Left/right exact functor "in nature" which is not a right/left adjoint 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.
 A: Inverse image of sheaves is exact, but generally only has a right adjoint (direct image). 
Also, for locally compact Hausdorff spaces, direct image with proper supports is left exact, but does not have any adjoints in general unless we pass to the derived category.
A: 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 $\mathrm{Hom}(V,-)$ (and it is easy to show that if the right adjoint exists, it must be 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.
A: 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).
A: $\newcommand{\C}{\mathbf{C}}
\newcommand{\AbGp}{\mathrm{AbGp}}
\newcommand{\Psh}{\mathrm{Psh}}
\newcommand{\Sh}{\mathrm{Sh}}$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.)
