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Let $\mathcal{C}$ and $\mathcal{D}$ be two categories and $F : \mathcal{C} \to \mathcal{D}$ and $G: \mathcal{D} \to \mathcal{C}$ be two functors such that $F$ is left adjoint to $G.$ Also assume that $G$ is an exact functor and the unit map $G \circ F$ is an isomorphism. Since $F$ is a left adjoint, therefore, one may easily observe that $F$ is a right exact functor.

My question is: on the basis of the above assumption can one concludes the functor $F$ is also exact.

Math
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1 Answers1

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No we can not in that generality. The dual situation is more common: a fully faithful functor $i:\mathcal E\to \mathcal D$ is right adjoint to a left exact functor $a:\mathcal D\to \mathcal E$. (If $\mathcal D$ is a presheaf category, it is more or less a characterization for $\mathcal E$ to be a Grothendieck topos.)

The question you are asking then is: does $i$ preserve finite colimits? A priori no: suppose $\mathcal D$ is the category of presheaves over the open sets of a topological space $X$ and $\mathcal E$ is the category of sheaves on $X$; then a morphism in $\mathcal E$ is an epi as soon as it is surjective at each stalk, while its image by $i$ is an epi when it is surjective on every open set. I let you cook an explicit example based on that. (Or go fish there.)

Pece
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  • For a finite group $G$, let us assume that my category $\mathcal{D}$ to be the diagram category $Func(\mathcal{B}G^{op}, \underline{Ab})$, where $\mathcal{B}_G$ is the category with objects are finite $G$-sets and $\mathcal{B}_G(S,T) = Mor_G(\mathbb{Z}[S], \mathbb{Z}[T])$ for finite $G$-sets $S$ and $T.$ Also, $\mathcal{C} = Func(\mathcal{B}{G/K}^{op}, \underline{Ab}) $ for a normal subgroup $K$ of $G.$ – Math Jul 19 '18 at 08:44
  • For a finite group $H$, let us assume that my category $\mathcal{D}$ to be the diagram category $Func(\mathcal{B}H^{op}, \underline{Ab})$, where $\mathcal{B}_H$ is the category with objects are finite $H$-sets and $\mathcal{B}_H(S,T) = Mor_H(\mathbb{Z}[S], \mathbb{Z}[T])$ for finite $H$-sets $S$ and $T.$ Also, $\mathcal{C} = Func(\mathcal{B}{H/K}^{op}, \underline{Ab}) $ for a normal subgroup $K$ of $H$ and for the projection $\psi : H \to H/K,$ my $G$ is defined to be $ G(M) = M \circ \psi.$ Then $G$ preserves all finite colimits. – Math Jul 19 '18 at 08:52