Can a contravariant functor be adjoint to a covariant one? I am a bit confused about the definition of adjoint functors, since everywhere the definitions found (see example wikipedia https://en.wikipedia.org/wiki/Adjoint_functors) seem to not specify if we are working with covariant / contravariant / no-matter-the-kind-of functors.
I know that of course every contravariant functors is essentially covariant - is it enough to work in the opposite category? Sometimes this solution does not satisfy completely, especially when I have to do some explicit computation.
This is the case in an exercise I am trying to solve, where I have to prove that if $F$ is left-adjoint to $G$, then $F$ is right exact while $G$ left exact (I'll omit details because I know it is a well-known result).
I actually had no problem in doing this exercise, but I assumed that $F$ and $G$ were covariant. My proof seems still working in the case are both contravariant, but the case $F$ covariant + $G$ contravariant (as well as for the converse) is interesting. Here my proof seems to show that this case is not possibile, leading to the following conclusion:

If $F$ and $G$ are two adjoint functors, they are both contravariant or they are both covariant.

Is this claim true, or am I missing some point?
Thank you.
 A: Let $F\colon\mathcal{C}\to\mathcal{D}$ and $G\colon\mathcal{D}\to\mathcal{C}$ be (covariant) functors. Then $G$ is a left adjoint to $F$ if there exists a natural isomorphism
$$
\hom_{\mathcal{C}}(G(X),Y)
\xrightarrow{\sim}
\hom_{\mathcal{D}}(X,F(Y)).
$$
For contravariant functors the same definition applies once we make them covariant. There are two ways for doing so, let's examine them.
So, let $F\colon\mathcal{C}\to\mathcal{D}$ and $G\colon\mathcal{D}\to\mathcal{C}$ be contravariant functors. We can get back to the previous situation by considering
$$\def\op{^{\mathrm{op}}}
F'\colon\mathcal{C}\to\mathcal{D}\op,\qquad
G'\colon\mathcal{D}\op\to\mathcal{C}
$$
and being $G'$ a left adjoint to $F'$ means there is a natural isomorphism
$$
\hom_{\mathcal{C}}(G'(X),Y)
\xrightarrow{\sim}
\hom_{\mathcal{D}\op}(X,F'(Y)).
$$
That's the same as saying there is a natural isomorphism
$$
\hom_{\mathcal{C}}(G(X),Y)
\xrightarrow{\sim}
\hom_{\mathcal{D}}(F(Y),X)
$$
(the functors are, in this case, adjoint on the left).
On the contrary, we could consider
$$
F''\colon\mathcal{C}\op\to\mathcal{D},\qquad
G''\colon\mathcal{D}\to\mathcal{C}\op
$$
and, being $G''$ a left adjoint to $F''$ means there is a natural isomorphism
$$
\hom_{\mathcal{C}\op}(G''(X),Y)
\xrightarrow{\sim}
\hom_{\mathcal{D}}(X,F''(Y))
$$
which is the same as saying there is a natural isomorphism
$$
\hom_{\mathcal{C}}(Y,G(X))
\xrightarrow{\sim}
\hom_{\mathcal{D}}(X,F(Y))
$$
(the functors are adjoint to the right).
You can notice that, in case the categories are abelian, contravariant functors which are adjoint on the right are left exact; they are right exact in case they are adjoint on the left. I use the usual convention that the contravariant functor $F$ is called left exact if it transforms the exact sequence $0\to A\to B\to C\to 0$ into the exact sequence $0\to F(C)\to F(B)\to F(A)$.
There is no way to sensibly pair up $F\colon\mathcal{C}\to\mathcal{D}$ and $G\colon\mathcal{D}\to\mathcal{C}$ if one is covariant and the other one is contravariant, because making them both covariant messes up either the domain or the codomain.
You may try all positions of $F$ and $G$ in the hom sets, but you'll not be able to compose maps in order to define “being natural”, whereas this is obviously possible for pairs of covariant or contravariant functors.
