Lemma 89.11.4. Let $L: \text{Mod}^{fg}_ R \to \textit{Sets}$, resp. $L: \text{Mod}_ R \to \textit{Sets}$ be a functor. Suppose $L(0)$ is a one element set and $L$ preserves finite products. Then there exists a unique $R$-linear functor $\widetilde{L} : \text{Mod}^{fg}_ R \to \text{Mod}_ R$, resp. $\widetilde{L} : \text{Mod}^{fg}_ R \to \text{Mod}_ R$, such that

\[ \vcenter { \xymatrix{ & \text{Mod}_ R \ar[dr]^{\text{forget}} & \\ \text{Mod}^{fg}_ R \ar[ur]^{\widetilde{L}} \ar[rr]^{L} & & \textit{Sets} } } \quad \text{resp.}\quad \vcenter { \xymatrix{ & \text{Mod}_ R \ar[dr]^{\text{forget}} & \\ \text{Mod}_ R \ar[ur]^{\widetilde{L}} \ar[rr]^{L} & & \textit{Sets} } } \]

commutes.

**Proof.**
We only prove this in case $L: \text{Mod}^{fg}_ R \to \textit{Sets}$. Let $M$ be a finitely generated $R$-module. We define $\widetilde{L}(M)$ to be the set $L(M)$ with the following $R$-module structure.

Multiplication: If $r \in R$, multiplication by $r$ on $L(M)$ is defined to be the map $L(M) \to L(M)$ induced by the multiplication map $r \cdot : M \to M$.

Addition: The sum map $M \times M \to M: (m_1, m_2) \mapsto m_1 + m_2$ induces a map $L(M \times M) \to L(M)$. By assumption $L(M \times M)$ is canonically isomorphic to $L(M) \times L(M)$. Addition on $L(M)$ is defined by the map $L(M) \times L(M) \cong L(M \times M) \to L(M)$.

Zero: There is a unique map $0 \to M$. The zero element of $L(M)$ is the image of $L(0) \to L(M)$.

We omit the verification that this defines an $R$-module $\widetilde{L}(M)$, the unique such that is $R$-linearly functorial in $M$.
$\square$

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