Prove that 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 $A$.
My solution : Let us take the inclusion maps $i_1:A→F_1$ and $i_2:A→F_2$. Then by universal property of modules we have $i_2=f_1*i_1$ and $i_1=f_2*i_2$, where the maps $f_1$ and $f_2$ are assigned by $f_1:F_1→F_2$ and $f_2:F_2→F_1$ and $*$ denotes the composition of two maps. Then from above argument we get $f_2*f_1$ is an identity map in $A$, so $f_1$ is invertible and hence isomorphism. But how can I show that $f_1$ is the identity only.
Please help me to solve this. I can't proceed further.
Hint: you need to use the uniqueness part of the universal property. Let $F$ be the free-module functor from Set to Mod, and $G$ the forgetful functor from Mod to Set; you've so far only used that "for every map $A \to GX$ there is a lift $FA \to X$". You need to use "for every map there is a unique lift".