Let $\phi\colon M\to N$ be a homomorphism of left $R$-modules. A transpose of $\phi$ is a homomorphism $\phi^T\colon N^* \to M^*$ of right $R$-modules (where $N^*$ and $M^*$ are respective dual modules, that is, right modules of left module homomorphisms $N\to R$ an $M\to R$) which maps a linear functional $\psi\colon N\to R$ to $\psi\circ\phi$.

In Blyth's "Module Theory: An Approach to Linear Algebra", the following exercise (p.$105$, ex.$9.5$) is given:

Prove that a homomorphism of $R$-modules is surjective if and only if its transpose is injective.

It's easy to prove that if a homomorphism $\phi\colon M\to N$ is surjective, then $\phi^T$ is injective. Indeed, if $\psi\circ\phi = \phi^T(\psi) = \phi^T(\theta) = \theta\circ\phi$, then $\psi = \theta$ as $\phi$ is an epimorphism of modules.

$(1)$ I can't think of an argument to prove the converse direction. But if we knew that $\phi$ injective $\Longrightarrow$ $\phi^T$ surjective (I don't know if it's true) and if $M,N$ were finitely generated projective modules, then we could prove it by duality and resorting to $\phi^{TT}$, using the fact that mapping a module to its double dual is natural and that the canonical homomorphisms $M\to M^{**}$ and $N\to N^{**}$ are in fact isomorphisms for finitely generated projective modules.

$(2)$ That said, is the dual statement

a homomorphism of $R$-modules is injective if and only if its transpose is surjective.

true? Assuming the initial proposition, it should be true for at least finitely generated projective modules by implying double transpose duality.

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    $\begingroup$ The converse doesn't hold: e.g. $\phi = \cdot 2: \mathbb Z \to \mathbb Z$ is not surjective but has an injective transpose, and so does $\phi: \mathbb Z/2 \to \mathbb Z/4$ (in both cases considered as left $\mathbb Z$-modules)... I believe $\phi$ has to split for this to work. $\endgroup$ – user Jun 3 at 20:25

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