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This is a follow up question to my previous question here.

I'm confused about the following: in Atiyah-Macdonald they state that there exists a unique isomorphism $M \otimes N \to N \otimes M$, $m \otimes n \mapsto n \otimes m$.

I'm not sure why AM write unique: If I already know the map, $m \otimes n \mapsto n \otimes m$, then there is no other map that is exactly the same. So I think I get uniqueness for free and all I have to show is that $m \otimes n \mapsto n \otimes m$ is an isomorphism.

On the other hand, and that's the way I understood the proposition in the book, if I want to show that there exists a unique isomorphism (without knowing what it is in advance) then I can use the universal property of the tensor product to get uniqueness and the map will turn out to be $m \otimes n \mapsto n \otimes m$ which is an isomorphism.

What am I missing?

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up vote 6 down vote accepted

You are misreading Atiyah-Macdonald.

You are reading: "There is only one linear map $M \otimes N \rightarrow N \otimes M$. Furthermore this map has the property that $m \otimes n \mapsto n \otimes m$.

They mean: "Consider the set of all linear maps $M \otimes N \rightarrow N \otimes M$ which satisfy the property $m \otimes n \mapsto n \otimes m$. There is only one such map."

The former can't possibly be right because you could precompose with any endomorphism of $M \otimes N$. The only times you'll have a unique linear map between those two spaces is if one of M and N is the zero vector space (otherwise you have this map and the zero map).

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Not every element of $M \otimes N$ is of the form $m \otimes n$. So we've only defined our function on some of $M \otimes N$, and the point is that there's a unique way to extend this function to an everywhere-defined isomorphism.

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So we use the universal property of the tensor product to show that this map extends uniquely? Is the other way around also correct: assume we don't know what it looks like, apply the universal property and see that it is the map given in the proposition? – Rudy the Reindeer Jun 8 '12 at 15:18

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