I'm reading about Sheaf Theory from the point of view of categories and I have the following question:

Suppose we have two small categories $\mathcal{C}_1, \mathcal{C}_2$ and $\alpha:\mathcal{C}_2\to \mathcal{C}_1$ a functor that admits a left adjoint $\beta:\mathcal{C}_1\to \mathcal{C}_2$.

Now if $F:\mathcal{C}^o_1\to \mathbf{Set}$ is a functor (presheaf) on $\mathcal{C}_1$ taking values in the category of sets (here $\mathcal{C}^o_1$ is the opposite category), then we can use $\alpha$ to define $\alpha_* F:=F\alpha^o:\mathcal{C}^o_2 \to \mathbf{Set}$.

Then we have a functor $\alpha_*:\mathbf{Fun}(\mathcal{C}^o_1,\mathbf{Set})\to \mathbf{Fun}(\mathcal{C}^o_2,\mathbf{Set})$

Question: Is it true that $\alpha_*$ admits a left adjoint too?

I guess that the answer must be YES by using somehow the functor $\beta_*:\mathbf{Fun}(\mathcal{C}^o_2,\mathbf{Set})\to \mathbf{Fun}(\mathcal{C}^o_1,\mathbf{Set})$ defined in the same way.

In order to prove that we need to stablish an isomorphism for every $F\in \mathbf{Fun}(\mathcal{C}^o_1,\mathbf{Set})$ and $G\in \mathbf{Fun}(\mathcal{C}^o_2,\mathbf{Set})$:

$$\mathbf{Nat}(\beta_*G,F) \cong \mathbf{Nat}(G,\alpha_* F) $$

But I don't know how to do it in a natural way. First of all, I tried to associate to a natural transformation $\varphi:G\to \alpha_* F$ another one $\psi:\beta_* G\to F$, but I don't see how do it. Secondly, I looked at this thread: How to show two functors form an adjunction , but I don't get how to construct in this case the unit and counit.

Thank you in advance for your help or references.


If $\alpha : C_2 \to C_1$ is any functor between small categories, then $\alpha_* : \widehat{C_1} \to \widehat{C_2}$ has a left adjoint $\alpha^*$. This can be seen directly from Freyd's Adjoint Functor Theorem. Explicitly, the left adjoint maps a presheaf $G \in \widehat{C_2}$ to the left Kan extension of $G : C_2^{op} \to \mathsf{Set}$ along $\alpha^{op}$, i.e. $\alpha^*(G)(x)=\mathrm{colim}_{x \to \alpha(y)} G(y)$.

Notice that $\alpha^*$ cannot be of the form $\beta_*$ since $\beta_*$ is continuous, but $\alpha^*$ is usually not continuous.

Example. Let $f : X \to Y$ be a continuous map of topological spaces. Then we get a functor $\alpha : \mathrm{Open}(Y) \to \mathrm{Open}(X)$, $V \mapsto f^{-1}(V)$, hence $\alpha_* = f_* : \mathrm{PSh}(X) \to \mathrm{PSh}(Y)$ (pushforward of presheaves) given by $(f_* F)(V)=F(f^{-1}(V))$ with left adjoint $\alpha^* = f^* : \mathrm{PSh}(Y) \to \mathrm{PSh}(X)$ (pullback of presheaves) given by $f^*(G)(U) = \mathrm{colim}_{f(U) \subseteq V} G(V)$.

  • $\begingroup$ Thank you very much. Finally it was naive of me to think that the adjoint must be related with the adjoint of the original functor. $\endgroup$ – Peter C Mar 3 '14 at 14:31
  • 2
    $\begingroup$ Actually, it is related to the original adjoint, because $[(-)^\mathrm{op}, \mathbf{Set}]$ is a 2-functor $\mathfrak{Cat}^\mathrm{coop} \to \mathfrak{CAT}$, and adjunctions in $\mathfrak{Cat}^\mathrm{coop}$ are the same as adjunctions in $\mathfrak{Cat}$. $\endgroup$ – Zhen Lin Mar 3 '14 at 16:40

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