[proofplan] We prove uniqueness and existence separately. Uniqueness follows from the fact that the pure tensors $m \otimes n$ generate $M \otimes_R N$ as an $R$-module, so any $R$-linear map on $M \otimes_R N$ is determined by its values on pure tensors. For existence, we define a map $\tilde{h}$ on the free $R$-module $\mathcal{F} = R^{\oplus(M \times N)}$ by sending each basis element $e_{(m,n)}$ to $f(m,n)$, and show that the bilinearity of $f$ forces $\tilde{h}$ to vanish on the submodule $K$ of bilinearity relations. Hence $\tilde{h}$ descends to a well-defined $R$-module homomorphism $h: \mathcal{F}/K = M \otimes_R N \to L$. [/proofplan]

[step:Prove uniqueness: any $R$-linear map on $M \otimes_R N$ is determined by its values on pure tensors] Suppose $h, h': M \otimes_R N \to L$ are two $R$-module homomorphisms satisfying $h(m \otimes n) = f(m,n) = h'(m \otimes n)$ for all $m \in M$, $n \in N$. By the construction of $M \otimes_R N = \mathcal{F}/K$, where $\mathcal{F} = R^{\oplus(M \times N)}$ is the free $R$-module on the set $M \times N$ and $K$ is the submodule of bilinearity relations, every element of $M \otimes_R N$ is a finite $R$-linear combination of pure tensors $m \otimes n$. Since $h$ and $h'$ are $R$-linear and agree on all pure tensors, they agree on all $R$-linear combinations of pure tensors, hence on all of $M \otimes_R N$. Therefore $h = h'$. [/step]

[step:Define $\tilde{h}: \mathcal{F} \to L$ on the free module by $e_{(m,n)} \mapsto f(m,n)$ and extend $R$-linearly]Recall the construction: $\mathcal{F} = R^{\oplus(M \times N)}$ is the free $R$-module with basis $\{e_{(m,n)}\}_{(m,n) \in M \times N}$, and $K \subseteq \mathcal{F}$ is the submodule generated by the bilinearity relations \begin{align*} &e_{(m_1 + m_2, n)} - e_{(m_1, n)} - e_{(m_2, n)}, \\ &e_{(m, n_1 + n_2)} - e_{(m, n_1)} - e_{(m, n_2)}, \\ &e_{(rm, n)} - r \cdot e_{(m, n)}, \\ &e_{(m, rn)} - r \cdot e_{(m, n)}, \end{align*} for all $m, m_1, m_2 \in M$, $n, n_1, n_2 \in N$, and $r \in R$. The tensor product is $M \otimes_R N = \mathcal{F}/K$, the canonical bilinear map is $i_{M \otimes N}: M \times N \to M \otimes_R N$, $(m,n) \mapsto e_{(m,n)} + K = m \otimes n$, and the pure tensor $m \otimes n$ denotes the coset $e_{(m,n)} + K$. By the universal property of the free $R$-module $\mathcal{F}$, the set-theoretic function $M \times N \to L$ defined by $(m,n) \mapsto f(m,n)$ extends uniquely to an $R$-module homomorphism \begin{align*} \tilde{h}: \mathcal{F} &\to L \\ e_{(m,n)} &\mapsto f(m,n). \end{align*} Explicitly, $\tilde{h}\bigl(\sum_{i} r_i \, e_{(m_i, n_i)}\bigr) = \sum_{i} r_i \, f(m_i, n_i)$.[/step]

[guided]The construction of the tensor product proceeds in two stages. First, we build the "maximally free" $R$-module $\mathcal{F}$ with one generator $e_{(m,n)}$ for each pair $(m,n) \in M \times N$. This module is too large — it treats every pair independently with no algebraic relations. Second, we quotient by the submodule $K$ generated by all bilinearity relations to enforce that $m \otimes n$ behaves bilinearly in $(m,n)$. To construct $h$, we reverse-engineer: start with a map on the big free module $\mathcal{F}$ and show it descends to the quotient $\mathcal{F}/K$. By the universal property of the free $R$-module $\mathcal{F}$, any function from the basis $\{e_{(m,n)}\}$ to an $R$-module $L$ extends uniquely to an $R$-linear map. We use $(m,n) \mapsto f(m,n)$ to define \begin{align*} \tilde{h}: \mathcal{F} &\to L \\ e_{(m,n)} &\mapsto f(m,n), \end{align*} extended $R$-linearly: $\tilde{h}\bigl(\sum_i r_i \, e_{(m_i,n_i)}\bigr) = \sum_i r_i \, f(m_i, n_i)$. The key question is: does $\tilde{h}$ respect the quotient, i.e., does $K \subseteq \ker(\tilde{h})$?[/guided]

