Let $A$ be a (not necessarily commutative) ring, $E$ a left $A$-module and $E^*$ the dual module of $E$. For an automorphism $u$ of $E$, let $\check{u}$ be the transpose of $u^{-1}$.
In Algebra, Chapter 2, Bourbaki makes the following statement:
The mapping $u\mapsto\check{u}$ is an isomorphism of the linear group $\mathbf{GL}(E)$ onto a subgroup of the linear group $\mathbf{GL}(E^*)$.
Maybe it's obvious, but I don't understand why this mapping is injective.
It is easy to see that $\check{u}=\text{id}_{E^*}$ is equivalent to $(u-\text{id}_E)(E)\subset M$, where $M$ is the orthogonal submodule to $E^*$. Is it known that $M=\{0\}$?
Can someone help me along?
Edit: (at the request of rschwieb)
Let $E,F$ be left $A$-modules, $u$ a linear mapping of $E$ to $F$. The transpose of $u$ is the mapping of $F^*$ into $E^*$ defined by $y^*\mapsto y^*\circ u$. It is a homomorphism of right $A$-modules.