[proofplan] The proof is a direct unpacking of the universal property defining the [tensor product](/page/Tensor%20Product). We first check that postcomposition with the canonical bilinear map $\tau$ really sends linear maps $M \otimes_R N \to P$ to bilinear maps $M \times N \to P$. Then the universal property supplies the inverse assignment, sending a bilinear map $b$ to the unique [linear map](/page/Linear%20Map) $\bar b$ with $\bar b \circ \tau = b$. Finally, uniqueness proves that the two assignments are inverse, and functoriality under postcomposition proves naturality in $P$. [/proofplan] [step:Send linear maps to bilinear maps by composing with $\tau$] Fix an $R$-module $P$. Define \begin{align*} \Phi_P: \operatorname{Hom}_R(M \otimes_R N, P) &\to \operatorname{Bilin}_R(M,N;P) \\ \ell &\mapsto \ell \circ \tau. \end{align*} Let $\ell: M \otimes_R N \to P$ be an $R$-linear map. Since $\tau: M \times N \to M \otimes_R N$ is $R$-bilinear and $\ell$ is $R$-linear, the composite \begin{align*} \ell \circ \tau: M \times N &\to P \end{align*} is $R$-bilinear. Therefore $\Phi_P$ is well-defined. [/step] [step:Use the universal property to construct the inverse map] Define \begin{align*} \Psi_P: \operatorname{Bilin}_R(M,N;P) &\to \operatorname{Hom}_R(M \otimes_R N, P) \end{align*} as follows. For an $R$-bilinear map \begin{align*} b: M \times N &\to P, \end{align*} the universal property of $M \otimes_R N$ gives a unique $R$-linear map \begin{align*} \bar b: M \otimes_R N &\to P \end{align*} such that \begin{align*} \bar b \circ \tau = b. \end{align*} Set $\Psi_P(b) := \bar b$. [guided] We need to reverse the construction from the previous step. Starting from an $R$-bilinear map \begin{align*} b: M \times N &\to P, \end{align*} the defining universal property of the tensor product says exactly that $b$ factors uniquely through the canonical bilinear map \begin{align*} \tau: M \times N &\to M \otimes_R N. \end{align*} That is, there exists a unique $R$-linear map \begin{align*} \bar b: M \otimes_R N &\to P \end{align*} for which \begin{align*} \bar b \circ \tau = b. \end{align*} This unique linear map is the only possible candidate for the inverse image of $b$ under $\Phi_P$, because any $\ell$ satisfying $\Phi_P(\ell)=b$ must satisfy $\ell \circ \tau=b$. Thus we define \begin{align*} \Psi_P: \operatorname{Bilin}_R(M,N;P) &\to \operatorname{Hom}_R(M \otimes_R N, P) \end{align*} by $\Psi_P(b):=\bar b$. [/guided] [/step] [step:Verify that the two assignments are inverse] Let $\ell: M \otimes_R N \to P$ be $R$-linear. Put $b := \Phi_P(\ell)=\ell \circ \tau$. By definition, $\Psi_P(b)$ is the unique $R$-linear map $\bar b: M \otimes_R N \to P$ satisfying $\bar b \circ \tau=b$. The map $\ell$ also satisfies \begin{align*} \ell \circ \tau = b, \end{align*} so uniqueness gives $\Psi_P(\Phi_P(\ell))=\ell$. Conversely, let $b: M \times N \to P$ be $R$-bilinear. By construction, $\Psi_P(b)=\bar b$, where $\bar b$ is the unique $R$-linear map satisfying $\bar b \circ \tau=b$. Hence \begin{align*} \Phi_P(\Psi_P(b))=\Phi_P(\bar b)=\bar b \circ \tau=b. \end{align*} Therefore $\Phi_P$ and $\Psi_P$ are inverse maps, so $\Phi_P$ is a bijection. [guided] First take an $R$-linear map \begin{align*} \ell: M \otimes_R N &\to P. \end{align*} Applying $\Phi_P$ gives the bilinear map $b:=\ell \circ \tau$. Applying $\Psi_P$ to this $b$ means: choose the unique $R$-linear map \begin{align*} \bar b: M \otimes_R N &\to P \end{align*} such that $\bar b \circ \tau=b$. But $\ell$ itself has this property, since $b$ was defined by $b=\ell \circ \tau$. The uniqueness clause in the universal property therefore forces $\bar b=\ell$, so \begin{align*} \Psi_P(\Phi_P(\ell))=\ell. \end{align*} Now take an $R$-bilinear map \begin{align*} b: M \times N &\to P. \end{align*} By definition, $\Psi_P(b)=\bar b$, where $\bar b$ is the unique $R$-linear map with \begin{align*} \bar b \circ \tau=b. \end{align*} Applying $\Phi_P$ to $\bar b$ gives \begin{align*} \Phi_P(\Psi_P(b))=\Phi_P(\bar b)=\bar b \circ \tau=b. \end{align*} Thus each assignment undoes the other, and $\Phi_P$ is a bijection. [/guided] [/step] [step:Check naturality under postcomposition] Let $P$ and $Q$ be $R$-modules, and let \begin{align*} f: P &\to Q \end{align*} be an $R$-linear map. Define postcomposition maps \begin{align*} f_*: \operatorname{Hom}_R(M \otimes_R N, P) &\to \operatorname{Hom}_R(M \otimes_R N, Q),& f_*(\ell)&:=f\circ \ell, \end{align*} and \begin{align*} f_*^{\operatorname{bil}}: \operatorname{Bilin}_R(M,N;P) &\to \operatorname{Bilin}_R(M,N;Q),& f_*^{\operatorname{bil}}(b)&:=f\circ b. \end{align*} For every $R$-linear map $\ell: M \otimes_R N \to P$, associativity of composition gives \begin{align*} \Phi_Q(f_*(\ell)) &= \Phi_Q(f\circ \ell) \\ &= (f\circ \ell)\circ \tau \\ &= f\circ(\ell\circ \tau) \\ &= f_*^{\operatorname{bil}}(\Phi_P(\ell)). \end{align*} Thus the bijections $\Phi_P$ commute with postcomposition by $R$-linear maps $P \to Q$, which is precisely naturality in $P$. [/step]