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We have a free module $F$ with basis $X$.

If $f: F \to B$ and $\alpha: A \to B$ are $R$-linear and $\alpha$ is surjective, to find an $R$-linear map $g:F \to A$ we would have to use the universal property of free modules, so to this end we're first defining a set map $\beta:X \to A$ defined by $\beta(x)=a_x$ where $a_x$ is such that $\alpha(a_x)=f(x)$. Then it follows there exists a unique $g$ such that $\alpha g=f$.

I want to know how is $\beta$ even a well-defined map to begin with? It's not clear that $\beta$ is well-defined, unless $\alpha$ is also injective, which it isn't.

Mark
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The point is that you choose an $a_x$ such that $\alpha(a_x)=f(x)$ for each $x$. There might be more than one $a$ such that $\alpha(a)=f(x)$, but you pick one and call it $a_x$ (doing this for every $x\in X$ simultaneously requires the axiom of choice in general). Then, for those specific $a_x$ you have chosen, you define $\beta(x)=a_x$.

Eric Wofsey
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