As a follow-up to this question, I am trying to prove the following:
If $F$ be a free $R$-module on a set $S$ via the function $i:S \longrightarrow F$ then $i$ is necessarily injective.
Attempted Proof: This means that for every $R$-module $X$ and every function $f:S \longrightarrow X$ there exists a unique $R$-linear function $\tilde{f}:F \longrightarrow X$ such that $\tilde{f} \circ i = f$. So, consider the function $f:S \longrightarrow X$ defined by $f(x) := 0_X \;\forall\; x \in S$ where $0_X$ denotes the additive identity in $X$. Then, by definition of a free module,
$ \tilde{f} \circ i = f \implies \tilde{f} \circ i = 0_X $ But, this means that $\tilde{f}$ is a left-inverse for $i$ and $i$ has a left-inverse if and only if it is injective. Therefore, $i$ must be injective.
So, my question is, does this proof work?