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Let $M,N$ be left $R$-modules with the standard $R$-module structures (making them bi-modules). The proof that $M \otimes N$ is isomorphic to $N \otimes M$ over a commutative ring $R$ in Dummit and Foote says that,

1) the map $(m,n) \mapsto n \otimes m$ is $R$-balanced and so induce a unique homomorphism $f(m \otimes n) = n \otimes m$.

2) Likewise, the map $(n,m) \mapsto m \otimes n$ is $R$-balanced and so induce a unique homomorphism $g(n \otimes m) = m \otimes n$.

Is it enough at this point to say that $f \circ g = $ id, and $g \circ f =$ id on the elementary tensors $n \otimes m$ and $m \otimes n$ respectively, and so they are isomorphisms? Or are there more to the proof that I am missing?

Jean Valjean
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    Having mutually inverse homomorphisms is all you need for an isomorphism. Yup, you're done. — And of course a homomorphisms that is the identity on elementary tensors, is also the identity on the whole tensor product. – Christoph Apr 26 '14 at 21:03
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    Or draw the diagram for the universal property defining the tensor product. The differnce between $N\otimes M$ and $M\otimes N$ is only the positioning of the arrows on paper, nothing "really" essential. – Hagen von Eitzen Apr 26 '14 at 21:18

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Yes it is!

The key here is to note that every element of $M \otimes N$ is a finite sum of elementary tensors $m \otimes n$.

However, not every element of $M \otimes N$ is an elementary tensor, but that's okay since $f$ linear implies $$\displaystyle f \bigg(\sum m_k \otimes n_k\bigg) = \sum f(m_k \otimes n_k).$$

This means it suffices to show that $g \circ f = \text{id}_{M \otimes N}$ and $f \circ g = \text{id}_{N \otimes M}$ on simple tensors (we just extend by linearity to the entire tensor product).

Robert Cardona
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