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In [1] it is said that the category $\mathbf{FinVect}$ of finite vector spaces is a tensor category. I am trying to convince myself that this is indeed the case.

One of the properties that $\mathbf{FinVect}$ must satisfy for this to hold is that it is indecomposable. I cannot think of a way to show that. Here are the definitions which are relevant for this problem:

Def: Let $\mathbb C$ be a locally finite $\mathbb{K}$-linear abelian rigid monoidal category. $\mathbb{C}$ is a multitensor category over $\mathbb{K}$ if $\otimes:\mathbb{C}\times\mathbb{C}\rightarrow \mathbb{C}$ is bilinear on morphisms. A multitensor category $\mathbb{C}$ is decomposable if it is equivalent to a direct sum of nonzero multitensor categories. If $\mathbb{C}$ is an indecomposable multitensor category and $\text{End}(1)\cong \mathbb{K}$ (as vector spaces), then $\mathbb{C}$ is a tensor category.

[1] Pavel Etingof et al. Tensor Categories. American Mathematical Society, 2015.

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    $\begingroup$ Doesn't it follow from $\mathbb K$ being indecomposable ? Indeed take if $\mathbf{FinVect} \simeq C_1\times C_2$ then with $\mathbb K$ corresponding to, say $(a,b)$ we would have $\hom(\mathbb{K,K}) \cong \hom(a,a)\times \hom(b,b)$. So one of $a$ or $b$ is $0$ and then it should follow from there (and the generating character of $\mathbb K$) that $C_1$ or $C_2$ is $0$ $\endgroup$ – Max Sep 1 '19 at 13:15
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If $\mathbf{FinVect}\simeq C\times D$ where $C$ and $D$ are both nonzero, let $c$ and $d$ be nonzero objects of $C$ and $D$, respectively. Then $(c,0)$ and $(0,d)$ are both nonzero objects of $C\times D$ (because their identity map is not equal to their zero map), but $\operatorname{Hom}((c,0),(0,d))\cong \operatorname{Hom}(c,0)\times\operatorname{Hom}(0,d)\cong 0$. This is a contradiction, since any two nonzero vector spaces have a nonzero linear map between them.

For some similar arguments for other categories, see my answer at Is Set "prime" with respect to the cartesian product? and its comments.

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