Theorem 1

Let $M$ be a finite-dimensional free $A$-module and $u$ an endomorphism of $M$. The following conditions are equivalent: (a) $u$ is bijective; (b) $u$ is right invertible (II, § 1, no. 9, Corollary 1 to Proposition 15); (c) $u$ is left invertible (II, § 1, no. 9, Corollary 2 to Proposition 15); (d) $u$ is surjective; (e) $\det u$ is invertible in $A$.

Let $(e_i)_{1 \leq i \leq n}$ be a basis of $M$. If $x_i = u(e_i)$ for $1 \leq i \leq n$, then

By § 7, no. 9, Theorem 2 a necessary and sufficient condition for the $x_i$ to form a basis of $M$ is that $\det u$ be an invertible element of $A$; this proves the equivalence of (a) and (e). Observe that (a) obviously implies each of conditions (b), (c) and (d); it remains to prove that each of conditions (b), (c) and (d) implies (e). Now, if there exists an endomorphism $v$ of $M$ such that $v \circ u = 1_M$ or $u \circ v = 1_M$, then $(\det v)(\det u) = 1$ and hence $\det u$ is invertible in $A$. If $u$ is surjective, so is $\bigwedge^n(u)$ (§ 7, no. 2, Proposition 3), in other words the homothety of ratio det $u$ in $A$ is surjective, which immediately implies that det $u$ is invertible.