A morphism of free modules that is an isomorphism upon dualization.

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.

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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.

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 $$(c_0,c_1,c_2,\dots)\mapsto(c_0+pc_1+p^2c_2+\dots,\text{ } c_1+pc_2+p^2c_3+\dots,\text{ } c_2+pc_3+p^2c_4+\dots,\text{ }\dots).$$