What can we say about the dual of an $R$-module homomorphism? Suppose $R$ is some ring (not necessarily commutative) and let $M,\,N$ be $R$-modules. Now let $f:M\to N$ be an $R$-module homomorphism. If ${}^\ast$ denotes duality, then we can also consider the dual homomorphism $f^\ast:N^\ast\to M^\ast$. What can we now say about $f^\ast$?


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*If $f$ is injective (resp. surjective), is it true for modules (I know this is true for vector spaces) that $f^\ast$ is surjective (resp. injective)? If not, what conditions do we need to make this true?

*Suppose $N$ is an $R$-$R$ bimodule that is self-dual, i.e. $N^\ast\cong N$; is $f^\ast\circ f$ necessarily a non-trivial homomorphism?
 A: If $f$ is surjective, it is immediate from the definition that $f^*$ is injective.  On the other hand, if $f$ is injective, $f^*$ need not be surjective.  For instance, consider $R=\mathbb{Z}$ and $f:\mathbb{Z}\to\mathbb{Z}$ given by multiplication by $2$.  Then $\mathbb{Z}^*\cong \mathbb{Z}$, and $f^*$ is also multiplication by $2$, which is not surjective.  In general, if $f$ is injective, there is a long exact sequence $$0\to (N/M)^*\to N^*\to M^*\to \operatorname{Ext}^1(N/M,R)\to \operatorname{Ext}^1(N,R)\to\dots$$
Thus $f^*$ is surjective iff the induced map $\operatorname{Ext}^1(N/M,R)\to \operatorname{Ext}^1(N,R)$ is injective.  In particular, this is true if $\operatorname{Ext}^1(N/M,R)=0$.  Another sufficient condition for $f^*$ to be surjective is if $M$ is a direct summand of $N$ (in which case $N^*\cong M^*\oplus (N/M)^*$ and $f^*$ is just the projection).
If $N$ is a bimodule and $N\cong N^*$, it still does not make sense to talk about composing $f$ and $f^*$ unless you have chosen a specific isomorphism between $N$ and $N^*$, which is typically highly non-canonical.  In any case, even over a field, you can choose such an isomorphism such that $f^*\circ f=0$ even for some nonzero maps $f$.  For instance, let $M=N=R^2$ with basis $\{e_1,e_2\}$, let $f$ be given by $f(e_1)=f(e_2)=e_1$, and choose the isomorphism $N\cong N^*$ that sends $e_1$ to $e^2$ and $e_2$ to $e^1$, where $\{e^1,e^2\}$ is the dual basis of $N^*$.
