I understand that the dual of a surjective linear map is injective for general modules and that the dual of an injective linear map is surjective for vector spaces.
For the converse, however, I suppose that a linear map is surjective if its dual is injective. It is easy to verify for finite dimensional vector spaces using, for example, dual bases. Now is there a nice argument to prove this claim — provided it remains true — for general infinite dimensional vector spaces or even modules without having to use bases?
Thank you.
Best Answer
This is quite false for modules. A simple counterexample over $R = \mathbb{Z}$ is the inclusion $f : \mathbb{Z} \to \mathbb{Q}$, whose dual (in the sense of applying $\text{Hom}(-, \mathbb{Z})$) is the map $0 \to \mathbb{Z}$, so it is injective. But $f$ is not surjective. The problem here, of course, is that "linear functionals," in the sense of homomorphisms $A \to \mathbb{Z}$ for an abelian group $A$, quite badly fail to separate points.
For vector spaces we can say the following. It will be easier to work with the contrapositive. Suppose $f : V \to W$ is a linear map which is not surjective, so its image does not contain some nonzero $w \in W$. If linear functionals on the quotient $W/\text{im}(f)$ separate points, then we can find a linear functional $\varphi \in W^{\ast}$ such that $\varphi(f(v)) = 0$ for all $v \in V$ but such that $\varphi(w) \neq 0$. Then $f^{\ast}(\varphi) = 0$, so $f^{\ast}$ is not injective. This gives:
The reason I explicitly state this hypothesis on linear functionals is that its proof in general requires the axiom of choice to choose a basis of $W/\text{im}(f)$. In the absence of the axiom of choice it is not necessarily always true so it is an extra hypothesis that needs to be added to the above. It's actually consistent with ZF that there exist infinite-dimensional vector spaces with trivial dual. An example is $\mathbb{R}$ as a vector space over $\mathbb{Q}$; there is a model of ZF constructed by Shelah where there are no nonzero homomorphisms $\mathbb{R} \to \mathbb{Q}$, so the $\mathbb{Q}$-linear dual of $\mathbb{R}$ is trivial, and then the inclusion $f : \mathbb{Q} \to \mathbb{R}$ would be a counterexample to the desired statement over $\mathbb{Q}$.
An example over $\mathbb{R}$ where I think the dual could be trivial but I don't know is the inclusion $f : \bigoplus_n \mathbb{R} \to \prod_n \mathbb{R}$ of the infinite direct sum of copies of $\mathbb{R}$ into the infinite product. $f$ is definitely not surjective. But I believe it's independent of ZF whether $f^{\ast}$ is injective, and specifically I believe it's consistent with ZF that $\prod_n \mathbb{R}/\bigoplus_n \mathbb{R}$ has trivial dual (but I don't have a proof), and that the dual of $\prod_n \mathbb{R}$ is just $\bigoplus_n \mathbb{R}$.