This is only a partial answer because I'm having trouble reconstructing something I think I figured out seven years ago...
It would seem the Dual Cantor-Bernstein implies Countable Choice. In a post in sci.math in March 2003 discussing the dual of Cantor-Bernstein, Herman Rubin essentially points out that if the dual of Cantor-Bernstein holds, then every infinite set has a denumerable subset; this is equivalent, I believe, to Countable Choice.
Let $U$ be an infinite set. Let $A$ be the set of all $n$-tuples of elements of $U$ with $n\gt 0$ and even, and let $B$ be the set of all $n$-tuples of $U$ with $n$ odd. There are surjections from $A$ onto $B$ (delete the first element of the tuple) and from $B$ onto $A$ (for the $1$-tuples, map to a fixed element of $A$; for the rest, delete the first element of the tuple). If we assume the dual of Cantor-Bernstein holds, then there exists a one-to-one function from $f\colon B\to A$ (in fact, a bijection). Rubin writes that "a 1-1 mapping from $B$ to $A$ quickly gives a countable subset of $U$", but right now I'm not quite seeing it...
To add the proof for my claim in Todd's answer, which essentially repeats Läuchli's original [1] arguments with minor modifications (and the addition that the resulted model satisfies $DC_\kappa$).
We will show that it is consistent to have a model in which $DC_\kappa$ holds, and there is a vector space over $\mathbb F_2$ which has no linear functionals.
Assume that $M$ is a model of $ZFA+AC$ and that $A$, the set of atoms has $\lambda>\kappa$ many atoms, where $\lambda$ is a regular cardinal. Endow $A$ with a structure of a vector space over $\mathbb F=\mathbb F_2$. Now consider the permutation model $\frak M$ defined by the group of linear permutations of $A$, and by ideal of supports generated by subsets of dimension $\le\kappa$.
Denote by $\operatorname{fix}(X)$ the permutations which fix every element of $X$, by $\operatorname{sym}(X)$ the permutations that fix $X$ as a set, and by $[E]$ the span of $E$ as a subset of $A$. We say that $E\subseteq A$ is a support of $X$ if $\pi\in\operatorname{fix}(E)\Rightarrow\pi\in\operatorname{sym}(X)$.
Final word of terminology, since $A$ will play both the role of set of atoms as well the vector space, given $U\subseteq A$ the complement will always denote a set complement, whereas the direct complement will be used to refer to a linear subspace which acts as a direct summand with $U$ in a decomposition of $A$.
Claim 1: If $E$ is a subset of $A$ then $\operatorname{fix}(E)$ is the same as $\operatorname{fix}([E])$.
Proof: This is obvious since all the permutations considered are linear. $\square$
From this we can identify $E$ with its span, and since (in $M$) the $[E]$ has the same cardinality of $E$ we can conclude that without loss of generality supports are subspaces.
Claim 2: $\frak M$$\models DC_\kappa$.
Proof: Let $X$ be some nonempty set, and $\lt$ a binary relation on $X$, both in $\frak M$. In $M$ we can find a function $f\colon\kappa\to X$ which witness $DC_\kappa$ in $V$.
Since $\frak M$ is transitive, we have that $\alpha,f(\alpha)\in\frak M$ and thus $\langle\alpha,f(\alpha)\rangle\in\frak M$. Let $E_\alpha$ be a support for $\lbrace\langle\alpha,f(\alpha)\rangle\rbrace$ then $\bigcup_{\alpha<\kappa} E_\alpha$ is a set of cardinality $<\kappa^+$ and thus in our ideal of suports. It is simple to verify that this is a support of $f$, therefore $f\in\frak M$ as wanted. $\square$
Claim 3: If $x,y\in A$ are nonzero (with respect to the vector space) then in $M$ there is a linear permutation $\pi$ such that $\pi x=y$ and $\pi y=x$.
Proof: Since $x\neq y$ we have that they are linearly independent over $\mathbb F$. Since we have choice in $M$ we can extend this to a basis of $A$, and take a permutation of this basis which only switches $x$ and $y$. This permutation extends uniquely to our $\pi$.
Claim 4: If $U\subseteq A$ and $U\in\frak M$ then either $U$ is a subset of a linear subspace of dimension at most $\kappa$, or a subset of the complement of such space.
Proof: Let $E$ be a support of $U$, then every linear automorphism of $A$ which fixes $E$ preserves $U$. If $U\subseteq [E]$ then we are done, otherwise let $u\in U\setminus [E]$ and $v\in A\setminus [E]$, we can define (in $M$ where choice exists) a linear permutation $\pi$ which fixes $E$ and switches $u$ with $v$. By that we have that $\pi(U)=U$ therefore $v\in U$, and so $U=A\setminus[E]$ as wanted. $\square$
Claim 5: If $U\subseteq A$ is a linear proper subspace and $U\in\frak M$ then its dimension is at most $\kappa$.
Proof: Suppose that $U$ is a subspace of $A$ and every linearly independent subset of $U$ of cardinality $\le\kappa$ does not span $U$, we will show $A=U$. By the previous claim we have that $U$ is the complement of some "small" $[E]$.
