I was reading the proof of the following proposition:
Proposition. Let $G$ and $H$ be covariant or contravariant functors from $A$-mod to $B$-mod, and let $\theta: G \rightarrow H$ be a morphism of functors. If $\theta(A)$ : $G(A) \rightarrow H(A)$ is an isomorphism and both $G$ and $H$ are covariant and rightexact or both are contravariant and left-exact then $\theta(M): G(M) \rightarrow H(M)$ is an isomorphism for every finitely presented A-module $M$.
The proof starts as follows:
Proof. We consider the case when $G$ and $H$ are contravariant and left-exact. Since the functors are additive, they commute with finite direct sums. Therefore $\theta(F): G(F) \rightarrow H(F)$ is an isomorphism for every finitely generated free $A$-module $F$...
I am having trouble understanding why $\theta(F)$ would be an isomorphism.
By the definition of a morphism, we have that for every $A$-mod morphism $f$ and $B$-mod morphism $g,$ the following diagram is commutative:
$$\require{AMScd} \begin{CD} G(A) @>{\theta(A)}>> H(A)\\ @V{f}VV @V{g}VV \\ G(F) @>{\theta(F)}>> H(F) \end{CD}$$
We have that $\theta(A)$ is an isomorphism by assumption.
How do they use this and the fact that $G,H$ "commute with finite direct sums" to say that $\theta(F)$ is an isomorphism? Do we have to choose a $f,g$ accordingly?
Notation:
$A$-mod is the category of $A$-modules, where $A$ is a commutative ring.
