If there is a bijection from a subset $S$ of a group $G$ onto $X$ then $F(X)$ isomorphic to $\langle S \rangle$, Where $F(X)$ free group on X

Question

Let $\phi: G \to F(X)$ be a group homomorphism suppose that $\phi$ maps a subset $S$ of $G$ bijectively onto $X$. Then $F(X) $ is isomorphic to $\langle S\rangle$, where $F(X)$ free group with basis $X$.

Idea of the proof : $\phi:S\to X$ is a bijection . $\phi^{-1}:X\to S\subset G$ is injective . Then by definition of free group there exist an unique group homomorphism $\bar{\phi}^{-1}: F(X)\to G$ which is also injective. I am unable to understand why the induced homomorphism is also injective?

Answer 1

The idea with that proof seems to be noting that $\phi \circ \bar{\phi}^{-1}$ is the identity $F(X) \to F(X)$. This is just because each element of $X$ maps to itself and the unique group homomorphism that does that is the identity.

Answer 2

Becaulse inverse of homomorphism is homomorphism and homomorphism is injection if only if $ker \phi^{-1} = \{ 0 \} $ . If it has more elements than 0.then $ \phi^{-1}(x)=0 $ then $x= \phi (0) =0$ so kerium of inverse of injection is singleton wich implies injectivity of inverse