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"?

• Can you prove dualities $c_0^* = l_1$ and $l_1^* = l_\infty$ in ZF? Certainly the inclusion of $c_0$ into $l_\infty$ is not surjective. – Gerald Edgar Jan 12 '18 at 1:41
• @GeraldEdgar It was my understanding that they are talking about the algebraic dual space, rather than the topological dual space. – Pace Nielsen Jan 12 '18 at 4:28
• @Gerald: It is consistent that $\ell_1$ is in fact the dual of $\ell_\infty$, because the dual of $\ell_\infty/c_0$ is trivial. – Asaf Karagila Jan 12 '18 at 4:58