6
$\begingroup$

Given two categories $\mathcal{C}$ and $\mathcal{D}$, we can describe the following category $\mathcal{E}$. It is the initial category whose object set contains $\mathrm{Obj}(\mathcal{C}) \times \mathrm{Obj}(\mathcal{D})$ and which is equipped with

  • A strict monoidal structure $\otimes : \mathcal{E} \times \mathcal{E} \to \mathcal{E}$
  • For each $c \in \mathcal{C}$ a functor $F_c : \mathcal{D} \to \mathcal{E}$ with $F_c(d) = (c, d)$.
  • For each $d \in \mathcal{D}$ a functor $G_d : \mathcal{C} \to \mathcal{E}$ with $G_d(c) = (c, d)$.

$\mathcal{E}$ can be explicitly constructed in the usual syntactic way, but it is a bit painstaking to describe and I think maybe obscures the idea anyway. On the other hand, if you haven't seen this kind of thing much before it's not immediately clear that the above category exists. Is there a better way to describe $\mathcal{E}$ to make it clearer that it exists, or is perhaps belief in the existence of such things an atomic mental widget?

It seems to me the clearest thing to say is the above definition followed by a few examples of morphisms, but I would be happy if there were an elegant way of easily describing the syntactic construction which yields $\mathcal{E}$.

Improve this question
$\endgroup$
13
  • 3
    $\begingroup$ Isn't it precisely just the free strict monoidal category on $\mathcal C \times \mathcal D$ ? $\endgroup$
    – Adrien
    Nov 10, 2016 at 18:03
  • 3
    $\begingroup$ In that case, it's probably the free strict monoidal category on the "funny tensor product" $C\Box D$. $\endgroup$
    – Mike Shulman
    Nov 10, 2016 at 18:40
  • 2
    $\begingroup$ When I read "for each $c\in \mathcal C$", I assume you mean "naturally in $c$". Similarly, when I read "$F_c(d) = (c,d)$", I assume you mean such an equation to hold both at the object and morphism level. I take from your comment that you do not mean such things? $\endgroup$
    – Theo Johnson-Freyd
    Nov 11, 2016 at 2:39
  • 1
    $\begingroup$ The reason that you should mean such things is that otherwise you will write down constructions that are not well-behaved under equivalences of categories. $\endgroup$
    – Theo Johnson-Freyd
    Nov 11, 2016 at 2:39
  • 2
    $\begingroup$ I think your construction itself is not invariant under equivalence, because the funny tensor product is not. If $C$ and $D$ are the terminal category, then $C\Box D$ is also terminal; but if $C$ and $D$ are the walking isomorphism $g:0\cong 1$, then $C\Box D$ contains $(g,1)\circ (1,g)$ and $(1,g) \circ (g,1)$ which are parallel but not equal, hence is not equivalent to the terminal category. I wouldn't be as dogmatic as Theo about saying what you should do, but it's certainly good to at least be aware of when you violate equivalence-invariance. $\endgroup$
    – Mike Shulman
    Nov 11, 2016 at 22:39

1 Answer 1

Reset to default
1
$\begingroup$

If anyone is curious, I have come up with a definition which I think is satisfactory and non-evil. Weber characterizes $C \Box D$ as the pushout of

$$C \times D_0 \gets^{i_C \times \mathrm{id}} C_0 \times D_0 \to^{\mathrm{id} \times i_D} C_0 \times D$$ where $C_0, D_0$ are the discrete categories of $C$ and $D$ and $i_C, i_D$ are the inclusions. This is evil (or at least, potentially so) since taking the discrete category is evil. However, if we instead consider the pushout of $$C \times \mathrm{Core}(D) \gets^{i_C \times \mathrm{id}} \mathrm{Core}(C) \times \mathrm{Core}(D) \to^{\mathrm{id} \times i_D} \mathrm{Core}(C) \times D$$ then I believe the resulting category (call it $C \circledast D$) will satisfy $C \circledast D \simeq C' \circledast D$ when $C \simeq C'$ and similarly for $D$.

We can then take the free symmetric monoidal category on $C \circledast D$ to get what I want. I'm not sure, but I suspect this can be obtained by taking $C \Box D$ and modding out by $(\mathrm{id}, f) \circ (g, \mathrm{id}) = (g, \mathrm{id}) \circ (\mathrm{id}, f)$ when $f$ or $g$ is an iso. Do this definition or this construction make sense?

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .