2
$\begingroup$

I asked this question on Math Stack Exchange and did not receive any answers or comments.

Suppose $M$ and $N$ are monoidal categories and let $M\times{N}$ denote the associated product category. $M\times{N}$ comes equipped with two natural projection functors $\pi_{M}:M\times{N}\rightarrow{M}$ and $\pi_{N}:M\times{N}\rightarrow{N}$. I am interested in the pushout of the diagram of $\pi_{M}$ along $\pi_{N}$. A pushout of the aforementioned diagram is a monoidal category $P$ along with two monoidal Functors $\phi_{M}:M\rightarrow{P}$ and $\phi_{N}:N\rightarrow{P}$ such that the diagram commutes. I am familiar with push-outs in set theory but I'm not sure how to go about finding the push-out of one monoidal functor along another. Any elucidation of what monoidal category would be the push-out of this diagram would be very helpful.

$\endgroup$
2

1 Answer 1

3
$\begingroup$

The answer is that it depends (immensely) on what you mean by "category of monoidal categories".

Let me start with a general statement:

Suppose $C$ is a pointed category.

Suppose further that $C$ has finite products, and let $M,N\in C$; and $P$ with maps $p: M\to P, q: N\to P$ such that $p\circ \pi_M = q\circ \pi_N$.

Compose with the canonical map $i_M : M\simeq M\times * \to M\times N$ to get $p = q\circ 0 = 0$, where $0 : M\to N$ is the $0$ morphism, i.e. the unique morphism $M\to *\to N$.

Similarly, $q = 0$.

It follows that the pushout is trivial.

Now, whether or not the category of monoidal categories is pointed sort of depends on your definition of the latter; you have to specify a few things:

1- are you really considering the $1$-category of monoidal categories or some higher version of it where you include monoidal natural isomorphims (or even monoidal natural transformations ? - although in this case I suppose you'll have to specify what kind of "2-pushout" you mean);

2- are you considering lax monoidal functors or strong monoidal functors ? In the latter case, and if you're really interested in the $1$-category, does "strong" require the unit to be mapped to the unit strictly, or do you only require an isomorphism $1\to f(1)$ (as part of the data) ?

To see how the answers might differ:

  • if you're considering the $1$-category of monoidal categories with strictly unit-preserving strong monoidal functors, then it is pointed and the above argument applies, and shows that the pushout is the trivial monoidal category.

  • If you're considering "not-necessarily-strictly unit preserving strong monoidal functors" (but still as a $1$-category), then it is no longer pointed and you have to adapt the argument. It is easy to see that $p,q$ are constant, by the same argument, but they must be mapped to the same thing ($p\circ\pi_M(1_M,1_N) = p(1_M)$).

But now it's not too hard to get from there that a pushout simply does not exist: if it existed, it would have a distinguished object ($p(1_M)=q(1_N)$) which isn't the unit, and $p,q$ are constant at that object. It follows that there can in general be several functors out of $P$ with the appropriate value on $p(1_M)=q(1_N)$, simply by altering the value of the unit or something. So in this second case, there is no pushout. This is because we've used a notion of morphisms that is sensible to isomorphisms within the categories, but our $1$-category isn't.

I'll leave the "lax monoidal functors" case as an exercise to work out.

  • Now what happens if we change and go to the $(2,1)$-category of monoidal categories ? Well, if we take strong monoidal functors as our morphisms, and monoidal natural isomorphisms as our $2$-morphisms, then this is exactly the category of $E_1$-algebras in the $(2,1)$-category of categories, and in particular it is pointed (in the $(2,1)$-categorical sense).

But does the argument above work for "higher" categories ? Not quite as written, we've only proved that $p,q$ are trivial but there might be some higher homotopical stuff hidden. And in fact, there is: in the $(2,1)$-category of monoidal categories, you don't impose $p\circ\pi_M = q\circ\pi_N$, but you add the data of an equivalence $p\circ\pi_M\simeq q\circ\pi_N$. If they're both trivial, this is imposing the data of a self equivalence of the trivial functor $M\times N\to P$, i.e. a monoidal morphism $M\times N\to Aut(1_P)$, i.e. a monoid map $\pi_0(M)\times\pi_0(N)\to Aut(1_P)$, where by $\pi_0$ I mean here something stronger than usual, namely the quotient of the object set by the equivalence relation generated by "there exists an arrow $x\to y$".

By the Eckmann-Hilton argument, $Aut(1_P)$ is commutative, so this factors through the abelianization $(\pi_0(M)\times\pi_0(N))^{ab}\cong \pi_0(M)^{ab}\times\pi_0(N)^{ab}$, and in fact the group-completion thereof.

So the homotopy pushout should have a single object (the unit), but as automorphisms of the unit, it has $\pi_0(M)^{ab,grp}\times \pi_0(N)^{ab,grp}$ (this does have a monoidal structure). Let me call this group $A$ and the associated monoidal category $BA$. Then the structure maps are given by the trivial morphism $M\to BA$, the trivial morphism $N\to BA$, and the equivalence between $M\times N\to M\to BA$ and $M\times N\to N\to BA$ is given, on the object $(m,n)$, by the corresponding class of $(m,n)$ in $A = Aut(1_{BA})$.

The details are not super important, but what this discussion shows is that the specifics of what you mean by "category of monoidal categories" will impact the answer a lot !

$\endgroup$
9
  • $\begingroup$ What is a "strictly unit-preserving strong monoidal functor"? $\endgroup$ Jun 26, 2021 at 15:51
  • $\begingroup$ @MikeShulman : a strong monoidal functor such that the structural morphism $1\to f(1)$ is the identity $\endgroup$ Jun 26, 2021 at 15:58
  • $\begingroup$ (which,arguably, is a weird class of functors to consider; but then again so is the notion of a pushout in the $1$-category of (monoidal) categories) $\endgroup$ Jun 26, 2021 at 16:05
  • $\begingroup$ Are you certain that the 1-category of such is pointed? It's not obvious to me that every monoidal category admits such a functor from the terminal monoidal category. $\endgroup$ Jun 26, 2021 at 19:14
  • 1
    $\begingroup$ Of course you're right. I'm not sure what mixed me up. In general, of course, it's error-prone to make 1-categorical statements about categories like this, but it seems that in this case you're right. $\endgroup$ Jun 26, 2021 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.