Due to the contravariance of the dual space functor on vector spaces, one might expect the pullback of an injection to be a surjection, and the pullback of a surjection to be an injection. Indeed, for finite dimensional vector spaces, this is the case.
In the general case, it seems clear that surjectivity of the linear map implies injectivity of the pullback. Suppose $f\colon V\to W$ is a linear surjection, and $\alpha \in W^*$. If $f^*(\alpha)=0$, then for all $v \in V$, $f^*(\alpha)(v) = \alpha(f(v)) = 0$. Since $f$ is surjective, $\alpha$ vanishes on all of $W$, and so $\alpha = 0$.
However the other way doesn't seem as nice. Suppose $f$ is injective. Given an $\alpha \in V^*$, can we find some $\beta \in W^*$ such that $f^*(\beta)=\alpha$? You can define a map $\beta\colon Im(f) \subseteq W\to V$ which takes $w=f(v)\mapsto \alpha(v)$. If we can extend the domain of $\beta$ to all of $W$, then $f^*$ is surjective. But to extend the linear functional from $Im(f) \subseteq W$ to all of $W$ seems to require at the very least a choice of basis, i.e. an invocation of the axiom of choice. In other words, there may in fact be models of set theory where the pullback of a linear injection is not a surjective map of dual spaces.
So I have two questions.
- Have I messed up the argument or is it really the case that the pullback of a linear injection need not be a surjection on the dual space in the absence of AC? And in ZFC, at best we can say this surjection may not be very "natural"? Note: my first question has been answered below by Asaf Karagila. The argument is correct; without AC it is consistent that there exist a vector space with trivial dual, which will violate badly the surjectivity of the pullback of an inclusion. I'm leaving the question open for the second question.
- Assuming the argument is correct, how can I understand the algebra of the surprising result? I expected a the result to hold generally, and be in some sense "natural". Is there some property of covariant functors that says when they take monomorphisms to monos, and when they take epimorphisms to epis? And similarly, when contravariant functors take epis to monos, etc. What is the defect of the dual space functor that it does one, but not the other? Note: my second question has been answered below by myself and tkr; the algebraic property which the dual space functor has (assuming AC) is right-exactness, which is equivalent to the statement that all vector spaces are injective modules (assuming AC).
Best Answer
The assertion "Every vector space has a basis" implies the axiom of choice in ZF. This means that without the axiom of choice there are spaces that have no basis.
It is open at the moment whether or not this implies that there is a space with a trivial dual as well. However, it is consistent that there is a space whose dual is trivial. That is to say that the only linear functional is indeed $0$.
The usual construction yields a vector space $W$ with the following properties:
Let us consider this case: Let $V$ be $\mathbb R^n$ for some $n$, and let $W$ be as described above (with the field being $\mathbb R$).
Choose $n$ vectors, $w_1,\ldots,w_n$, which are linearly independent (this process requires no axiom of choice since it can be described in finitely many steps). Let $W'$ be the span of $w_1,\ldots,w_n$.
There is a natural map $f$ from $V$ into $W'$, and hence into $W$. Consider $f^*$, its domain is $W^*=\{0\}$. It is far far from being onto $V$.
Wait, it gets worse. Consider $f\colon V\to M=W\oplus \mathbb R^{n-1}$ as the map which maps $e_1\to w\in W$ (which is nonzero) and $e_2,\ldots,e_n$ into the $\mathbb R^{n-1}$ part.
Every functional on $M$ is $0$ on $W$ and a usual functional on $\mathbb R^{n-1}$. However there is no surjective map from $M^*$ onto $V^*$, simply since $\dim(M^*)=n-1<\dim(V^*)=n$.