Lemma 10.11.2. Let $M, N$ be $R$-modules. Then there exists a pair $(T, g)$ where $T$ is an $R$-module, and $g : M \times N \to T$ an $R$-bilinear mapping, with the following universal property: For any $R$-module $P$ and any $R$-bilinear mapping $f : M \times N \to P$, there exists a unique $R$-linear mapping $\tilde{f} : T \to P$ such that $f = \tilde{f} \circ g$. In other words, the following diagram commutes:

\[ \xymatrix{ M \times N \ar[rr]^ f \ar[dr]_ g & & P\\ & T \ar[ur]_{\tilde f} } \]

Moreover, if $(T, g)$ and $(T', g')$ are two pairs with this property, then there exists a unique isomorphism $j : T \to T'$ such that $j\circ g = g'$.

**Proof.**
We first prove the existence of such $R$-module $T$. Let $M, N$ be $R$-modules. Let $T$ be the quotient module $P/Q$, where $P$ is the free $R$-module $R^{(M \times N)}$ and $Q$ is the $R$-module generated by all elements of the following types: ($x\in M, y\in N$)

\begin{align*} (x + x', y) - (x, y) - (x', y), \\ (x, y + y') - (x, y) - (x, y'), \\ (ax, y) - a(x, y), \\ (x, ay) - a(x, y) \end{align*}

Let $\pi : M \times N \to T$ denote the natural map. This map is $R$-bilinear, as implied by the above relations when we check the bilinearity conditions. Denote the image $\pi (x, y) = x \otimes y$, then these elements generate $T$. Now let $f : M \times N \to P$ be an $R$-bilinear map, then we can define $f' : T \to P$ by extending the mapping $f'(x \otimes y) = f(x, y)$. Clearly $f = f'\circ \pi $. Moreover, $f'$ is uniquely determined by the value on the generating sets $\{ x \otimes y : x\in M, y\in N\} $. Suppose there is another pair $(T', g')$ satisfying the same properties. Then there is a unique $j : T \to T'$ and also $j' : T' \to T$ such that $g' = j\circ g$, $g = j'\circ g'$. But then both the maps $(j\circ j') \circ g$ and $g$ satisfies the universal properties, so by uniqueness they are equal, and hence $j'\circ j$ is identity on $T$. Similarly $(j'\circ j) \circ g' = g'$ and $j\circ j'$ is identity on $T'$. So $j$ is an isomorphism.
$\square$

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