid u(f) \circ \phi = \phi ' \} \end{matrix}$$

We sometimes drop the subscript ${}^ u$ from the notation and we simply write $\mathcal{I}_ V$. We will use these categories to define a left adjoint to the functor $u^ p$. Before we do so we prove a few technical lemmas.

Lemma 7.5.1. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories. Suppose that $\mathcal{C}$ has fibre products and equalizers, and that $u$ commutes with them. Then the categories $(\mathcal{I}_ V)^{opp}$ satisfy the hypotheses of Categories, Lemma 4.19.8.

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 $\phi '' : V \to u(U'')$. $\square$

Lemma 7.5.2. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories. Assume

1. the category $\mathcal{C}$ has a final object $X$ and $u(X)$ is a final object of $\mathcal{D}$ , and

2. the category $\mathcal{C}$ has fibre products and $u$ commutes with them.

Then the index categories $(\mathcal{I}^ u_ V)^{opp}$ are filtered (see Categories, Definition 4.19.1).

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 $\phi _2 : V \to u(U_2)$. $\square$

Given $g : V' \to V$ in $\mathcal{D}$ we get a functor $\overline{g} : \mathcal{I}_ V \to \mathcal{I}_{V'}$ by setting $\overline{g}(U, \phi ) = (U, \phi \circ g)$ on objects. Given a presheaf $\mathcal{F}$ on $\mathcal{C}$ we obtain a functor

$\mathcal{F}_ V : \mathcal{I}_ V^{opp} \longrightarrow \textit{Sets}, \quad (U, \phi ) \longmapsto \mathcal{F}(U).$

In other words, $\mathcal{F}_ V$ is a presheaf of sets on $\mathcal{I}_ V$. Note that we have $\mathcal{F}_{V'} \circ \overline{g} = \mathcal{F}_ V$. We define

$u_ p\mathcal{F}(V) = \mathop{\mathrm{colim}}\nolimits _{\mathcal{I}_ V^{opp}} \mathcal{F}_ V$

As a colimit we obtain for each $(U, \phi ) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}_ V)$ a canonical map $\mathcal{F}(U)\xrightarrow {c(\phi )}u_ p\mathcal{F}(V)$. For $g : V' \to V$ as above there is a canonical restriction map $g^* : u_ p\mathcal{F}(V) \to u_ p\mathcal{F}(V')$ compatible with $\mathcal{F}_{V'} \circ \overline{g} = \mathcal{F}_ V$ by Categories, Lemma 4.14.8. It is the unique map so that for all $(U, \phi ) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}_ V)$ the diagram

$\xymatrix{ \mathcal{F}(U) \ar[r]^{c(\phi )} \ar[d]_{\text{id}} & u_ p\mathcal{F}(V) \ar[d]^{g^*} \\ \mathcal{F}(U) \ar[r]^{c(\phi \circ g)} & u_ p\mathcal{F}(V') }$

commutes. The uniqueness of these maps implies that we obtain a presheaf. This presheaf will be denoted $u_ p\mathcal{F}$.

Lemma 7.5.3. There is a canonical map $\mathcal{F}(U) \to u_ p\mathcal{F}(u(U))$, which is compatible with restriction maps (on $\mathcal{F}$ and on $u_ p\mathcal{F}$).

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

Note that any map of presheaves $\mathcal{F} \to \mathcal{F}'$ gives rise to compatible systems of maps between functors $\mathcal{F}_ V \to \mathcal{F}'_ V$, and hence to a map of presheaves $u_ p\mathcal{F} \to u_ p\mathcal{F}'$. In other words, we have defined a functor

$u_ p : \textit{PSh}(\mathcal{C}) \longrightarrow \textit{PSh}(\mathcal{D})$

Lemma 7.5.4. The functor $u_ p$ is a left adjoint to the functor $u^ p$. In other words the formula

$\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(\mathcal{F}, u^ p\mathcal{G}) = \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{D})}(u_ p\mathcal{F}, \mathcal{G})$

holds bifunctorially in $\mathcal{F}$ and $\mathcal{G}$.

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 $u_ p\beta : u_ p\mathcal{F} \to u_ p u^ p\mathcal{G}$. We claim that the functor $u^ p\mathcal{G}_ Y$ on $\mathcal{I}_ Y$ has a canonical map to the constant functor with value $\mathcal{G}(Y)$. Namely, for every object $(X, \phi )$ of $\mathcal{I}_ Y$, the value of $u^ p\mathcal{G}_ Y$ on this object is $\mathcal{G}(u(X))$ which maps to $\mathcal{G}(Y)$ by $\mathcal{G}(\phi ) = \phi ^*$. This is a transformation of functors because $\mathcal{G}$ is a functor itself. This leads to a map $u_ p u^ p \mathcal{G}(Y) \to \mathcal{G}(Y)$. Another trivial verification shows that this is functorial in $Y$ 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. $\square$

Remark 7.5.5. Suppose that $\mathcal{A}$ is a category such that any diagram $\mathcal{I}_ Y \to \mathcal{A}$ has a colimit in $\mathcal{A}$. In this case it is clear that there are functors $u^ p$ and $u_ p$, defined in exactly the same way as above, on the categories of presheaves with values in $\mathcal{A}$. Moreover, the adjointness of the pair $u^ p$ and $u_ p$ continues to hold in this setting.

Lemma 7.5.6. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories. For any object $U$ of $\mathcal{C}$ we have $u_ ph_ U = h_{u(U)}$.

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

$\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{D})}(u_ ph_ U, \mathcal{G}) = \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ U, u^ p\mathcal{G}) = u^ p\mathcal{G}(U) = \mathcal{G}(u(U))$

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

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