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What is a natural isomorphism?

I have often encountered the phrase "There is a natural [way/map/etc.]..." when describing say isomorphisms, maps, etc. What exactly does "natural" mean?

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Usually it's not really well-defined, but just means informally that there's a single definition that is much more straightforward and less arbitrary than other possible ones. There's also a quite technical definition of "natural transformation" in category theory, which generalizes many of the intuitive uses of "natural". –  Henning Makholm Sep 13 '11 at 13:05
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You can get a good understanding by replacing the word 'natural' with 'coordinate-free' whenever you see it. A nice example of a 'natural' isomorphism is the demonstration that every vector space $V$ over field $F$ is isomorphic to its double dual, via the mapping $\phi:V\to V^{**}$ defined by $\phi(v)(\alpha) = \alpha(v)$ for each $v\in V$, $\alpha\in V^*$. –  Chris Taylor Sep 13 '11 at 13:12
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@Chris: ... given certain restrictions on $V$ (finite dimension for algebraic duals, something more involved for continuous duals). –  Henning Makholm Sep 13 '11 at 14:50
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marked as duplicate by joriki, Asaf Karagila, J. M., t.b., Willie Wong Sep 13 '11 at 15:58

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As the above comments allude to, mathematicians often use the word informally. It's often used in the sense that the way/map/whatever the author is describing is the one that most people reader the paper would expect it to be (though this can be a little frustrating if one disagrees). It can also mean that the way/map/whatver is independent of any choices that are implicit in its definition. An example would be the isomorphism between a vector space and its double dual, which doesn't require specifying a basis.

There is also a technical usage of the term, coming from category theory. See http://en.wikipedia.org/wiki/Natural_transformation. This definition is essentially a way of making the 'independence-from-choices' notion rigorous.

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