Let $F:\mathcal{C}\to\mathcal{D}$ be a functor between small categories. We define the functor
\begin{align*}
f:\hat{\mathcal{D}}&\longrightarrow\hat{\mathcal{C}} \\
G&\longmapsto G\circ F^{\mathrm{op}},
\end{align*}
where $\hat{\mathcal{C}}=[\mathcal{C}^{\mathrm{op}},Sets]$ and $\hat{\mathcal{D}}=[\mathcal{D}^{\mathrm{op}},Sets]$ are the presheaf categories of $\mathcal{C}$ and $\mathcal{D}$ respectively.
I want to show that $f$ has both left and right adjoints.
For the right adjoint, defining a functor $f^{*}:\hat{\mathcal{C}}\to\hat{\mathcal{D}}$ by setting
\begin{equation*}
f^{*}(H)(D):=\mathrm{Hom}_{\hat{\mathcal{C}}}(f(y_{D}),H)$
\end{equation*}
for each presheaf $H\in\hat{\mathcal{C}}$ and each object $D\in\mathcal{D}$, we get the desired right adjoint since by the Yoneda lemma we get that
\begin{equation*}
f^{*}(H)(D)\cong \mathrm{Hom}_{\hat{\mathcal{C}}}(y_{D},f_{*}(H)).
\end{equation*}
However, I have a problem finding the left adjoint. I have a strong feeling that
the desired map is the functor $f_{*}:\hat{\mathcal{C}}\to\hat{\mathcal{D}}$ which is induced by the composition arrow
\begin{equation}
{\mathcal{C}}\xrightarrow{F}{\mathcal{D}}\xrightarrow{y_{\mathcal{D}}}\hat{\mathcal{D}}
\end{equation} via the universal property of the Yoneda embedding $y_{\mathcal{C}}:{\mathcal{C}}\to\hat{\mathcal{C}}$, i.e. the unique colimit preserving functor that makes the diagram
commute. It is known that this functor has a right adjoint. I want to prove that this right adjoint is isomorphic to $f$.
I am having trouble showing this. I have started doubting that this map is the desired one. Any help?
Best Answer
Any right adjoint $R$ to a cocontinuous functor $L:\widehat{\mathcal{C}}\to \mathcal E$ from a presheaf category is defined by $R(e)(c)=\mathcal E(L(y_c),e)$. So the right adjoint $R$ of $f_*$ is defined by $$R(H)(c)=\widehat{\mathcal{D}}(f(y_c),H)=\widehat{\mathcal D}(y_{F(c)},H)=H(F(c)),$$ showing $R$ coincides with $f$, as desired.
By the way, your notation choices could be improved, as there's no notation connection between $F$ and $f$. A common options are to call $f$, instead, $F^*$, to give the impression of "pullback of presheaves along $F$"; of course, one could also just write this as $(-)\circ F$ to be quite transparent. The right Kan extension is sometimes denoted by $F_*$, close to the notation you chose for the left, in which case the left Kan extension would be denoted $F_!$. Again, there is also the more explicit option of $\mathrm{Ran}_F$ and $\mathrm{Lan}_F$, respectively.