Is it consistent with ZF that $V\to V^{\ast \ast}$ is always surjective? In a comment to a recent question, Jeremy Rickard asked whether it is consistent with ZF that the map $V \to V^{**}$ from a vector space to its double dual is always surjective.  We know that "always injective" is consistent (since that's what happens in ZFC) and Jeremy Rickard's argument shows that "always an isomorphism" is not consistent.  But what about "always surjective"?

Gro-Tsen points out that Harry West has already showed this is impossible elsewhere on MO. I am missing one step in West's answer. I thought I'd write up the issue here, and someone can explain to me what I am missing.
First of all, for any field $F$ and any set $X$, we can form the free vector space with basis $X$, call it $FX$, and the vector space of functions from $X$ to $F$, call it $F^X$. It is easy to see that $(FX)^{\ast} \cong F^X$ so, if $V \to V^{\ast \ast}$ is always surjective, then the obvious injection $FX \to (F^X)^{\ast}$ must always be an isomorphism. So we may and do assume:
Key Consequence For every set $X$, the obvious injection $FX \to (F^X)^{\ast}$ is an isomorphism.
Now, suppose that $\alpha$ is an ordinal with cofinality $>\omega$.
Lemma 1 Let $X \subset \alpha$ have the property that $X \cap \beta$ is finite for every $\beta < \alpha$. Then $X$ is finite.
Proof If not, then $X$ is an infinite well-ordered set (by restricting the order from $\alpha$) so it contains a copy of $\omega$. By the hypothesis on $\alpha$, there is some $\beta_0 < \alpha$ containing this copy of $\omega$. But then $X \cap \beta_0$ is infinite. $\square$
Let $V$ be the subspace of $F^{\alpha}$ consisting of functions which are supported on $F^{\beta}$ for some $\beta< \alpha$.
Lemma 2 $V^{\ast} = F \alpha$.
Proof: Let $\phi \in V^{\ast}$. We can restrict $\phi$ to $F^{\beta}$ for each $\beta < \alpha$ and, by the definition of $V$, the functional $\phi$ is determined by the list of these restrictions. On each $F^{\beta}$, by the Key Consequence, $\phi|_{F_{\beta}}$ coincides with some unique vector from $F \beta$. Let the support of that vector be $X_{\beta}$ and let $X = \bigcup_{\beta} X_{\beta}$. Then $X \cap \beta= X_{\beta}$ for each $\beta < \alpha$ so, by Lemma 1, $X$ is finite. The functional $\phi$ is then induced by a functional in $FX \subset F \alpha$. $\square$.
But then $V^{\ast \ast} = (F \alpha)^{\ast} = F^{\alpha}$, whereas $V$ is a proper subspace of $F^{\alpha}$. QED

My only issue is, in ZF, are we sure that there are ordinals of cofinality $>\omega$? In ZFC, one simply takes the first uncountable ordinal, $\omega_1$. If we had a cofinal sequence $0=x_0$, $x_1$, $x_2$, \dots, in $\omega_1$, then $\omega_1$ would be the union of the countable intervals $[x_i, x_{i+1})$, and would hence be countable.
But in ZF, a countable union of countable sets doesn't have to be countable. I tried some tricks to get around this and failed; please let me know what I missed.
 A: $\def\fin{\text{finite}}\def\count{\text{countable}}$This is a CW answer to write up the proof which Harry West gave on another answer, as pointed out by Gro-Tsen. Thanks to Prof. West for the clever solution and Gro-Tsen for pointing it out. I'm just rewriting it to record details which took me a while to follow. The answer is "no", there is always some vector space with $V^{\ast \ast}$ not spanned by the image of $V$.
Let $F$ be a field. For any set $X$, the vector space $F^X$ is the vector space of functions $X \to F$. It contains the subspace $F^X_{\fin}$ of finitely supported functions, which is isomorphic to the free vector space with basis $X$. It is straightforward that $(F^X_{\fin})^{\ast} = F^X$ so, if $(F^X)^{\ast} \supsetneq F^X_{\fin}$, then we are done already. Thus, we may make the key assumption that $(F^X)^{\ast} = F^X_{\fin}$ for every set $X$.
Now, let $X$ be a well-ordered uncountable set, such as the first uncountable ordinal $\omega_1$. The only place that we will use the well-order on $X$ is to be sure that every infinite subset of $X$ contains a copy of $\omega$ (in other words, infinite subsets of $X$ can't be Dedekind finite).
Let $F^X_{\count}$ be the subspace of $F^X$ consisting of countably supported functions.
Claim: $(F^X_{\count})^{\ast} = F^X_{\fin}$.
Proof: Let $\phi$ be any linear function $\phi : F^X_{\count} \longrightarrow F$. For any countable subset $Y$ of $X$, we have $F^Y \subset F^X_{\count}$. So the restriction of $\phi$ to $F^Y$ is a vector in $(F^Y)^{\ast}$ and, by the key assumption, $(F^Y)^{\ast} = F^Y_{\fin}$. So, for each $Y$, there is some unique finitely supported vector $\phi_Y$ in $F^Y$ such that $\phi(\psi)$ is the dot product $\psi \cdot \phi_Y$ whenever $\psi$ is supported on $Y$. By the uniqueness of $\phi_Y$, if $Y_1 \subseteq Y_2 \subset X$ are finite sets, then $\phi_{Y_1}$ is the restriction of $\phi_{Y_2}$ to $Y_1$.
Let $S = \bigcup_{X \subset Y,\ \count} \text{Support}(\phi_Y)$. For every countable $Y$, we know that $S \cap Y$ is finite. We claim that this forces $S$ to be finite; if $S$ is infinite then (using the well order on $X$) there is a countably infinite subset $Z$ of $S$ and $S \cap Z= Z$ is infinite, contradicting that $S \cap Y$ is finite for every countable $Y$.
So $S$ is finite, and $\phi$ is given by dot product with a vector in $F^S$, so $\phi$ is given by dot product with a vector in $F^X_{\fin}$. We have shown that $(F^X_{\count})^{\ast} = F^X_{\fin}$. $\square$
But then $(F^X_{\count})^{\ast \ast} = (F^X_{\fin})^{\ast} = F^X$, and $F^X_{\count}$ is a proper subspace of $F^X$. This completes the proof. $\square$
As Prof. West observes in his write up, one can identify a specific vector space $V$ for which this proof shows $V$ does not surject onto $V^{\ast \ast}$: The vector space $F^{\omega}_{\fin} \oplus F^{\omega_1}_{\count}$.
