proof that there is a unique isomorphism between two free modules I'm trying to figure out if there's any way to shorten the following proof of Corollary 7 in Dummit & Foote (p. 355) so that I don't have to write down the expressions of any specific elements like $x = r_1a_1 + \dots + r_na_n$. Is it possible to write out the proof without needing to examine specific elements, or is it necessary in order to show that any endomorphism from $F_1$ to $F_1$ that is the identity on $A$ must also be the identity on $F_1$?
Corollary 7. Let $F_1$ and $F_2$ be free modules on the same set $A$. Then there is a unique $R$-module isomorphism between $F_1$ and $F_2$ which is the identity on $A$.
Proof. We first note by the universal property of $F_1$ that there can exist at most one homomorphism (and hence isomorphism) from $F_1$ to $F_2$ that is the identity on $A$. Since there exist inclusions $A\hookrightarrow F_1$ and $A\hookrightarrow F_2$, it follows from the universal properties of $F_1$ and $F_2$ that there exist unique $R$-module homomorphisms $\Phi_1\colon F_1\to F_2$ and $\Phi_2\colon F_2\to F_1$ that are the identity on $A$. Now define the $R$-module homomorphism $\Phi\colon F_1\to F_1$ by $\Phi = \Phi_2\circ\Phi_1$. It is clear that $\Phi$ is the identity on $A$. But then for all $x = r_1a_1 + \dots + r_na_n\in A$,
    \begin{align*}
    \Phi(x) & = \Phi(r_1a_1 + \dots + r_na_n) \\
    & = r_1\Phi(a_1) + \dots + r_n\Phi(a_n) \\
    & = r_1a_1 + \dots + r_na_n \\
    & = x.
    \end{align*}
    Thus, $\Phi$ is the identity on $A$, so $\Phi_1$ is invertible and hence is an isomorphism.

You can use the universal property applied to $A \hookrightarrow F_1$. The function $\Phi: F_1 \rightarrow F_1$ is the identity on $A$, so it satisfies the universal property for the map $A \hookrightarrow F_1$. In other words $\Phi$ extends $A \hookrightarrow F_1$ to a map $F_1 \rightarrow F_1$. But there is already such a map, and that is the identity $F_1 \rightarrow F_1$. By uniqueness of the universal property, they must be the same.
