I know for a vector space $V$ over the real or complex numbers, there exists a canonical embedding into its double dual $V^{**}$, and if $V$ is given an inner product, then there is a canonical embedding from $V$ into $V^*$.
However, I was not sure if there is any canonical embedding from $V^*$ into $V^{**}$ that is compatible with the above emebeddings, i.e. if $$\sigma:V \rightarrow V^{**}, \sigma(v)(f)=f(v)$$ $$\iota:V \rightarrow V^*, \iota(v)(w)=\langle w,v \rangle$$ are injective linear map.
So my question is: Is there a canonical (does not depend on basis) injective linear map $\mu: V^* \rightarrow V^{**}$ such that $$\sigma = \mu \circ \iota$$ I try to build an inner product on $V^*$ based on the inner product on $V$ but to no avail.
When $V$ is finite dimensional, $\iota$ is an isomorphism, so you can just take $\mu=\sigma\circ \iota^{-1}$. In general, however, no such canonical map $\mu$ exists.
As evidence for this, consider the case that $V=\mathbb{R}^{\oplus\mathbb{N}}$ with the usual dot product as the inner product. Then $V^*$ can naturally be identified with $\mathbb{R}^{\mathbb{N}}$, and the image of $\iota$ is the subspace $\mathbb{R}^{\oplus \mathbb{N}}$ of sequences of finite support. Note now that it is consistent with ZF that the dual of $\mathbb{R}^\mathbb{N}$ is just $\mathbb{R}^{\oplus \mathbb{N}}$. In that case, there cannot exist any linear injection $V^*\to V^{**}$ at all, essentially since $V^{**}$ is countable-dimensional and $V^*$ is not (though it takes a bit of work to make this argument rigorous in the absence of choice). So, it is not possible to prove that such a map $\mu$ exists without using the axiom of choice.