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.

## 7.5 Functoriality of categories of presheaves

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

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

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.

**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

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

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

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

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

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

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

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

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

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