Okay so i was trying to prove that Sanov subgroup is free of rank 2 without using ping pong lemma. I'd like to prove it directly(if possible), using the universal property. So Sanov Subgroup is a subgroup of $SL_2(Z)$ generated by $\alpha=$ $ \begin{pmatrix} 1 & 2 \\ 0 & 1 \\ \end{pmatrix} $ and $\beta= \begin{pmatrix} 1 & 0 \\ 2 & 1 \\ \end{pmatrix}$
So to prove that the Sanov Subgroup $F$ is free of rank two i need that, calling $X=\{a,b\}$ e $i$ the inclusion $i:X->F$(let's say $i(a)=\alpha$ and $i(b)=\beta$) for every group $G$ and any map $f:X->G$ there exists a unique homomorphism $\phi:F->G$ so that $i\phi=f$.
That's what i did, even though i don't think this is totally correct. So suppose we have this map $f$ with, let's say, $f(a)=g_1$ and $f(b)=g_2$. Now the definition of our homomorphism $\phi$ is natural, with $\phi(\alpha)=g_1$ and $\phi(\beta)=g_2$. For the other elements $w \in F$, $w=\alpha^{k_1}\beta^{k_2}\alpha^{k_3}\beta^{k_4}....$ with $k_i \in Z$ we define $\phi(w)=g_1^{k_1}g_2^{k_2}g_3^{k_3}g_4^{k_4}...$
This is an omomorphism an it satisfies the above universal property. Now i should prove the uniquenes but, again, it's pretty simple. If we had such $\tau$ for the universal property it must be $\tau(\alpha)=g_1$ and $\tau(\beta)=g_2$ and then by the homomorphism property $\tau(wv)=\tau(w)\tau(v)$ it easily follows that $\tau=\phi$.
Now this proof doesn't convince me at all (most because i almost never used the structure of Sanov Subgroup, basically i could substitute those two matrices with any other two and use the same proof..)