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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.

  1. 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.
  2. 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).
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What if there is no functionals except zero? – Asaf Karagila Nov 28 '11 at 22:54
Doesn't the extension exist by Hahn-Banach? – Grasshopper Nov 29 '11 at 5:50
@Grasshopper: Hahn-Banach requires some choice; furthermore there are non-topological vector spaces on which Hahn-Banach won't work, in which case the extension would require a basis somewhere in the background (if not of the space, then of a larger space which embeds it). – Asaf Karagila Nov 29 '11 at 8:28
@ziggurism: Functors don't necessarily preserve much. Being a monomorphism is not an equational condition so may not be preserved. However, being a split monomorphism is an equational condition and so is preserved by all (covariant) functors. Thus, as soon as you are able to prove that every subspace is a direct summand, it will be true that dualisation takes monomorphisms to epimorphisms. – Zhen Lin Dec 1 '11 at 8:00
@Zhen: So in essence you're saying that the only way this would be true is if the axiom of choice holds? – Asaf Karagila Dec 1 '11 at 11:06
up vote 2 down vote accepted

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:

  1. $W$ is not spanned by any finite set (i.e. it does not have a finite dimension),
  2. Every proper subspace has a finite dimension,
  3. The only linear transformation from $W$ into itself are multiplication by scalars from the field. In particular this implies that there is only one linear functional: the zero functional.

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 $\dim(M^*)=n-1<\dim(V^*)=n$.

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This means that without the axiom of choice there are spaces that have no basis. How does this follows from the first sentence? – user13838 Nov 29 '11 at 13:51
@percusse: If the axiom of choice is equivalent to the assertion that every vector space has a basis, it means that if one is the negated then so is the other. In particular if the axiom of choice does not hold, somewhere in the universe there exists a vector space that has no basis. – Asaf Karagila Nov 29 '11 at 13:58
My problem is the first part of your comment. To conclude with negation, the initial assertion should hold. Let me ask you this way, since I am not fluent at these: What happens if "Every vector space has a basis" can be shown without invoking Axiom Choice? How should we interpret this result then? – user13838 Nov 29 '11 at 14:18
@percusse: If every vector space has a basis one can derive the axiom of choice. This is a theorem of ZF and in fact uses a bit less. If you want to replace ZF then I cannot assure you what will happen. – Asaf Karagila Nov 29 '11 at 14:27
Thank you Asaf. With regards to the axiom of choice questions, that is an excellent answer. So part of my question was: is this proof correct, in the sense that it doesn't invoke the axiom of choice unnecessarily? You've answered that in the affirmative, and given several examples showing the nonexistence of the surjective pullback. Very nice! – ziggurism Nov 30 '11 at 1:08

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