# Proof that $A \otimes B \cong B \otimes A$ [duplicate]

Possible Duplicate:
There exists a unique isomorphism $M \otimes N \to N \otimes M$

I want to show that for Abelian groups $A$ and $B$ that the tensor product $A \otimes B$ is isomorphic to $B \otimes A$. I believe that I have accomplished this and have posted my attempt as an answer to my own question. I would appreciate any constructive feedback.

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## marked as duplicate by Zhen Lin, Rudy the Reindeer, Jack Schmidt, Norbert, azarelJun 12 '12 at 20:19

Let's use the universal definition of $A\otimes B$: $(A\otimes B,\iota)$, where $\iota\colon A\times B\to A\otimes B$ is a bilinear (resp. $R$-bilinear) map, is the unique group (resp. $R$-module) with the property that for any bilinear map $\phi\colon A\times B\to C$ (resp. $R$-bilinear), where $C$ is any abelian group (rep. $R$-module), there exists a unique homomorphism $\Phi\colon A\otimes B\to C$ such that $\phi=\Phi\circ\iota$.

Let $s\colon A\times B\to B\times A$ be the map $s(a,b) = (b,a)$, and let $(B\otimes A,j)$ be the tensor product of $B$ and $A$. Note that $B\otimes A$ has the corresponding universal property relative to $B\times A$ and $j$.

Now, the map $js\colon A\times B\to B\otimes A$ is a bilinear map (composition of a homomorphism and a bilinear map). Therefore, there exists a unique $\mathcal{F}\colon A\otimes B\to B\otimes A$ such that $js = \mathcal{F}\iota$. Likewise, the map $\iota s^{-1}\colon B\times A\to A\otimes B$ is bilinear, so there exists a unique $\mathcal{G}\colon B\otimes A\to A\otimes B$ such that $\iota s^{-1} = \mathcal{G}j$.

Now consider $\mathcal{GF}\colon A\otimes B\to A\otimes B$. We have that $$\mathcal{GF}\iota = \mathcal{G}js = \iota s^{-1}s = \iota.$$ But there is supposed to be a unique map $f\colon A\otimes B\to A\otimes B$ such that $f\circ\iota = \iota$ (since $\iota$ is bilinear), and clearly $f=\mathrm{id}_{A\otimes B}$ works. Therefore, $\mathcal{GF}=\mathrm{id}_{A\otimes B}$.

Symmetrically, $\mathcal{FG}j = \mathcal{F}\iota s^{-1} = jss^{-1} = j$. But $j$ is a bilinear map $B\times A\to B\otimes A$, so there is supposed to be a unique map $g\colon B\otimes A\to B\otimes A$ such that $gj=j$. Since $\mathrm{id}_{B\otimes A}$ works, that is the unique function with the desired property. Since $\mathcal{FG}$ also works, $\mathcal{FG}=\mathrm{id}_{B\otimes A}$.

Thus, $\mathcal{FG}=\mathrm{id}_{B\otimes A}$ and $\mathcal{GF}=\mathrm{id}_{A\otimes B}$. Therefore, $\mathcal{F}\colon A\otimes B\to B\otimes A$ is an isomorphism, as desired.

Note that we don't need to know how we represent $A\otimes B$; we just need the universal property (and that a tensor product exists for any [ordered] pair of groups). Though one can likewise use the universal property to show that if $(M,\iota)$ is a tensor product for $A\times B$, then $(M,\iota s^{-1})$ is a tensor product for $B\times A$, so you just need to know $A\otimes B$ exists.

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$\newcommand{\xmap}[3]{#1 \colon #2 \longrightarrow #3}$ Define a biadditive function $\xmap{f}{A \times B}{B \otimes A}$ by $f(a,b) = b \otimes a$. By the universal property of the tensor product there exists a unique morphism of groups $$\xmap{\phi}{A \otimes B}{B \otimes A}$$ such that $$\phi( a \otimes b) = f(a,b).$$

Similarly, corresponding to the biadditive function $\xmap{g}{B \times A}{A \otimes B}$ defined by $g(b,a) = a \otimes b$, there is a unique morphism of groups $$\xmap{\psi}{B \otimes A}{A \otimes B}$$ such that $$\psi(b \otimes a) = g(b,a).$$ Then, $$(\psi \circ \phi)(a \otimes b) = \psi(\phi(a \otimes b)) = \psi(f(a,b)) = \psi(b \otimes a) = g(b, a) = a \otimes b$$ and $$(\phi \circ \psi)(b \otimes a) = \phi(\psi(b \otimes a)) = \phi(g(b,a)) = \phi(a \otimes b) = f(a, b) = b \otimes a$$ which shows that $\psi$ and $\phi$ are mutual inverses, thus proving the claim.

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A small suggestion: to be completely rigorous (and it seems this is what you are going for), you should mention that it suffices to check that $\phi$ and $\psi$ are inverse on "pure tensors," i.e. those of the form $a \otimes b$ and $b \otimes a$. The cleanest way to say this is that by the universal property, the identity on $A \otimes B$ is the unique homomorphism which sends $a \otimes b \mapsto a \otimes b$, and similarly for $B \otimes A$. – Justin Campbell Jun 12 '12 at 19:00
Yes, your work looks good, and if you incorporate Justin's advice I would grade it as 100%! – rschwieb Jun 12 '12 at 19:04
Thanks for the tips – ItsNotObvious Jun 12 '12 at 19:21