[proofplan]
This is a standard "universal objects are unique up to unique isomorphism" argument. We use the universal properties of both $(M \otimes_R N, i_{M \otimes N})$ and $(T, j)$ to produce maps in both directions, then invoke the uniqueness clause of each universal property to show the composites are the respective identity maps.
[/proofplan]
[step:Apply the universal property of $M \otimes_R N$ to the bilinear map $j$ to obtain $\varphi: M \otimes_R N \to T$]
The map $j: M \times N \to T$ is $R$-bilinear by hypothesis. By the [Universal Property of Tensor Product](/theorems/2908) applied to the pair $(M \otimes_R N, i_{M \otimes N})$, there exists a unique $R$-module homomorphism
\begin{align*}
\varphi: M \otimes_R N \to T
\end{align*}
such that $\varphi \circ i_{M \otimes N} = j$. Concretely, $\varphi(m \otimes n) = j(m, n)$ for all $m \in M$, $n \in N$.
[/step]
[step:Apply the universal property of $(T, j)$ to the bilinear map $i_{M \otimes N}$ to obtain $\psi: T \to M \otimes_R N$]
The map $i_{M \otimes N}: M \times N \to M \otimes_R N$ is $R$-bilinear (this is part of the construction of the tensor product). Since $(T, j)$ also satisfies the universal property of the tensor product, applying it to $i_{M \otimes N}$ yields a unique $R$-module homomorphism
\begin{align*}
\psi: T \to M \otimes_R N
\end{align*}
such that $\psi \circ j = i_{M \otimes N}$. Concretely, $\psi(j(m, n)) = m \otimes n$ for all $m \in M$, $n \in N$.
[/step]
[step:Show $\psi \circ \varphi = \operatorname{id}_{M \otimes_R N}$ and $\varphi \circ \psi = \operatorname{id}_T$ using the uniqueness clauses]
Consider the composite $\psi \circ \varphi: M \otimes_R N \to M \otimes_R N$. On the canonical bilinear map:
\begin{align*}
(\psi \circ \varphi) \circ i_{M \otimes N} = \psi \circ (\varphi \circ i_{M \otimes N}) = \psi \circ j = i_{M \otimes N}.
\end{align*}
The identity map $\operatorname{id}_{M \otimes_R N}$ also satisfies $\operatorname{id}_{M \otimes_R N} \circ \, i_{M \otimes N} = i_{M \otimes N}$. By the uniqueness clause of the [Universal Property of Tensor Product](/theorems/2908), there is exactly one $R$-module homomorphism $M \otimes_R N \to M \otimes_R N$ that composes with $i_{M \otimes N}$ to give $i_{M \otimes N}$. Therefore $\psi \circ \varphi = \operatorname{id}_{M \otimes_R N}$.
By the same argument with the roles of $(M \otimes_R N, i_{M \otimes N})$ and $(T, j)$ interchanged: the composite $\varphi \circ \psi: T \to T$ satisfies
\begin{align*}
(\varphi \circ \psi) \circ j = \varphi \circ (\psi \circ j) = \varphi \circ i_{M \otimes N} = j.
\end{align*}
Since $\operatorname{id}_T \circ \, j = j$ and the universal property of $(T, j)$ guarantees uniqueness, we conclude $\varphi \circ \psi = \operatorname{id}_T$.
Hence $\varphi$ is an $R$-module isomorphism with inverse $\psi$.
[guided]
This is the canonical argument that any two objects satisfying the same universal property are uniquely isomorphic. The structure is always the same: build maps both ways using the two universal properties, then use uniqueness to identify the round-trip composites with identity maps.
We have already constructed $\varphi: M \otimes_R N \to T$ with $\varphi \circ i_{M \otimes N} = j$ and $\psi: T \to M \otimes_R N$ with $\psi \circ j = i_{M \otimes N}$.
To show $\psi \circ \varphi = \operatorname{id}_{M \otimes_R N}$, we check that both maps $M \otimes_R N \to M \otimes_R N$ satisfy the same universal condition. Compute:
\begin{align*}
(\psi \circ \varphi) \circ i_{M \otimes N} = \psi \circ j = i_{M \otimes N}.
\end{align*}
The identity also satisfies $\operatorname{id} \circ \, i_{M \otimes N} = i_{M \otimes N}$. The universal property of $(M \otimes_R N, i_{M \otimes N})$ says there is **exactly one** $R$-module homomorphism $h: M \otimes_R N \to M \otimes_R N$ with $h \circ i_{M \otimes N} = i_{M \otimes N}$. Both $\psi \circ \varphi$ and $\operatorname{id}$ qualify, so they must be equal: $\psi \circ \varphi = \operatorname{id}_{M \otimes_R N}$.
The symmetric argument (using the universal property of $(T, j)$) gives $\varphi \circ \psi = \operatorname{id}_T$:
\begin{align*}
(\varphi \circ \psi) \circ j = \varphi \circ i_{M \otimes N} = j = \operatorname{id}_T \circ \, j.
\end{align*}
By uniqueness in the universal property of $(T, j)$, $\varphi \circ \psi = \operatorname{id}_T$.
Since $\varphi$ has a two-sided inverse $\psi$, it is an $R$-module isomorphism. Why is it unique? Any isomorphism $\varphi': M \otimes_R N \xrightarrow{\sim} T$ with $\varphi' \circ i_{M \otimes N} = j$ is in particular an $R$-module homomorphism satisfying the condition that the universal property of $(M \otimes_R N, i_{M \otimes N})$ governs. By the uniqueness clause, $\varphi' = \varphi$.
[/guided]
[/step]
