Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: Coproduct question

Mike Stay (May 26 2020 at 19:41):

Suppose I've got a 2-sorted Lawvere theory with

• two sorts $L, M$
• one function symbol $a:L\times M\to M$

Mike Stay (May 26 2020 at 19:45):

What does the coproduct $X_1 + X_2 = (L, M, a)$ of two models $X_1=(L_1, M_1, a_1)$ and $X_2=(L_2, M_2, a_2)$ look like? I thought it would have

• $L = L_1 + L_2$
• if $m\in M_1$ then $m \in M$
• if $m\in M_2$ then $m \in M$
• if $l \in L_1$ and $m \in M_1$ then $a(l,m) = a_1(l,m)$
• if $l \in L_2$ and $m \in M_2$ then $a(l,m) = a_2(l,m)$
• otherwise, adjoin a formal element $a(l,m)$ to $M$

That is, whenever we know how an element $l$ acts on $m,$ we use the result, but whenever we don't, we just add a formal place holder.

Call this proposal $Z=(L_Z, M_Z, a_Z).$

This is similar to how in the coproduct of monoids, whenever we juxtapose two elements from the same monoid, we can replace them with their product; but whenever we have elements from different monoids next to each other, we have a formal product of the two. The end result is that elements of the coproduct of two monoids have normal forms that are strings of alternating elements.

Mike Stay (May 26 2020 at 19:49):

A map $X \to X'$ of these gadgets is a pair $(\lambda:L\to L', \mu:M\to M')$ such that $\mu \circ a = a' \circ (\lambda \times \mu).$ There are obvious inclusion maps from $X_1$ and $X_2$ into $Z.$

Mike Stay (May 26 2020 at 19:50):

But now consider $Y=(L_Y, M_Y, a_Y)$ to be $Z$ with one more element $\bot \in M_Y$ such that for all $l \in L, a_Y(l, \bot) = \bot.$ There are the obvious inclusions from $X_1$ and $X_2$ into $Y.$

Mike Stay (May 26 2020 at 19:54):

If my proposal $Z$ really is the coproduct, then there should be a unique map from $Z$ to $Y$ making the triangles in the universal property diagram commute. But I can think of two! The first is the inclusion of $Z$ into $Y,$ where $\bot$ is not in the image. The second maps the formal elements $a_Z(l,m)$ to $\bot.$ In both cases, the maps are the identities on the images of $X_1$ and $X_2$, so the diagrams commute.

Mike Stay (May 26 2020 at 19:55):

I've been told that the category of models of any multisorted Lawvere theory is always complete and cocomplete, so the coproduct has to exist. But if it's not my proposal, I have no idea what it could be.

Christian Williams (May 26 2020 at 19:59):

Hm, yeah that's a very reasonable guess for the coproduct, and also a convincing reason why it cannot be.

Christian Williams (May 26 2020 at 20:00):

I've also seen that the category of models is complete and cocomplete. Limits can be computed pointwise, because they commute with products and the result will still be a product-preserving functor.

Christian Williams (May 26 2020 at 20:01):

Sifted colimits can also be computed pointwise, because these are the colimits which commute with products. But coproducts are not sifted.

Christian Williams (May 26 2020 at 20:02):

I've had some trouble finding a good reference for how to construct all colimits.

Christian Williams (May 26 2020 at 20:05):

Aha! I think it's actually not true, and that might be a typo on nLab and elsewhere.

Christian Williams (May 26 2020 at 20:05):

From Algebraic Theories, Adamek and Rosicky:

2.6 Example. Coproducts are not sifted colimits. In fact, for almost no algebraic theory T is Alg T closed under coproducts in Set T . Concrete example: if T ab is the theory of abelian groups (1.6), then binary coproducts in Alg T ab are products and in Set T ab they are disjoint union.

Mike Stay (May 26 2020 at 20:07):

OK, whew! I thought I was going crazy!

Christian Williams (May 26 2020 at 20:08):

You are not!

Mike Stay (May 26 2020 at 20:08):

So for this theory $T,$ Alg($T$) doesn't have coproducts.

Mike Stay (May 26 2020 at 20:08):

Thanks!

