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

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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?


1 Answer 1


That is indeed the right answer: the left adjoint is determined by what it does on (the Yoneda image of) $\mathcal{C}$, since it is a colimit-preserving functor, and there we have $$\mathrm{Hom}_{\hat{\mathcal{C}}}(y(c), F^* \phi)\cong\phi(Fc) \cong \mathrm{Hom}_{\hat{\mathcal{D}}}(y(Fc),\phi),$$ where $F^*$ is the functor given by composition with $F$.


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