A morphism of free modules that is an isomorphism upon dualization. 
Let $R$ be a commutative ring and let $M,N$ be free modules over $R$ and suppose we have a map $f: M \rightarrow N$ such that upon taking $\text{ Hom}(-,R)$ we get an isomorphism $f^* : N^* \rightarrow M^*.$ Must $f$ be an isomorphism as well?  

This is clear if $M,N$ are free and of finite dimension, since then double dualization can be applied. But in general, what is true?
Edit. In the comments egreg has sketched an example which seems fine. This seems to involve cardinality issues, so what happens if I assume that $M$ and $N$ are both countably generated?
 A: As pointed out by Jeremy Rickard, this answer is incorrect as originally stated and actually shows only that $f$ is an isomorphism iff $f^*$ is a topological isomorphism.  His answer gives an example where $f^*$ is an isomorphism but not a topological isomorphism and so $f$ is not an isomorphism.

For any $M$, the dual module $M^*$ can be given the topology of pointwise convergence (that is, the product topology considering $M^*$ as a subset of $R^M$ where $R$ has the discrete topology).  Note furthermore that for any homomorphism $f:M\to N$, the induced homomorphism $f^*:N^*\to M^*$ is continuous.  Finally, I claim that $M$ is the continuous dual of $M^*$ with respect to its topology: that is, every continuous homomorphism $M^*\to R$ is evaluation at an element of $M$.
To prove this, suppose $f:M^*\to R$ is a continuous homomorphism.  Since $f$ is continuous, $\ker(f)$ is an open neighborhood of $0$, which means that there are finitely many elements $x_1,\dots,x_n\in M$ such that for all $\alpha\in M^*$ with $\alpha(x_i)=0$ for $i=1,\dots,n$, $f(\alpha)=0$.  Let $M_0\subseteq M$ be a finitely generated free direct summand of $M$ containing each $x_i$ (for instance, pick a basis for $M$ and let $M_0$ be spanned by all the basis elements which have nonzero coefficient in some $x_i$).  Then $f$ vanishes on the kernel of the restriction map $M^*\to M_0^*$ and thus induces a homomorphism $g:M_0^*\to R$.  Since $M_0$ is finitely generated, $g$ is given by evaluation at some element $x\in M_0$.  It follows that $f$ is also given by evaluation at $x$.
It now follows easily that we can recover $f:M\to N$ from $f^*:N^*\to M^*$ by applying the continuous dual functor.  Thus $f$ is an isomorphism of $R$-modules iff $f^*$ is an isomorphism of topological $R$-modules.  However, beware that $f^*$ may be an isomorphism of $R$-modules without being an isomorphism of topological $R$-modules, since its inverse may not be continuous.  We do get a positive answer in the special case that $R$ is finite, in which case $N^*$ and $M^*$ are compact Hausdorff and so a continuous bijection between them automatically has continuous inverse.
A: Let $R=\mathbb{Z}_p$, the $p$-adic integers, and let $M=N$ be countably generated free modules. We can regard elements of $M$ as finite sequences of $p$-adic integers, and elements of $M^*$ as arbitrary sequences of $p$-adic integers.
Let $f$ be the map
$$(a_0,a_1,a_2,\dots)\mapsto(a_0,-pa_0+a_1,-pa_1+a_2,\dots),$$
so that $f^*$ is the map
$$(b_0,b_1,b_2,\dots)\mapsto(b_0-pb_1,b_1-pb_2,b_2-pb_3\dots).$$
Then $f$ is not invertible, since $(1,0,0\dots)$ is not in the image.
But $f^*$ is invertible, with inverse
$$(c_0,c_1,c_2,\dots)\mapsto(c_0+pc_1+p^2c_2+\dots,\text{  }
c_1+pc_2+p^2c_3+\dots,\text{  } c_2+pc_3+p^2c_4+\dots,\text{  }\dots).$$
