The Stacks project

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: ($x\in M, y\in N$)

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. $\square$