Here is the context: Consider two finitely generated projective modules $M$ and $N$ over the commutative algebra of (smooth or) continous functions on a manifold $\mathcal{M}$ (they are sections of vector bundles over $\mathcal{M}$). I known that $M$ and $N$ are isomorphic. I have a morphism $f : M \to N$ which is one-to-one.
Here is my question: Does this implies that $f$ is onto (so an isomorphism)?
It is easy to construct counter-examples for modules over rings (for instance $M = N = \mathbb{Z}$ over $\mathbb{Z}$ with $f(n) = 2n$) that is one-to-one but not onto… But what about my specific context?
Thanks in advance
Thierry
Let $A$ be a commutative ring and suppose $a\in A$, with $a$ neither invertible nor a zero divisor (in your case a function that only vanishes at a single point).
Then the map $f\colon A\to A$, $f(x)=ax$ is one-to-one, but not surjective.