The proof is given below:
But I do not understand the proof starting from the eighth line starting from "If now $G$ is ...." till the end of the proof, could anyone explain it to me in a simpler way? especially I do not understand when the author said at the end of the ninth line and the beginning of the tenth one "Then $E = \lambda^{-1}[ \mathscr{A}]$ is obviously a free basis of $G$", actually it is not obvious for me.
Thanks!
I am working from "Introduction to knot theory" of Richard H. Crowell and Ralph H. Fox
EDIT:
A group that has a free basis is called a free group.
A generating set $E$ of elements of a group $G$ a free basis if, given any group $H$, any function $\phi : E \rightarrow H$ can be extended uniquely to a homomorphism of $G$ into $H$.
Since $\lambda$ is an isomorphism, we get that $\lambda^{-1}[\mathscr{A}]$ is a generating set of $G$. Now take some map $f \colon \lambda^{-1}[\mathscr{A}] \rightarrow H$. We need to extend $f$ uniquely to a homomorphism $G \rightarrow H$. We can compose $f$ with $\lambda^{-1}$ to get a map from $[\mathscr{A}]$ to $H$, which induces a unique homomorphism $g \colon F[\mathscr{A}] \rightarrow H$. This means that $g \circ \lambda \colon G \rightarrow H$ is a homomorphism of groups. This homomorphism is indeed an extension of $f$ as $(g \circ \lambda) \mid_{\lambda^{-1}[\mathscr{A}]}(x) = g(\lambda(x)) = f \circ \lambda^{-1}(\lambda(x)) = f(x)$ (we used that $\lambda(x) \in [\mathscr{A}]$). Why is this homomorphism unique? Try to figure that out by yourself by using the similar arguments as above and that you get unique homomorphisms from $F[\mathscr{A}] \rightarrow H$ that extend maps from $[\mathscr{A}]$.