# Isomorphism with dual space

Let $V$ be a vector space over $F$ with basis $\{v_1,\cdots, v_n\}$, and for every $i$, let $f_{v_i}\colon V\rightarrow F$ be a linear map satisfying $f_{v_i}(v_j)=\delta_{ij}$. Then $\{f_{v_1}, \cdots, f_{v_n}\}$ forms a basis of the dual space $V^*$.

The map $T\colon v_i\mapsto f_{v_i}$ uniquely extends to a linear injective map from $V$ to $V^*$, and since these spaces have same dimension, $T$ is isomorphism. Thus $V$ and $V^*$ are isomorphic, when $dim(V)$ is finite. However there is a canonical isomorphism between $V$ and $V^{**}$ when $dim(V)$ is finite.

Question: Suppose that $dim(V)$ is finite. Then the above map $T$ is isomorphism which depends on choice of basis. But can we always say that there is no canonical isomorphism between $V$ and $V^*$?

My confusion lies here: I understood above discussion as:

if $dim(V)$ is finite, then there is isomorphism between $V$ and $V^*$, which can be obtained by considering a basis of $V$ and corresponding basis of $V^*$. But, eventhough, this isomorphism $T$ depends on choice of basis, it may happen that there is another isomorphism $V\rightarrow V^*$, which is independent of choice of basis. How can I conclude that there is no canonical isomorphism between $V$ and $V^*$?

This is in sharp contrast with the so called "evaluation map" $V\to V^{**}$ given by $v\mapsto(v^*\mapsto v^*(v))$. This map (always an isomorphism for finite-dimensional $V$) is "canonical" in the sense that it comes up naturally and there is no obvious alternative. It is simply the first non-trivial map $V\to V^{**}$ that one can possibly come up with.
What they say is simply that there is no way to exhibit an isomorphism $V\to V^*$ without first writing down a basis.