Does $f\otimes_A 1_{A/m}:M\otimes A/m\to N\otimes A/m$ injective for all maximal $m$ imply $f$ is an isomorphism? 
Let $A$ be a commutative ring. Suppose $f\colon M\to N$ is a morphism of free $A$-modules of equal, finite rank. If $f\otimes_A 1_{A/m}$ is injective for all maximal ideals of $A$, does this imply $f$ is an isomorphism? 

I suspect it is true, since it seems similar to statements about localizing at maximal ideals in commutative algebra. 
Fixing bases of $M$ and $N$, let $D$ be the determinant of $f$ with respect to these bases. Towards a contradiction, suppose $D$ is not a unit, so that $f$ is not invertible. Then $D$ is contained in some maximal ideal $m$. By hypothesis, 
$$
f\otimes_A 1_{A/m}\colon M\otimes_A A/m\to N\otimes_A A/m
$$
is injective, and these are now $A/m$-vector spaces of equal, finite dimension, so $f\otimes 1_{A/m}$ is an isomorphism. 
As a scalar, $D$ is now $0$ since it's in $m$, is there a contradiction to be found using $f\otimes 1_{A/m}$?
 A: Suppose first that $A$ is local with maximal ideal $\mathfrak m$. Then you want to show that if $\hat f:M/\mathfrak mM\to N/\mathfrak mN$ is injective $f$ is an isomorphism. Since this is an injective map of $k=A/\mathfrak m$ vector spaces of equal dimension it is also onto, and this means that $f$ itself is onto (use Nakayama). It suffices you show that an onto map $f:M\to N$ with $M,N$ free of equal rank must be an isomorphism. There is a lovely argument by Vasconcelos: $M=A^n$ is an $A[X]$ module by $Xm=f(m)$ and the hypothesis is that $\mathfrak aM=M$ for the ideal $(X)$ of $A[X]$. By Nakayama there is an element $u=X p(X)\in \mathfrak a$ such that $(1-Xp(X))M=0$. It follows that if $f(m)=Xm=0$ then $m=0$, so $f$ is a monomorphism. Note this works when $f:M\to M$ is an endomorphism with $M$ finitely generated.
This proves the claim for local rings. Can you try to prove the general case now?
A: A homomorphism $f$ of free modules of the same rank, say $n$, corresponds to an $n\times n$ matrix, say $M$.
Notice that $f$ is an isomorphism iff $M$ is invertible.
Since $f$ modulo every maximal ideal is injective, hence bijective, we have that $M$ is invertible modulo every maximal ideal, that is, $\det M\ne0$ for every maximal ideal. This shows that $\det M$ is an invertible element of $A$, hence $M$ is invertible.
