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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 $V^*$?

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This is hard to answer because the term "canonical isomorphism" is lacking a rigorous definition that would allow non-existence of such an isomorphism to be proven. You should understand this statement as "there is no natural way to define such an isomorphism".

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.

See also this question on MO regarding the definition of "canonical".

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