Injective map needed in definition of free groups

Question

I have the following definition:

Let $X$ be a set. The free group genereted by $X$ is a group $F$, if there exists an injection $i:X\to F$ s.t. for all Groups $G$ and (not neccesarly injectiv) morphisms $j: X\to G$ there is a unique grouphomomorphism $f:F\to G$ such that the diagram commutes.

I'm wondering about why we ask that $i$ has to be injective and why $j$ dont has to be. Is there a specific reason? What would happen if I would allow $i$ to be not injective? What would happen if I would allow just injective $j$s?

Answer 1

If $i$ is not injective but it turns out that $j$ is, then there is no group homomorphism $f:F\longrightarrow G$ such that $f\circ i=j$, since $f\circ i$ cannot possibly be injective.