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The cartesian product of two sets $A$ and $B$ can be seen as a tensor product.

Are there examples for the tensor product of two sets $A$ and $B$ other than the usual cartesian product ?

The context is the following: assume one has a set-valued presheaf $F$ on a monoidal category, knowing $F(A)$ and $F(B)$ how does one define $F(A \otimes B)$ ?

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    "Tensor product" has several possible meanings depending on which other structure you're interested in. What is the context for your question here?2012-06-20
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    Assume you have a set-valued presheaf $F$ on a monoidal category, if you know $F(A)$ and $F(B)$, how do you define $F(A \otimes B)$ ?2012-06-20
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    That should be part of the question. While you're editing, please add a "category-theory" tag.2012-06-20
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    @Henning : I've edited my question accordingly.2012-06-20
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    What makes you think $F(A \otimes B)$ should be determined by $F(A)$ and $F(B)$?2012-06-20
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    Both the product and the coproduct of a category can induce a monoidal structure on that category.2012-06-20
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    @QiaochuYuan: I thought the tensor product would be carried on by the functor $F$, ie $F(A \otimes B) = F(A) \otimes F(B)$.2012-06-21
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    @Alex: that is an additional condition you need to impose. It doesn't follow from the definition of a presheaf at all. (Consider the case that $C$ is a discrete category and the monoidal product is just some monoid operation on it.)2012-06-21

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Simply being in a monoidal category is a rather liberal condition on the tensor product; it tells you very little about what the tensor product actually looks like.

Here is a (perhaps slightly contrived?) example:

Let $C$ be the category of vector spaces over a finite field $\mathbb F_p$ with linear transformations. The vector space tensor product makes this into a monoidal category with $\mathbb F_p$ itself as the unit. $C^{op}$ is then also a monoidal category, and the ordinary forgetful functor is a Set-valued presheaf on $C^{op}$.

However, $F(A\otimes B)$ cannot be the cartesian product $F(A)\times F(B)$, because $F(A)\times F(B)$ has the wrong cardinality when $A$ and $B$ are finite.