[step:Verify $\tilde{h}$ vanishes on each generator of $K$ using the bilinearity of $f$]We must check that $\tilde{h}$ vanishes on the four types of generators of $K$. **Additivity in the first variable.** For $m_1, m_2 \in M$ and $n \in N$: \begin{align*} \tilde{h}(e_{(m_1 + m_2, n)} - e_{(m_1, n)} - e_{(m_2, n)}) &= f(m_1 + m_2, n) - f(m_1, n) - f(m_2, n) = 0, \end{align*} where the last equality holds because $f$ is $R$-bilinear and hence additive in the first argument. **Additivity in the second variable.** For $m \in M$ and $n_1, n_2 \in N$: \begin{align*} \tilde{h}(e_{(m, n_1 + n_2)} - e_{(m, n_1)} - e_{(m, n_2)}) &= f(m, n_1 + n_2) - f(m, n_1) - f(m, n_2) = 0. \end{align*} **$R$-homogeneity in the first variable.** For $r \in R$, $m \in M$, and $n \in N$: \begin{align*} \tilde{h}(e_{(rm, n)} - r \cdot e_{(m, n)}) &= f(rm, n) - r \cdot f(m, n) = 0, \end{align*} since $f(rm, n) = r \cdot f(m, n)$ by the $R$-bilinearity of $f$. **$R$-homogeneity in the second variable.** For $r \in R$, $m \in M$, and $n \in N$: \begin{align*} \tilde{h}(e_{(m, rn)} - r \cdot e_{(m, n)}) &= f(m, rn) - r \cdot f(m, n) = 0. \end{align*} Since $\tilde{h}$ is $R$-linear and vanishes on all generators of $K$, it vanishes on all of $K$: $K \subseteq \ker(\tilde{h})$.[/step]

[guided]This is the step where the bilinearity of $f$ is consumed. Each of the four types of generators of $K$ encodes one of the four bilinearity axioms: additivity in each variable and $R$-homogeneity in each variable. Since $\tilde{h}(e_{(m,n)}) = f(m,n)$ and $f$ satisfies all four axioms, $\tilde{h}$ evaluates to zero on each generator. The $R$-linearity of $\tilde{h}$ then extends this to all of $K$. We verify each type. For additivity in the first variable, with $m_1, m_2 \in M$, $n \in N$: \begin{align*} \tilde{h}(e_{(m_1 + m_2, n)} - e_{(m_1, n)} - e_{(m_2, n)}) &= f(m_1 + m_2, n) - f(m_1, n) - f(m_2, n). \end{align*} The bilinearity of $f$ gives $f(m_1 + m_2, n) = f(m_1, n) + f(m_2, n)$, so this is zero. The second variable version is identical with the roles of $M$ and $N$ swapped. For $R$-homogeneity in the first variable, with $r \in R$, $m \in M$, $n \in N$: \begin{align*} \tilde{h}(e_{(rm, n)} - r \cdot e_{(m, n)}) = f(rm, n) - r \cdot f(m, n) = 0, \end{align*} using $f(rm, n) = rf(m, n)$. The second variable version is analogous, using $f(m, rn) = rf(m, n)$. Since $K$ is generated as an $R$-module by elements on which $\tilde{h}$ vanishes, and $\tilde{h}$ is $R$-linear, we have $K \subseteq \ker(\tilde{h})$.[/guided]

[step:Descend $\tilde{h}$ to the quotient to obtain $h: M \otimes_R N \to L$ with the required property] Since $K \subseteq \ker(\tilde{h})$, the universal property of the quotient module gives a unique $R$-module homomorphism \begin{align*} h: M \otimes_R N = \mathcal{F}/K &\to L \end{align*} satisfying $h(x + K) = \tilde{h}(x)$ for all $x \in \mathcal{F}$. In particular, for any $m \in M$ and $n \in N$: \begin{align*} h(m \otimes n) = h(e_{(m,n)} + K) = \tilde{h}(e_{(m,n)}) = f(m, n). \end{align*} This means $h \circ i_{M \otimes N} = f$, as required. Combining with the uniqueness proved in the first step, $h$ is the unique $R$-module homomorphism $M \otimes_R N \to L$ satisfying $f = h \circ i_{M \otimes N}$. [/step]