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I have the following definition of free module:

A free module $F$ is an $R$-module together with a set map $i:X \to F$ such that given any other $R$-module $M$ and a set map $f:X \to M$ there exists a unique $R$-module homomorphism $\overline{f}:F \to M$ such that $\overline{f} \circ i = f$

However I see that very often $i$ is taken as an inclusion and $X$ as a subset of $F$.

Please can you clarify me what kind of identification is done here?

Note: this definition was taken from here

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This definition does not give new free modules when $i$ is not an inclusion. Say $i(a)=i(b)$ where $a,b\in X$. Let $M$ be a nonzero module and $f:X\rightarrow M$ be a map with $f(c)$ for $c\in X$ with $c\neq b$ and say $f(b)\neq 0$. Then $f$ does not extend to a map $g:F\rightarrow M$ since $f(a)=0$ but $f(b)\neq 0$.

In other words, there is no free module $(F,X,i)$ with $i$ is not an inclusion.

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  • $\begingroup$ what you mean is that necessarily $i$ is an inclusion and $X$ a subset of $F$ to have a good definition? $\endgroup$ – Javier Mar 8 '18 at 13:49
  • $\begingroup$ Your definition is also a definition of a free module. But as you say, it just hides the fact that there is no free module with $i$ is not an inclusion. $\endgroup$ – Levent Mar 8 '18 at 13:53
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    $\begingroup$ To be a bit more clear : your argument shows that for every free module $i$ must be injective. Once you know that, you can prove that WLOG one can assume that it actually is an inclusion. $\endgroup$ – Arnaud D. Mar 8 '18 at 13:59
  • $\begingroup$ @ArnaudD. Yes, of course. Thanks for pointing out. I used the terms inclusion and injective in a wrong way. $\endgroup$ – Levent Mar 8 '18 at 14:07

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