[proofplan] We use each universal property once to construct maps in opposite directions. The universal property of $(T,\tau)$ applied to $\tau'$ gives a unique $R$-[linear map](/page/Linear%20Map) $\varphi: T \to T'$, and the universal property of $(T',\tau')$ applied to $\tau$ gives a unique $R$-linear map $\psi: T' \to T$. We then compare the composites $\psi \circ \varphi$ and $\varphi \circ \psi$ with the relevant identity maps by checking that they agree after precomposition with the universal bilinear maps. Uniqueness in the two universal properties forces the composites to be identities, so $\varphi$ is an isomorphism; the same uniqueness also proves that no other compatible isomorphism can exist. [/proofplan] [step:Construct the comparison map from $T$ to $T'$] Since $\tau': M \times N \to T'$ is an $R$-bilinear map and $(T,\tau)$ satisfies the universal property for $R$-bilinear maps out of $M \times N$, there exists a unique $R$-linear map \begin{align*} \varphi: T &\to T' \end{align*} such that \begin{align*} \varphi \circ \tau = \tau'. \end{align*} [guided] The pair $(T,\tau)$ is universal among $R$-bilinear maps out of $M \times N$. To use this property, we must provide a target $R$-module and an $R$-bilinear map from $M \times N$ to that target. Here the target $R$-module is $T'$, and the relevant $R$-bilinear map is \begin{align*} \tau': M \times N &\to T'. \end{align*} By the universal property of $(T,\tau)$, there is a unique $R$-linear map \begin{align*} \varphi: T &\to T' \end{align*} making the factorisation through $\tau$ agree with $\tau'$, namely \begin{align*} \varphi \circ \tau = \tau'. \end{align*} This map $\varphi$ is the only possible candidate for an isomorphism compatible with the two tensor-product structure maps, because any such compatible map must satisfy exactly this equation. [/guided] [/step] [step:Construct the comparison map from $T'$ to $T$] Since $\tau: M \times N \to T$ is an $R$-bilinear map and $(T',\tau')$ satisfies the universal property for $R$-bilinear maps out of $M \times N$, there exists a unique $R$-linear map \begin{align*} \psi: T' &\to T \end{align*} such that \begin{align*} \psi \circ \tau' = \tau. \end{align*} [guided] We now apply the same argument with the roles of the two universal pairs reversed. The target $R$-module is $T$, and the $R$-bilinear map from $M \times N$ to this target is \begin{align*} \tau: M \times N &\to T. \end{align*} Since $(T',\tau')$ has the universal property, this bilinear map factors uniquely through $\tau'$. Therefore there exists a unique $R$-linear map \begin{align*} \psi: T' &\to T \end{align*} such that \begin{align*} \psi \circ \tau' = \tau. \end{align*} This gives the comparison map in the reverse direction. [/guided] [/step] [step:Show that the composite on $T$ is the identity] The composite \begin{align*} \psi \circ \varphi: T &\to T \end{align*} is $R$-linear because it is a composite of $R$-linear maps. Moreover, \begin{align*} (\psi \circ \varphi) \circ \tau &= \psi \circ (\varphi \circ \tau) \\ &= \psi \circ \tau' \\ &= \tau. \end{align*} The identity map \begin{align*} \operatorname{id}_T: T &\to T \end{align*} is also $R$-linear and satisfies \begin{align*} \operatorname{id}_T \circ \tau = \tau. \end{align*} By the uniqueness part of the universal property of $(T,\tau)$ applied to the bilinear map $\tau: M \times N \to T$, we conclude that \begin{align*} \psi \circ \varphi = \operatorname{id}_T. \end{align*} [guided] To prove that $\varphi$ is invertible, we first show that $\psi$ is a left inverse of $\varphi$. The map \begin{align*} \psi \circ \varphi: T &\to T \end{align*} is $R$-linear because both $\varphi$ and $\psi$ are $R$-linear. We compare it with $\operatorname{id}_T$ by precomposing both maps with the universal bilinear map $\tau$. Using associativity of composition and the defining equations for $\varphi$ and $\psi$, we obtain \begin{align*} (\psi \circ \varphi) \circ \tau &= \psi \circ (\varphi \circ \tau) \\ &= \psi \circ \tau' \\ &= \tau. \end{align*} The identity map \begin{align*} \operatorname{id}_T: T &\to T \end{align*} is $R$-linear and satisfies \begin{align*} \operatorname{id}_T \circ \tau = \tau. \end{align*} Thus both $\psi \circ \varphi$ and $\operatorname{id}_T$ are $R$-linear maps from $T$ to $T$ whose compositions with $\tau$ equal the same bilinear map $\tau: M \times N \to T$. The uniqueness clause in the universal property of $(T,\tau)$ therefore forces \begin{align*} \psi \circ \varphi = \operatorname{id}_T. \end{align*} [/guided] [/step] [step:Show that the composite on $T'$ is the identity] The composite \begin{align*} \varphi \circ \psi: T' &\to T' \end{align*} is $R$-linear because it is a composite of $R$-linear maps. Moreover, \begin{align*} (\varphi \circ \psi) \circ \tau' &= \varphi \circ (\psi \circ \tau') \\ &= \varphi \circ \tau \\ &= \tau'. \end{align*} The identity map \begin{align*} \operatorname{id}_{T'}: T' &\to T' \end{align*} is also $R$-linear and satisfies \begin{align*} \operatorname{id}_{T'} \circ \tau' = \tau'. \end{align*} By the uniqueness part of the universal property of $(T',\tau')$ applied to the bilinear map $\tau': M \times N \to T'$, we conclude that \begin{align*} \varphi \circ \psi = \operatorname{id}_{T'}. \end{align*} [/step] [step:Conclude that the compatible isomorphism is unique] The identities \begin{align*} \psi \circ \varphi = \operatorname{id}_T, \qquad \varphi \circ \psi = \operatorname{id}_{T'} \end{align*} show that $\varphi$ is an $R$-module isomorphism with inverse $\psi$. By construction, \begin{align*} \varphi \circ \tau = \tau'. \end{align*} It remains to prove uniqueness among compatible isomorphisms. Let \begin{align*} \theta: T &\to T' \end{align*} be an $R$-module isomorphism satisfying \begin{align*} \theta \circ \tau = \tau'. \end{align*} Then $\theta$ is an $R$-linear map from $T$ to $T'$ with the same compatibility equation as $\varphi$. By the uniqueness part of the universal property of $(T,\tau)$ applied to $\tau': M \times N \to T'$, we obtain \begin{align*} \theta = \varphi. \end{align*} Hence $\varphi$ is the unique $R$-module isomorphism $T \to T'$ satisfying $\varphi \circ \tau = \tau'$. [/step]