Let $V$ be a vector space over some field $\mathbb F$. The dual space is defined as the set of all linear maps $f : V \to \mathbb F$ which is again a vector space over $\mathbb F$ (with pointwise addition and scalar multiplication).
The canonical map $$i : V \to V^{**}, i(v)(f) = f(v)$$ is linear. It is well-known that it is always injective, but the standard proof depends on using a suitable basis of $V$. To get such a basis for an arbitrary $V$, we need the axiom of choice (AC).
Question:
Can the injectivity of $i$ for all $V$ be proved without AC? Or is the injectivity of $i$ for all $V$ equivalent to AC?