# Tensor product of arbitrary categories

I would like to consider a definition of tensor product in the category of (small, finite, whatever is needed) categories, analogous to the tensor product of vector spaces. I will first rewrite bilinear function in a suitable way that will be easily extendable to categories.

# Vector spaces

Consider (finite-dimensional) vector spaces $C$ and $D$. With any choice of base vectors, vectors are determined by coefficients $a_i$ and linear maps by $M_{ij}$. A bilinear map from $C\times D$ (or $C\oplus D$ if you prefer) to $A$ is linear in both arguments. One, more complicated, way to say this is to assign to each element $c$ of $C$, a linear map $M(c)$ from $D$ to $A$, and to each element $d$ of $D$, a linear map $N(d)$ from $C$ to $A$. Since linear maps form vector spaces, we require the corresponding assignments to respect the structure (be linear), so $M_{ij}=\alpha_{ijk}c_k$ and $N_{ij}=\beta_{ijk}d_k$. It is also needed that starting from a pair $(c,d)$ we end up with the same vector in $A$, namely $$(\alpha_{ijk}-\beta_{ikj})c_kd_j=0,$$ so $\alpha_{ijk}=\beta_{ikj}$, and our bilinear map is determined by coefficients $\alpha_{ijk}$, which can be interpreted as a linear map from $C\otimes D$ to $A.$

1) Has a similar thing been done for categories? If yes, where? If not, I give my try below.

# Categories

Substitute vector spaces $C$, $D$ and $A$ with categories, linear maps $M:D\rightarrow A$, $N:C\rightarrow A$, $\alpha:C\rightarrow A^D$ and $\beta:D\rightarrow A^C$ with functors. Here $A^X$ denotes category of functors from $X$ to $A$, so $M$ and $N$ are objects in $A^D$ and $A^C$, and $\alpha$ and $\beta$ map arrows to natural transformations. As in the case of vector space, not all pairs $(\alpha,\beta)$ provide a "bifunctorial" map (I am not calling it bifunctor, since it means something else). One should require that for each morphism in $C\times D$ $$(f,g):(c_s,d_s)\rightarrow (c_t,d_t)$$ the induced commuting diagrams in $A$, corresponding to natural transformations, are equal: $$\require{AMScd} \begin{CD} \alpha(c_s)(d_s) @>{\alpha(c_s)(g)}>> \alpha(c_s)(d_t)\\ @V{\alpha(f)(d_s)}VV @VV{\alpha(f)(d_t)}V \\ \alpha(c_t)(d_s) @>{\alpha(c_t)(g)}>> \alpha(c_t)(d_t) \end{CD} \hspace{1cm}=\hspace{1cm} \begin{CD} \beta(d_s)(c_s) @>{\beta(g)(c_s)}>> \beta(d_t)(c_s)\\ @V{\beta(d_s)(f)}VV @VV{\beta(d_t)(f)}V \\ \beta(d_s)(c_t) @>{\beta(g)(c_t)}>> \beta(d_t)(c_t) \end{CD}$$

In other words $\alpha(x)(y)=\beta(y)(x).$ Each given functor $\alpha$ defines a mapping $\beta$ by changing arguments. I haven't checked it thoroughly, but it seems that $\beta$s obtained in this way are always functors. So, just like in the vector space case, we need only one mapping.

Consider the "bifunctorial" mapping $\gamma:C\times D\rightarrow A$ given by $\gamma((c,d))=\alpha(c)(d)$ for objects, and $\gamma((f,g))=\alpha(c_t)(g)\circ \alpha(f)(d_s)$ for morphisms. As long as I can see, $\gamma$ does not necessarily preserve compositions!

EDIT: Functors $\gamma':C\times D\rightarrow A$ have the following commuting diagram in A: $$\require{AMScd} \begin{CD} \gamma'(c_s,d_s) @>{\gamma'(1,g)}>> \gamma'(c_s,d_t)\\ @V{\gamma'(f,1)}VV @VV{\gamma'(f,1)}V \\ \gamma'(c_t,d_s) @>{\gamma'(1,g)}>> \gamma'(c_t,d_t) \end{CD}$$

2) Can different $\alpha$s give the same $\gamma$? It is clear that they would have to map objects in the same way, but when one argument is morphism, it is not clear.

