
Lemma 7.19.3. Let $u : \mathcal{C} \to \mathcal{D}$ and $v : \mathcal{D} \to \mathcal{C}$ be functors of categories. Assume that $v$ is right adjoint to $u$. Then we have

1. $u^ ph_ V = h_{v(V)}$ for any $V$ in $\mathcal{D}$,

2. the category $\mathcal{I}^ v_ U$ has an initial object,

3. the category ${}_ V^ u\mathcal{I}$ has a final object,

4. ${}_ pu = v^ p$, and

5. $u^ p = v_ p$.

Proof. Proof of (1). Let $V$ be an object of $\mathcal{D}$. We have $u^ ph_ V = h_{v(V)}$ because $u^ ph_ V(U) = \mathop{Mor}\nolimits _\mathcal {D}(u(U), V) = \mathop{Mor}\nolimits _\mathcal {C}(U, v(V))$ by assumption.

Proof of (2). Let $U$ be an object of $\mathcal{C}$. Let $\eta : U \to v(u(U))$ be the map adjoint to the map $\text{id} : u(U) \to u(U)$. Then we claim $(u(U), \eta )$ is an initial object of $\mathcal{I}_ U^ v$. Namely, given an object $(V, \phi : U \to v(V))$ of $\mathcal{I}_ U^ v$ the morphism $\phi$ is adjoint to a map $\psi : u(U) \to V$ which then defines a morphism $(u(U), \eta ) \to (V, \phi )$.

Proof of (3). Let $V$ be an object of $\mathcal{D}$. Let $\xi : u(v(V)) \to V$ be the map adjoint to the map $\text{id} : v(V) \to v(V)$. Then we claim $(v(V), \xi )$ is a final object of ${}_ V^ u\mathcal{I}$. Namely, given an object $(U, \psi : u(U) \to V)$ of ${}_ V^ u\mathcal{I}$ the morphism $\psi$ is adjoint to a map $\phi : U \to v(V)$ which then defines a morphism $(U, \psi ) \to (v(V), \xi )$.

Hence for any presheaf $\mathcal{F}$ on $\mathcal{C}$ we have

\begin{eqnarray*} v^ p\mathcal{F}(V) & = & \mathcal{F}(v(V)) \\ & = & \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(h_{v(V)}, \mathcal{F}) \\ & = & \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(u^ ph_ V, \mathcal{F}) \\ & = & \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{D})}(h_ V, {}_ pu\mathcal{F}) \\ & = & {}_ pu\mathcal{F}(V) \end{eqnarray*}

which proves part (2). Part (3) follows by the uniqueness of adjoint functors. $\square$

Comment #1778 by Kiran Kedlaya on

Typo in the third paragraph: "calim".

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).