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    $\begingroup$ At C) The natural map goes in the other direction $V^*\otimes_k V\rightarrow Hom_k(V,V), \quad f\otimes v\mapsto f(-)\cdot v$ and has a base free definition. The problem is that for if $V$ is infinite-dimensional, the image of that map consists of all maps in $Hom_k(V,V)$ with finite-dimensional Range. $\endgroup$ Commented Jul 11, 2016 at 23:17
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    $\begingroup$ @Steven: I think part of what makes your question ambiguous is that much depends on the definition of "finite-dimensional" you use; if it's "has a finite basis" then there's a pretty tautological sense in which you can't make any claim about finite-dimensional vector spaces without referring to a choice of basis. I assume that's not interesting to you. Here is a different example where I don't even know what choices, if any, are required: if $A$ is a topological group then there's a natural map from $A$ to its double Pontryagin dual $[[A, S^1], S^1]$ which, if $A$ is LCH, is an iso. What... $\endgroup$ Commented Jul 12, 2016 at 6:31
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    $\begingroup$ Maybe what you want here is to consider the free monoidal closed abelian category generated by a single object $X$ and by $\hom(1,1) = k$, and then observe that $X \to X^{**}$ is not an isomorphism? (I'm not sure if there even exists a nonzero map $X^{**} \to X$) $\endgroup$ Commented Jul 12, 2016 at 6:49
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    $\begingroup$ It seems like more people want to show off their knowledge of fun facts about vector spaces than want to answer the question... $\endgroup$ Commented Jul 12, 2016 at 12:10
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    $\begingroup$ Here's a possible approach: I think the ur-example of a problem where you have to make an arbitrary choice is the following: "Let $A$ be non-empty and let $f:A\rightarrow B$. Is $B$ non-empty?". Clearly the only way to proceed is to choose an arbitrary $a\in A$ an consider $f(a)\in B$. Any other problem involving an arbitrary choice reduces to this one. For example we have a function from the set of finite bases of $V$ to the set of inverses to $V\rightarrow V^{**}$, but to show that $V\rightarrow V^{**}$ actually has an inverse we have to actually pick one of these bases. $\endgroup$ Commented Jul 12, 2016 at 14:18