I want to show that there is a unique isomorphism $M \otimes N \to N \otimes M$ such that $x\otimes y\mapsto y\otimes x$. (Prop. 2.14, i), Atiyah-Macdonald)
My proof idea is to take a bilinear $f: M \times N \to N \otimes M$ and then use the universal property of the tensor product to get a unique linear map $l : M \otimes N \to N \otimes M$. Then show that $l$ is bijective.
Can you tell me if my proof is correct:
Let $M,N$ be two $R$-modules. Let $(M \otimes N, b)$ be their tensor product.
Then $$ \varphi: M \times N \to N \otimes M$$ defined as $$ (m,n) \mapsto n \otimes m$$ and $$ (rm , n) \mapsto r(n \otimes m)$$ $$ (m , rn) \mapsto r(m \otimes n)$$
is bilinear. Hence by the universal property of the tensor product there exists a unique $R$-module homomorphism ($\cong$ linear map) $l: M \otimes N \to N \otimes M$ such that $l \circ b = \varphi$.
$l$ is bijective:
$l$ is surjective: Let $n \otimes m \in N \otimes M$. Then $l(m \otimes n) = l(b(m,n)) = \varphi (m,n) = n \otimes m$.
$l$ is injective: Let $l(m\otimes n) = l(b(m,n)) = 0 = \varphi(m,n) = n \otimes m$. Then $n \otimes m = 0$ implies that either $n$ or $m$ are zero and hence $m \otimes n = 0$.