Free groups and injective function

Question

We have the following about free groups:

Given a non-empty set $X$, it is possibly to find a group $G$ and a function $\sigma :X\to G$ such that $\sigma (X)$ algebraically generates $G$ and for every group $H$ and every function $f:X\to H$ there exists a homomorphism $\widetilde{f} : G\to H$ such that $\widetilde{f} \circ \sigma = f$.

Now, I want to show that $\sigma$ must be injective.

I'd want to give an injective function $f:X\to G$, because of this there would be a function $\widetilde{f}$ such that $\widetilde{f}\circ \sigma$ is injective.

But is it possibly to prove that cardinality of $X$ is equal or less than cardinality of $G$?

In your second line it must be "...there exists a unique homomorphism..." — DonAntonio, Apr 28 '13 at 18:59
@DonAntonio: not in this case because he assumes $\sigma(X)$ generates $G$. — Luiz Cordeiro, Apr 28 '13 at 18:59
So what is your question? Why is $\sigma,$ injective or why is $,|H|\le |G|,$ ? — DonAntonio, Apr 28 '13 at 19:00
@LuizCordeiro, the question's title is about free groups...and of course $,\sigma(X),$ generates $,G,$, otherwise all this, with or without the free part, is false. — DonAntonio, Apr 28 '13 at 19:01
So if I assume that $,\langle,\sigma(X),\rangle=G,$ then automatically $,G,$ is free? No, of course. What is then the additional condition that must be given? — DonAntonio, Apr 28 '13 at 19:05
No @LuizCordeiro, the universal property requires uniqueness, just as it is written in your link... — DonAntonio, Apr 28 '13 at 19:18

score 2 · Answer 1 · answered Apr 28 '13 at 19:10

Let $X$ be any set and $G=\mathscr{F}(X)$ be a free group as you defined, with corresponding function $\sigma:X\rightarrow G$. Let's show that $\sigma$ is injective.

Let $x\neq y$ in $X$, and let $f:X\rightarrow\mathbb{Z}$ be any function such that $f(x)\neq f(y)$ (for example, $f(x)=1$, $f(y)=2$ and $f(z)=0$ for any other $z$). Since $\mathbb{Z}$ is a group, there exists a homomorphism $\overline{f}:G\rightarrow \mathbb{Z}$ such that $\overline{f}\circ\sigma=f$, then $\overline{f}(\sigma(x))=f(x)\neq f(y)=\overline{f}(\sigma(y))$, and therefore $\sigma(x)\neq \sigma(y)$ (since $\overline{f}$ is a function).

We then conclude that $\sigma$ is injective.

Ohhh, good. Thank you very much, Luiz! – user73564 Apr 28 '13 at 19:22 — user73564, Apr 28 '13 at 19:22

Free groups and injective function

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