= Solution
The identity $T=S^{-1}(ST)$ shows that $S,ST$ generate the same <group>. Direct multiplication gives
$$
S^2=-I,\qquad ST=\begin{pmatrix}0&-1\\1&1\end{pmatrix},\qquad(ST)^3=-I.
$$
Consequently their images $\bar S,\overline{ST}$ in the <quotient group> $PSL(2,\mathbb Z)$ satisfy $\bar S^2=1$ and $\overline{ST}^{\,3}=1$. The <universal property of a group presentation> gives a <group homomorphism> taking $u\mapsto\bar S$, $v\mapsto\overline{ST}$ from the stated presentation. It is <surjective> because these images generate the target. Injectivity is not needed here.
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