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Let $C$ be an abelian monoidal category such that $1 \in C$ is simple, and each object is a finite sum of copies of $1$, i.e. isomorphic to $1^{\oplus n}$ for some $n \in \mathbb{N}$. It is well-known that $k:=\mathrm{End}(1)$ is a commutative ring . and that $C$ is $k$-linear. By Schur's Lemma it $k$ is even a field. Now, consider the functor $\mathrm{hom}(1,-) \hom(1,-) : C \to \mathsf{Mod}(k)$. It maps $1^{\oplus n}$ to $k^n$, thus factors as an essentially surjective functor $C \to \mathsf{Mod}_f(k)$. It is also fully faithful, because $\mathrm{hom}(1^{\oplus \hom(1^{\oplus n},1^{\oplus m}) \cong \prod_n \prod_m \mathrm{hom}(1,1) hom(1,1) = k^{n \times m}$. It has a canonical lax monoidal structure given by $\hom(1,x) \otimes \hom(1,y) \xrightarrow{\otimes} \hom(1 \otimes 1,x \otimes y) \cong \hom(1,x \otimes y)$. This is an isomorphism: Since both sides commute with finite direct sums in $x$ and $y$, it is enough to verify this for $x=y=1$, where it is clear. Thus, $C \cong \mathsf{Mod}_f(k)$ as monoidal categories.

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Let $C$ be an abelian monoidal category such that $1 \in C$ is simple, and each object is a finite sum of copies of $1$, i.e. isomorphic to $1^{\oplus n}$ for some $n \in \mathbb{N}$. It is well-known that $k:=\mathrm{End}(1)$ is a commutative ring. By Schur's Lemma it is even a field. Now, consider the functor $\mathrm{hom}(1,-) : C \to \mathsf{Mod}(k)$. It maps $1^{\oplus n}$ to $k^n$, thus factors as an essentially surjective functor $C \to \mathsf{Mod}_f(k)$. It is also fully faithful, because $\mathrm{hom}(1^{\oplus n},1^{\oplus m}) \cong \prod_n \prod_m \mathrm{hom}(1,1) = k^{n \times m}$.