3) Is there, and how to construct a category $C\otimes D$ through which $\gamma$ (or $\alpha$) factors uniquely?

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## migrated from mathoverflow.netJul 30 '14 at 7:34

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Your characterisation of tensor products of vector spaces is somewhat convoluted. If you do it the normal way, you will see that the tensor product of categories is precisely the cartesian product. (For an easier version of this, try to work out what the tensor product of partially ordered sets is.) – Zhen Lin Jul 29 '14 at 21:36
Hi Zhen, thank for the reply. If you say that the tensor product of categories is precisely the cartesian product, it means that you have some definition for the former in mind which is different from the latter, and then you prove that they are equal. Am I right? – Branko Nikolic Jul 29 '14 at 21:56
The cartesian product has all the properties you ask for of a "tensor product". There is a natural bijection between functors $\gamma:C\to A^D$ and functors $\alpha:C\times D\to A$; this is straightforward to verify. You may find it helpful to note that every map $(f,g)$ in $C\times D$ can be factored as $(f,1)\circ (1,g)$. – Eric Wofsey Jul 29 '14 at 23:03
@Eric: I am a bit new to category theory, so I cannot see the bijection right away. With your suggestion to factor (f,g) (see EDIT) I am able to construct functor $\alpha:C\rightarrow A^D$ from a given functor $\gamma':C\times D\rightarrow A$. Doing the other way around would require $\alpha(f)(d_s)=\alpha(f)(d_t)$, which I do not see why should be true for every $\alpha$... – Branko Nikolic Jul 30 '14 at 12:29
@Eric: I see now, the point is that $\gamma'(f,1_{d_s})$ does not have to be equal to $\gamma'(f,1_{d_t})$... Thanks! – Branko Nikolic Jul 30 '14 at 12:41

Zhen Lin is correct; the tensor product of categories agrees with the usual Cartesian product. The same goes for preordered sets, and this is especially easy to check. So lets go ahead and do that.

Definition. Given preordered sets $Z,Y,X,$ and given a function $$f : Z \leftarrow Y,X$$ lets declare the following terminology.

• $f$ is jointly order-preserving iff for all $y' \gtrsim y$ in $Y$ and $x' \gtrsim x$ in $X$, we have $f(y',x') \gtrsim f(y,x)$.

• $f$ is separately order-preserving iff:

• for all $y$ in $Y$ and all $x' \gtrsim x$ in $X$, we have $f(y,x') \gtrsim f(y,x)$.
• for all $y' \gtrsim y$ in $Y$ and all $x$ in $X$, we have $f(y',x) \gtrsim f(y,x)$

I claim that:

Proposition. $f$ is jointly-order preserving iff it is separately order-preserving.

Proof.

$(\Rightarrow)$ Assume $f$ is jointly order-preserving. Consider $y$ in $Y$ and $x' \gtrsim x$ in $X$. Then by reflexivity we have:

$$y \gtrsim y \qquad x' \gtrsim x$$

Hence $$f(y,x') \gtrsim f(y,x).$$

A similar approach works for the other argument.

$(\Leftarrow)$ Assume $f$ is separately order-preserving. Consider $y' \gtrsim y$ in $Y$ and $x' \gtrsim x$ in $X$. Then $$f(y',x') \gtrsim f(y,x'), \qquad f(y,x') \gtrsim f(y,x)$$

Hence by transitivity, we have:

$$f(y',x') \gtrsim f(y,x)$$

I always find it cool that the definition of preordered set almost seems perfectly rigged so that the above proof goes through!

Anyway, the universal property of the tensor product is that order-preserving maps out of $Y \otimes X$ are meant to be the same as separately order-preserving maps out of $Y,X$. And the universal property of the Cartesian product is that order-preserving maps out of $Y \times X$ are meant to be the same as jointly order-preserving maps out of $Y,X$. But since these are the same, we conclude:

Corollary. For preordered sets $Y$ and $X$, the tensor product $Y \otimes X$ always exists and agrees with $Y \times X$.

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