Isomorphic Free $R$-module has Universal Property

Question

Here is corollary 7 in chapter 10.3 of Dummit and Foote book on Algebra:

1. If $F_1$ and $F_2$ are free modules on the same set $A$, there is a unique isomorphism between $F_1$ and $F_2$ which is the identity map on $A$
2. If $F$ is any free $R$-module with basis $A$, then $F \simeq F(A)$. In particular, $F$ enjoys the same universal property with respect to $A$ as $F(A)$ does.

I already proved part 1., but I am having trouble with the 2., particularly the "in particular" part. Let $F$ is a free $R$-module. Now, from what I understand, $F(A)$ is the "canonical" construction of a free module over some set $A$, but it no more "free" over $A$ than $F$ is, so this means we can just take $F_1 = F$ and $F_2 = F(A)$ and apply part 1. But I am having trouble showing that $F$ has the universal property. Let $T : F(A) \to F$ denote an isomorphism between $F(A)$ and $F$. Given some $R$-module $M$ and mapping $\varphi : A \to M$, we need to find a unique homomorphism $\phi : F \to M$ such that $\phi (a) = \varphi (a)$ for all $a \in A$, somehow using $F(A)$'s universal property and the isomorphism $T$. It seems that this requires I cleverly choose some mapping $f: A \to M$ and apply $F(A)$'s universal property to get $\phi(a) = f(a)$ for all $a \in A$. I thought about taking $f$ to be some sort of composition with inverses, etc., so that the right cancellation would occur and I would end up with $\phi (a) = \varphi(a)$, but I don't see how to do this....

Answer 1

The isomorphism $T \colon F(A) \to F$ induces for every $R$-module $M$ a bijection \begin{align*} \operatorname{Hom}_R(F, M) &\to \operatorname{Hom}_R(F(A), M), \\ \phi &\mapsto \phi \circ T, \\ \phi' &\leftarrow\!\shortmid \phi' \circ T^{-1}, \end{align*} i.e. homomorphisms $F \to M$ correspond uniquely to homomorphisms $F(A) \to M$.

The isomorphism $T$ satisfies (and is uniquely determined by) $T(a) = a$ for every $a \in A$. So for every homomorphisms $\phi \colon F \to M$ we have for the corresponding homomorphism $\phi' = \phi \circ T \colon F(A) \to M$ that $\phi'(a) = \phi(a)$ for all $a \in A$.

It follows that $\phi(a) = \varphi(a)$ for all $a \in A$ if and only if $\phi'(a) = \varphi(a)$ for all $a \in A$, and that $\phi$ is the unique homomorphism $F \to M$ with this property if and only if $\phi$ is the unique homomorphism $F(A) \to M$ with this property.

Since we know that there exists a unique homomorphism $\phi' \colon F(A) \to M$ with $\phi'(a) = \varphi(a)$ for all $a \in A$ (by the universal property of $F(A)$), it follows that $\phi = \phi' \circ T^{-1} \colon F \to M$ is the unique homomorphism $F \to M$ with $\phi(a) = \varphi(a)$ for all $a \in A$.