I'm proving some properties about the free groups using the categorical definition, which is the following:
Let $X$ be a set, $L$ a group and $i:X\to L$ a map. The pair $(L,i)$ is free on $X$ if for every group $H$ and every map $f:X\to H$ there exists a unique group homomorphism $\varphi:G\to H$ such that $\varphi\circ i=f$.
I've already proved that
(1) $i$ must be injective and that
(2) if $(L_1,i_2)$ and $(L_2,i_2)$ are free on $X$ then there exists $\varphi:L_1\to L_2$ such that $\varphi\circ i_1=i_2$.
Now I'm stuck with the following:
If $|X|=|Y|$ and $(L_1,i_1)$ and $(L_2,i_2)$ are free on $X$ and $Y$ respectively, then $L_1\cong L_2$.
I tried using a biyection $g:X\to Y$ and (2) to find $\varphi:L_1\to L_2$ which I can prove that is surjective. However, I cannot show that it is inyective outside $i_1(X)$, what can I do?
Best Answer
Fix a bijection $g: X\rightarrow Y$ with inverse $h: Y\rightarrow X$. Then $f_1= i_2\circ g: X\rightarrow L_2$ defines a $\varphi_1: L_1\rightarrow L_2$ and similarly $f_2= i_1\circ h: Y\rightarrow L_1$ defines a $\varphi_2: L_2\rightarrow L_1$. The composition $\varphi_2\circ \varphi_1: L_1\rightarrow L_1$ is the homomorphisms corresponding to the mapping $id_X: X\rightarrow X$. As this map "extends" uniquely to a homomorphism $L_1\rightarrow L_1$ and the identity $id_{L_1}$ is obviously such a homomorphism, we have $\varphi_2\circ \varphi_1= id_{L_1}$. Similarly $\varphi_1\circ \varphi_2= id_{L_2}$, thus $\varphi_1$ is the desired isomorphism.