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{\mathrm{Mor}}\nolimits _\mathcal {D}(u(U), V) = \mathop{\mathrm{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{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(h_{v(V)}, \mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(u^ ph_ V, \mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{D})}(h_ V, {}_ pu\mathcal{F}) \\ & = & {}_ pu\mathcal{F}(V) \end{eqnarray*}
which proves part (4). Part (5) follows by the uniqueness of adjoint functors.
\square
Comments (1)
Comment #1778 by Kiran Kedlaya on
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