Adjoint pair between Category of Chain Complexes I have to show that if (L,R) is an adjoint pair between two abelian category A and B, then  L and R induces
L:  Ch(A)->Ch(B)
R:  Ch(B) ->Ch(A)
(where Ch(A) is the category of chain complexes in A)
such that (L,R) is an adjoint pair.
I know I have to use the naturality of the isomorphism given by (L,R) but I'am a bit confused..Thanks for the help!!
 A: Since $L$ is a functor, if $f: M \to N$ is a morphism in A then $L(f): L(M) \to L(N)$ is a morphism in B. Thus, if we have a chain complex 
$$ (M_\bullet, d) =(\cdots \to M_i \to M_{i-1} \to \cdots) $$
the only reasonable candidate for $\mathbf{L}(M_\bullet, d^M)$ is  
$$(L(M_\bullet), L(d)) = (\cdots \to L(M_i) \to L(M_{i-1}) \to \cdots),$$
where of course I'm applying $L$ to the all of the morphisms $d_i: M_i \to M_{i-1}$, too. We have to check is  this again a chain complex and that composition of morphisms works, but this is straightforward:

* by virtue of being a functor, $L(d)^2 = L(d^2)$, and since $L$ is additive we know $L(0) = 0$. 

* composition of chain maps works fine essentially because chain maps in Ch(A) are built out of morphisms in A, and $L$ is a functor on A. It's a routine check, but do it if it's not totally clear. 
As you may guess, $\mathbf{R}$ is obtained in the same way. 
For the adjunction, try starting with a chain map $\phi \in \mathrm{Hom}_\mathbf{Ch(B)}(\mathbf{L}(M_\bullet, d), (N_\bullet, d'))$, use that this gives us a sequence $\phi_i \in \mathrm{Hom}_{\mathbf{B}}(L(M_i), N_i)$, use this to get $\psi_i \in \mathrm{Hom}_{\mathbf{A}}(M_i, R(N_i))$ by the original adjunction, and then prove that $\psi = (\dots, \psi_i, \dots)$ link up together to give a chain map $(M_\bullet, d) \to (R(N_\bullet), R(d'))$. 
