# Given $\phi \in \mathrm{End}(M)$, when does $\phi$ injective imply $\phi^*$ surjective and $\phi^*$ injective imply $\phi$ surjective?

Let $$M$$ be a module over a commutative ring (with unity) $$R$$. Let $$\phi : M \to M$$ be an $$R$$-module homomorphism. Then we have a dual map $$\phi^* : M^* \to M^*$$ given by $$\phi^*(f)=f\circ \phi, \forall f \in M^*$$. My questions are:

(1) If $$\phi$$ is surjective, then $$\phi^*$$ is injective. Under what conditions on $$R$$ or $$M$$, can we say that $$\phi$$ is injective $$\implies \phi^*$$ is surjective ?

(2) If $$M$$ is torsion-less, i.e., $$\bigcap_{f\in M^*}\ker f=0$$, then $$\phi^*$$ surjective $$\implies \phi$$ is injective. Under what conditions on $$R$$ or $$M$$, can we say that $$\phi^*$$ is injective $$\implies \phi$$ is surjective ?

[NOTE: Here, $$M^*:=\mathrm{Hom}_R(M,R)$$]

• I am guessing this is not the sort of condition that you're interested in (hence why this is only a comment), but if $M$ is Artinian and finitely generated, then any injective endomorphism is necessarily an isomorphism. This gives one situation where (1) holds. – Alex Wertheim May 16 at 23:16
• @AlexWertheim: sure ... but that is too strong ... I am looking for some condition to deal with taking the dual ... – user102248 May 16 at 23:23
• The question is too broad. For example, if $R\to R$ is multiplication by a non-zero divisor which is a non-unit, then it is is injective, but the dual is not surjective. So, unless you have some restrictions, it is difficult to answer in this open ended fashion. – Mohan May 16 at 23:36