Now let $v\in A$ and $u\in U$ both nonzero vectors. If $u+v\in U$ then $v\in U$. If $u+v\in [E]$ then $v\in U$ since otherwise $u=u+v+v\in[E]$. Therefore $v\in U$ and so $A\subseteq U$, and thus $A=U$ as wanted.$\square$
Claim 6: If $\varphi\colon A\to\mathbb F$ a linear functional then $\varphi = 0$.
Proof: Suppose not, for some $u\in A$ we have $\varphi(u)=1$, then $\varphi$ has a kernel which is of co-dimension $1$, that is a proper linear subspace and $A=\ker\varphi\oplus\lbrace 0,u\rbrace$. However by the previous claim we have that $\ker\varphi$ has dimension $\kappa$ at most, and without the axiom of choice $\kappa+1=\kappa$, thus deriving contradiction to the fact that $A$ is not spanned by $\kappa$ many vectors.
Aftermath: There was indeed some trouble in my original proof, after some extensive work in the past two days I came to a very similar idea. However with the very generous help of Theo Buehler which helped me find the original paper and translate parts, I studied Läuchli's original proof and concluded his arguments are sleek and nicer than mine.
While this cannot be transferred to $ZF$ using the Jech-Sochor embedding theorem (since $DC_\kappa$ is not a bounded statement), I am not sure that Pincus' transfer theorem won't work, or how difficult a straightforward forcing argument would be.
Lastly, the original Läuchli model is where $\lambda=\aleph_0$ and he goes on to prove that there are no non-scalar endomorphisms. In the case where we use $\mathbb F=\mathbb F_2$ and $\lambda=\aleph_0$ we have that this vector space is indeed amorphous which in turn implies that very little choice is in such universe.
Bibliography:
- Läuchli, H. Auswahlaxiom in der Algebra. Commentarii Mathematici Helvetici, vol 37, pp. 1-19.
Best Answer
Let me give an alternative proof of the result in Ron Maimon's answer, that the canonical map $V\to V^{**}$ can consistently, in the absence of the axiom of choice, be surjective. The only technical advantage of my proof is that it uses only the consistency of ZF plus countable choice plus "all sets (in Polish spaces) have the Baire property". That consistency was proved (by Shelah) relative to just ZF, whereas the consistency of "all sets are Lebesgue measurable" needs an inaccessible cardinal. I think my proof is also a bit simpler than the one using measure. (Also, I don't need to mention ultrafilters, which some people might consider an advantage.)
Let $f:V^*\to\mathbb R$ be a linear map, and let me identify $V^*$ with the space of infinite sequences of reals. Topologize $V^*$ with the product topology, where each factor $\mathbb R$ has the usual topology of the reals. That makes $V^*$ a Polish space, so I can use the assumption about Baire category. In particular, if I partition $\mathbb R$ into intervals of length 1, then the inverse images of these intervals under $f$ have the Baire property, and they can't all be meager, by the Baire category theorem. So at least one of them, call it $f^{-1}(I)$, differs by a meager set from a nonempty open set. Inside that nonempty open set, I can find a basic open set of the form `B=$\prod_iU_i$ where, for some $n$, the first $n$ of the factors $U_i$ are intervals of some length $\delta$ and the later factors $U_i$ are $\mathbb R$. For the first $n$ indices $i$, let $U'_i$ be the interval with the same midpoint as $U_i$ but only half the length, and let $B'$ be the product that is like $B$ except using the $U'_i$ instead of the $U_i$ for the first $n$ factors. Consider an arbitrary $z\in V^*$ whose first $n$ components are smaller than $\delta/2$ in absolute value. Then translation by $z$ in $V^*$ maps $B'$ homeomorphically to another subset of $B$. So the two sets $f^{-1}(I)\cap B'$ and $\{x\in B':z+x\in f^{-1}(I)\}$ are both comeager in $B'$ and therefore must intersect. Let $x$ be in their intersection. Both $f(x)$ and $f(z+x)$ are in the interval $I$ of length 1. Subtracting (and remembering that $f$ is linear), we get that $|f(z)|\leq 2$.
Summarizing, we have that $|f(z)|\leq 2$ for all $z\in V^*$ whose first $n$ components are smaller than $\delta/2$ in absolute value. By linearity, if the first $n$ components of $z$ are smaller than $\alpha\delta/2$ in absolute value, for some positive $\alpha$, then $|f(z)|\leq2\alpha$. In particular, if the first $n$ components of $z$ are zero, then so is $f(z)$. That is, the kernel of $f$ includes the subspace $N$ consisting of those $z\in V^*$ whose first $n$ components vanish. So $f$ factors through the quotient $V^*/N$, which is finite-dimensional (in fact, $n$-dimensional). Knowing what linear functionals on finite-dimensional spaces look like, we immediately conclude that $f$ is given by inner product with a member of $V$ (having non-zero components in at most the first $n$ positions).