Around these parts, the aphorism "A gentleman never chooses a basis," has become popular.



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> **Question.** Is there a gentlemanly way to prove that the natural map from $V$ to $V^{\*\*}$ is surjective if $V$ is finite-dimensional?
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As in life, the exact standards for gentlemanliness are a bit vague. Some arguments seem to be implicitly picking a basis. I'm hoping there's an argument which is unambiguously gentlemanly.



"A gentleman never chooses a basis."
Following up on Qiaochu's query, one way of distinguishing a finite-dimensional $V$ from an infinite one is that there exists a space $W$ together with maps $e: W \otimes V \to k$, $f: k \to V \otimes W$ making the usual [triangular equations](http://ncatlab.org/nlab/show/triangle+identities) hold. The data $(W, e, f)$ is uniquely determined up to canonical isomorphism, namely $W$ is canonically isomorphic to the dual of $V$; the $e$ is of course the evaluation pairing. (While it is hard to write down an explicit formula for $f: k \to V \otimes V^\*$ without referring to a basis, it is nevertheless *independent* of basis: is the same map no matter which basis you pick, and thus canonical.) By swapping $V$ and $W$ using the symmetry of the tensor, there are maps $V \otimes W \to k$, $k \to W \otimes V$ which exhibit $V$ as the dual of $W$, hence $V$ is canonically isomorphic to the dual of its dual. 


Just to be a tiny bit more explicit, the inverse to the double dual embedding $V \to V^{\*\*}$ would be given by 


$$V^{\ast\ast} \to V \otimes V^\* \otimes V^{\ast\ast} \to V$$ 


where the description of the maps uses the data above. 


