Exercise 3.D.9

Proof. Suppose $\mathit{ST}$ is invertible. Since $\mathit{ST}$ is surjective, it follows that $S$ is also surjective and, thus, invertible.

Let $u,v\in V$ such that $\mathit{Tv}=\mathit{Tu}$. We have that $\mathit{STv}=\mathit{STu}$. But $\mathit{ST}$ is injective, hence $v=u$, proving $T$ is injective, and thus, invertible.

The converse is the same as Exercise 1. □