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Let $\mathcal{F}$ and $\mathcal{G}$ be abelian categories. It is well-known that if a functor $\phi : \mathcal{F} \rightarrow \mathcal{G}$ has a right-adjoint (so $\phi$ is itself a left-adjoint to some other functor), then $\phi$ is right exact. Similarly, if $\phi$ has a left-adjoint (so $\phi$ is itself a right-adjoint to some other functor), then $\phi$ is left-exact.

I was going through the list of left- and right-exact functors I know, and they all were covered by the above condition. Can someone give me examples that appear "in nature" (so aren't too artificial) of left- or right-exact functors that are not adjoints?

I'm aware of the adjoint functor theorem, but its conditions are much stronger than just being left- or right-exact.

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The functor $\hom(P, -): \mathrm{Mod}_R \to \mathrm{Ab}$ for $P$ a projective $R$-module preserves finite colimits, but it doesn't preserve all colimits unless $P$ is finitely generated. –  Todd Trimble Feb 16 '13 at 16:07
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Related question -- mathoverflow.net/questions/93716/… –  Leonid Positselski Feb 16 '13 at 19:13
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What about the torsion functor $\Gamma_I$, where $I$ is an ideal? –  Liran Shaul Feb 16 '13 at 19:24
    
@Liran: The $I$-torsion functor, considered as taking values in the category of $I$-torsion $R$-modules, is right adjoint to the inclusion of the category of $I$-torsion $R$-modules in the category of $R$-modules. But considered as taking values in the category of $R$-modules, you are of course right. –  Fred Rohrer Feb 17 '13 at 7:16
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up vote 12 down vote accepted

I would disagree that the hypotheses of the adjoint functor theorem are much stronger than exactness. Left exactness is equivalent to preserving all finite limits, and the hypotheses of the adjoint functor theorem are existence of all limits, preserving all limits, and a smallness condition that usually is easy to verify. Furthermore, to know that a left exact functor preserves all limits, it suffices to know that it preserves arbitrary products (since any limit can be expressed as a kernel of an appropriate map between products). So in most typical applications, the only difference between being left exact and having a left adjoint is whether a functor preserves infinite products.

This also shows how to find a counterexample: find a left exact functor that does not preserve infinite products. For instance, if $M$ is any flat module over a commutative ring $R$, tensoring with $M$ is left exact, but will not preserve infinite products unless $M$ has nice finiteness properties (if $R$ is Noetherian, the condition is that $M$ is finitely generated). In particular, if $R=\mathbb{Z}$ you could take $M=\mathbb{Q}$, or if $R$ is a field you could take $M$ to be any infinite-dimensional vector space.

As Todd notes in his comment, you can similarly get an example for right exactness instead of left exactness by Homming out of a projective module that is not finitely generated.

You can get a more artificial sort of counterexample by taking abelian categories with a size restriction on their objects that prevents the adjoint from existing (because you don't have all (co)limits). For instance, take the category of countable-dimensional vector spaces over some field, and consider the endofunctor given by tensoring with a countably infinite dimensional space $V$. This is right exact, and it ought to have a right adjoint given by $\mathrm{Hom}(V,-)$. But this right adjoint is undefined because (for instance) $\mathrm{Hom}(V,V)$ is uncountable-dimensional and so it doesn't exist in our category.

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The last paragraph doesn't prove that the adjoint does not exist. –  Martin Brandenburg Feb 16 '13 at 17:36
    
Very enlightening. Thanks! –  Brad Feb 17 '13 at 1:55
    
"...a smallness condition which is usually easy to verify.". I think I once tried to verify it when trying to use the special adjoint functor theorem to prove the existence of free groups from the forgetful functor from $(Grp)$ to $(Set)$ -- and the smallness condition turned out to be not true! –  user30035 Feb 17 '13 at 22:56
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That's true--the conditions of the special adjoint functor theorem are a bit more, well, special. But the solution set condition of the general adjoint functor theorem is usually quite easy; for "forgetful functors", it just amounts to verifying that an object generated by a set has cardinality bounded in terms of the size of that set. –  Eric Wofsey Feb 18 '13 at 2:42
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There are important functors like this associated with coalgebras or corings. E.g., let $C$ be a coassociative coalgebra with counit over a field and $N$ be a right $C$-comodule. Then the cotensor product with $N$ over $C$ is a covariant left exact functor $N\square_C{-}\colon C{-}comod \to k{-}vect$ from the category of of left $C$-comodules to the category of $k$-vector spaces. This functor preserves infinite direct sums, but not infinite products, so it cannot be an adjoint (in general).

Similarly, let $M$ be a left $C$-comodule. Then the Cohom from $M$ over $C$ is a covariant right exact functor $Cohom_C(M,{-})\colon C{-}contra\to k{-}vect$ from the category of left $C$-contramodules to the category of $k$-vector spaces. This functor preserves infinite products, but not infinite direct sums, so it cannot be an adjoint (in general).

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There are many adjoints which preserve direct sums but not products, and the other way round. –  Fernando Muro Feb 16 '13 at 20:11
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Let me repeat again: the cotensor product functor is left, but not right exact. And it preserves infinite direct sums, not products. Similarly, the Cohom functor is right, but not left exact. And it preserves infinite products (in the second argument), but not direct sums. Functors with such properties can be neither left, not right adjoint. –  Leonid Positselski Feb 16 '13 at 22:59
    
I see, I misunderstood the fact that sides don't match. Sorry. –  Fernando Muro Feb 17 '13 at 0:51
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$\newcommand{\C}{\mathbf{C}} \newcommand{\AbGp}{\mathrm{AbGp}} \newcommand{\Psh}{\mathrm{Psh}} \newcommand{\Sh}{\mathrm{Sh}}$Let $\C$ be any site (i.e. small category equipped with a Grothendieck topology). Then the sheafification functor $\AbGp(\Psh(\C)) \to \AbGp(\Sh(\C))$ preserves finite limits, but not in general all limits; so it is a left exact functor without a left adjoint. (It does, however, have a right adjoint — the forgetful functor.)

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