# Contravariant right exact functor?

There are many examples of exact functors, and also there are standard examples for contravariant/covariant left exact (e.g. hom functor) and covariant right exact (e.g. tensor product). Is there any example of contravariant right exact functor which is in general not left exact, especially the one with explicit description (i.e. not just like "adjoint to some functor in some category") or the one which is found to be useful? Also, I'd like to find some general abstract explanation on why there is no easy example for contravariant right exact functor while the other three types have typical examples.

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Can't you just take a covariant left exact functor and then take the opposite of the target category? –  Qiaochu Yuan Mar 9 '12 at 21:18
If you take a category with finite projective dimension (or possibly injective?) $d$ then $Ext^d(_,X)$ should satisfy this condition. –  Will Sawin Mar 9 '12 at 21:30
@Gjujin: I don't agree with the premise of your question that there are few such functors. @Will: When $X$ has injective dimension $\leq d$, then $\mathrm{Ext}^d(-,X)$ is an example. –  Martin Brandenburg Mar 9 '12 at 21:54

There is a natural functor with such property in the theory of coalgebras and co/contramodules over them. Given a (coassociative, counital) coalgebra $C$ over a field $k$, a left comodule $M$ over $C$ is a $k$-vector space with a structure (coaction) map $M\to C\otimes_k M$. A left contramodule $P$ over $C$ is a $k$-vector space with the structure (contraaction) map $Hom_k(C,P)\to P$. (The appropriate co/contraassociativity and counit axioms have to be satisfied in both cases.)
Given a left $C$-comodule $M$ and a left $C$-contramodule $P$, the $k$-vector space of cohomomorphims $Cohom_C(M,P)$ is defined as the quotient space of the vector space $Hom_k(M,P)$ by the image of the difference of two maps $Hom_k(C\otimes_k M,P)\rightrightarrows Hom_k(M,P)$, one of which is induced by the $C$-coaction in $M$ and the other one by the $C$-contraaction in $P$. The $Cohom$ is a right exact functor in both of its arguments, contravariant in the first (comodule) argument and covariant in the second one.
If $T$ is a contravariant cohomological $\delta$-functor with $T^{d+1}=0$, then $T^d$ is an example of a contravariant right-exact functor. There is no real conceptual difference between "contravariant" and "covariant" because of the duality of abelian categories; a contravariant functor is just a functor on the dual category. Thus you also get examples from covariant cohomological $\delta$-functors. There are lot's of specific examples in the context of derived functors (Will has mentioned Ext = derived Hom).