Dual spaces isomorphic, implies vector spaces itself are isomorphic? When I have two vector spaces $W, V$ over $k$ a field. And I know that the algebraic dual spaces of $V$ and $W$ are isomorphic. Can I conclude, (in the infinite dimensional case) that $V$ and $W$ are isomorphic?
I am saying algebraic dual, because i don't want to be confused with the dual of continuous linear functionals. But I am just using the definition of dual space, that everyone uses in linear algebra.
Say I know, that I have a linear map $\varphi: V \to W$ such that its dual map is an isomorphism, can I then conclude it was an isomorphism all along? I know this is true if $\varphi$ is assumed to be injective, because then the injectivity of the dual map, implies surjectivity of the original map. Can I say something, if my map $\varphi$ was assumed to be surjective and the dual map is an isomorphism?
 A: For the question in the first paragraph, the answer is "no" in the general case, at least in ZFC. However, as Mark Saving notes in the comments, depending on your field $k$ the answer may be independent of ZFC, and indeed depend on your Set Theory.
If $V$ is a vector space of infinite dimension $\kappa$, then we know that $\dim(V^*)\gt\dim(V)$. Moreover, we know that $\dim(V^*)=|k|^{\dim(V)}$, by a theorem of Kaplansky and Erdős. See for example this answer and this answer, the latter with citation for this theorem.
Now take a field $k$ of cardinality $2^{\mu}$, with $\mu\gt 2^{\aleph_0}$, and take a $k$-vector space $V$ of dimension $\aleph_0$, and a $k$-vector space $W$ of dimension $2^{\aleph_0}$.
Then
$$\dim(V^*) =|k|^{\aleph_0} = (2^{\mu})^{\aleph_0} = 2^{\mu\aleph_0} = 2^{\mu},$$
and
$$\dim(W^*) = |k|^{2^{\aleph_0}} = (2^{\mu})^{2^{\aleph_0}}= 2^{\mu 2^{\aleph_0}} = 2^{\mu}.$$
Thus, $V^*$ and $W^*$ are isomorphic (they have the same dimension), but $V$ and $W$ are not isomorphic, since one is countably-dimensional and the other is uncountably-dimensional.
A: Yes. If $\phi^\lor$ is an isomorphism, then $\phi$ is an isomorphism.
The most straightforward way of seeing this is through category theory. We must show that the dual functor is faithful. That is, if we have vector spaces $W, V$ and linear maps $f, g : W \to V$ such that $f^\lor = g^\lor$, then $f = g$.
Indeed, suppose we have such $W, V, f, g$, and consider some $x \in W$. Suppose $f(x) \neq g(x)$. Then take some basis $B$ of $V$ such that $f(x) - g(x) \in B$, and consider the unique linear map $h : V \to k$ such that for all $b \in B$,
$$h(b) = \begin{cases}
  1 & b = f(x) - g(x) \\
  0 & otherwise
\end{cases}$$
Then $h \circ f = f^\lor(h) = g^\lor(h) = h \circ g$, so in particular, $h(f(x)) = h(g(x))$. So $h(f(x) - g(x)) = 0$; contradiction.
Now that we’ve established the dual functor is faithful, we note that the category of vector spaces is balanced - a linear map which is both a monomorphism and an epimorphism is an isomorphism. It follows from faithfulness that since $\phi^\lor$ is an isomorphism, it must be an epimorphism, and hence $\phi$ is a monomorphism. Similarly, it follows that $\phi^\lor$ is a monomorphism, and hence $\phi$ is an epimorphism. Therefore, $\phi$ is an isomorphism.
To replicate the previous paragraph without category theory, we first show $\phi : V \to W$ is injective. Consider some $x \in V$ such that $\phi(x) = 0$. Then consider the unique linear map $f : k \to V$ such that $f(1) = x$. We see that $\phi \circ f = 0$, so $f^\lor \circ \phi^\lor = (\phi \circ f)^\lor = 0^\lor = 0$. Take the equation $f^\lor \circ \phi^\lor = 0$ and compose $(\phi^\lor)^{-1}$ on the right to get $f^\lor = 0 = 0^\lor$. It follows from faithfulness that $f = 0$, so $x = f(1) = 0$. Thus, $\phi$ is injective. You said in your question that you can take it from there.
Finally, this argument is not sufficient to show that if $V^\lor$ is isomorphic to $W^\lor$, then $V \cong W$. In general, dual spaces are far larger than their underlying spaces, and there are many maps between them that don’t arise as a dual map. I don’t know whether this implication holds, but I have a strong sense it doesn’t (or may even be independent from ZFC). An analogous question arises in set theory: if there is a bijection between $P(S)$ and $P(R)$, is there a bijection between $S$ and $R$? It turns out the question cannot be answered in ZFC.
But I think it’s likely there’s a well-known answer somewhere. You just have to look harder.
