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In an attempt to understand dual spaces and adjoints (in linear algebra) I came accross this video, which mentions "natural isomorphisms". Not knowing what a natural isomorphism is, I tried to look it up, and almost everything I found talked about category theory, or like in here claimed that a natural isomorphism is one that does not depend on the choice of a basis.

This made absolutely no sense to me. An isomorphism is a bijective linear function from a vector space to another. It takes a vector and spits another vector. How is a basis related to this anyhow?

So then I found this unanswered question before writing this, where one of the comment says that what depends of the basis is the construction of the isomorphism. But what does it mean for the construction of something to depend on a basis? Can a rigorous definition be given without needing to learn a lot of category theory?

Cora_
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    Well, an isomorphism, or any function, needs to be defined. The point here is that the definition depends on the choice of basis. – lulu Aug 05 '19 at 10:44
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    @lulu can you make "the definition depends on the choice of basis" more rigorous? – Cora_ Aug 05 '19 at 10:45
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    Well, I can give you an example. We know, for instance, that $\mathbb R^2$ is isomorphic to the space of linear polynomials with real coefficients. If you take the standard basis of $\mathbb R^2$ then the map $(a,b)\mapsto ax+b$ defines an isomorphism. However, you could choose a different basis and get a different isomorphism. – lulu Aug 05 '19 at 10:48
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    I think you are expecting something deeper than there really is here. Take another statement. "All $n-$dimensional real vector spaces are isomorphic". Clearly true since if ${v_i}{i=1}^n$ is a basis for the first and ${w_i}{i=1}^n$ is a basis for the second we can define the isomorphism $v_i\mapsto w_i$. But that clearly depends on the choice of bases. That dependence is explicit...I wrote the map down using the bases. That's what we are talking about. – lulu Aug 05 '19 at 10:53
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    I wrote a bit about naturality and basis independence here. In particular, yes, you can define basis independence for a morphism $f : X → X$ by saying that it commutes with automorphisms of $X$ (in Vect this translates to basis independence of matrix of $f$), and from there you can sensibly extend it to morphisms $f : FX → GX$ for functors $F$, $G$. Then naturality formally implies basis independence, but is in general a significantly stronger condition. – user54748 Aug 06 '19 at 17:01

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Yes, it is the construction (i.e. the actual formal definition) of the linear map that depends on the basis.

There is an important thing lacking from the context here, namely that in the situations when such a question arises, we want to define this mapping (isomorphism) for all vector spaces at once, basically by the same formula.
(And here is the place where category theory turns into the picture: consider all (finite dimensional) vector spaces as vertices of a big graph, and consider linear maps as edges between them. The extra thing that makes it a category is composition of consecutive edges.)

The classical examples are the isomorphisms $\varphi:X\to X^*$ and $\psi:X\to X^{**}$ for finite dimensional vector spaces $X$, where the construction of $\varphi$ depends on a choice of basis (but besides that / after that, it uses the same formula for every $X$), whereas the construction of $\psi$ does not.

Berci
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