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Suppose I constructed a linear map $\phi$ without choosing a basis, but in order to check that $\phi$ is an isomorphism, I chose a basis. Is $\phi$ still considered a natural isomorphism?

Edit: The problem is asking to construct a map for $V^{*}\otimes W \to Hom(V,W)$ where $V$ is finite dimensional, which I wrote down $\lambda \otimes w \overset{\phi}{\mapsto} \lambda(\bullet) w$. Now to show that this is an actual isomorphism, I had to select a dual basis for $V$.

I understand that you can do a change of basis and $\phi$ will still be both injective and surjective, but something about choosing a basis in the proof irks me when the problem specifies that the map to be natural.

beeflavor
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    The fact of $\phi$ being an isomorphism doesn't depend on the technique of proof of this fact. (Presumably, your definition of $\phi$ is basis-independent.) Given what you've written, "yes, $\phi$ is a natural isomorphism". – Andrew D. Hwang Jul 26 '13 at 15:35
  • I though every linear map is determined by its action on a basis, ergo regardless of one's choice, the basis "influence" is always there. – Vishesh Jul 26 '13 at 16:32
  • Thank you for your answers. The problem is asking to construct an isomorphism for $V^{*}\otimes W \to Hom(V,W)$, which I wrote down $\lambda \otimes w \mapsto \lambda(\bullet) w$. Now to show that this is an actual isomorphism, I had to select a dual basis for $V$. I will put this in the original post. – beeflavor Jul 26 '13 at 17:14
  • This is a natural morphism of (bi)functors, which is an isomorphism when $V$ is finite dimensional. – egreg Jul 26 '13 at 17:32

1 Answers1

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This question has been answered in a comment:

The fact of $\phi$ being an isomorphism doesn't depend on the technique of proof of this fact. (Presumably, your definition of $\phi$ is basis-independent.) Given what you've written, "yes, $\phi$ is a natural isomorphism". – user86418 Jul 26 '13 at 15:35

You are allowed to use any tool you want for proving that the map is an isomorphism, for example bases (and you have to in this case, because finite dimensionality is equivalent to having a finite basis). Naturality is in the map, not in the proof.

egreg
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Jessica B
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