Christian Williams (May 26 2020 at 20:09):

Probably, yeah. I don't know what characterizes the categories of models which have coproducts. The category of groups has coproducts, why is it special?

Christian Williams (May 26 2020 at 20:09):

But yeah, we should probably fix that on the nLab.

Jem (May 26 2020 at 20:12):

Why is it a homomorphism? Don't we have to take the formal elements of Z to the formal elements on Y?

Let $X_1 = (L_1, M_1, a_1) = (\mathbf 0, \mathbf 1, \pi_2)$ and $X_2 = ( L_2, M_2, a_2) = (\mathbf 1, \mathbf 0, \pi_2)$. Then $Z$ looks like the naturals - $L_Z$ is $\mathbf 1$, $M_Z$ is (up to iso) $\mathbb N$, and $a(-,n) = n+1$. The inclusion from $M_1$ selects $0$.

Then the homomorphism you're claiming would take $0$ to $0_Y$ and take every positive integer $n$ to $\bot$. But then $\bot = \mu(1) = \mu(a_Y(u,0)) = a_Z(\lambda(u),\mu(0)) = a_Z(u,0) = 1$, which is clearly false.

Mike Stay (May 26 2020 at 20:22):

Sorry, what are $\mathbf {0, 1}?$ The zero- and one-element sets?

Jem (May 26 2020 at 20:22):

Yep.

Nathanael Arkor (May 26 2020 at 20:23):

Christian Williams said:

From Algebraic Theories, Adamek and Rosicky:

2.6 Example. Coproducts are not sifted colimits. In fact, for almost no algebraic theory T is Alg T closed under coproducts in Set T . Concrete example: if T ab is the theory of abelian groups (1.6), then binary coproducts in Alg T ab are products and in Set T ab they are disjoint union.

This says simply that you cannot compute binary coproducts directly from the coproducts in the functor category. The binary coproducts do exist, but they're computed differently (i.e. in this case by taking products).

Nathanael Arkor (May 26 2020 at 20:24):

The category of models for a Lawvere theory does have limits and colimits, as proven in Corollary 5.2.2 of http://web.science.mq.edu.au/~street/MitchB.pdf.

Nathanael Arkor (May 26 2020 at 20:25):

(Note that that paper uses the original definition: taking categories with coproducts, not products, so you'll need to insert an op to get the result you expect.)

Jem (May 26 2020 at 20:26):

That includes multi-sorted finite product theories too, not just single sorted ones, right?

Nathanael Arkor (May 26 2020 at 20:27):

Yes, though I'm not sure of a reference for that precise result.

Nathanael Arkor (May 26 2020 at 20:27):

Generally the facts about one-sorted algebraic theories carry over to S-sorted algebraic theories.

Jens Hemelaer (May 26 2020 at 20:27):

In this case, the original guess for the coproduct is correct, right?

Mike Stay (May 26 2020 at 20:29):

Jem said:

Then the homomorphism you're claiming would take $0$ to $0_Y$ and take every positive integer $n$ to $\bot$. But then $\bot = \mu(1) = \mu(a_Y(u,0)) = a_Z(\lambda(u),\mu(0)) = a_Z(u,0) = 1$, which is clearly false.

Oops! Yes, that demolishes my argument. Thanks! (Though I think you swapped $Y$ and $Z$.)

Christian Williams (May 26 2020 at 20:31):

Ah, sorry about that. That's good news then.

John Baez (May 26 2020 at 20:33):

Christian Williams said:

From Algebraic Theories, Adamek and Rosicky:

2.6 Example. Coproducts are not sifted colimits. In fact, for almost no algebraic theory $T$ is $Alg T$ closed under coproducts in $Set^T$ . Concrete example: if $T_{ab}$ is the theory of abelian groups (1.6), then binary coproducts in $Alg T_{ab}$ are products and in $Set^{T_{ab}}$ they are disjoint union.

This doesn't say $Alg T$ doesn't have coproducts - I know it does. It's just saying these coproducts rarely arise from coproducts in the functor category $Set^T$! And that's kind of obvious, as in their abelian group example.

When you take a coproduct of abelian groups, you "get a bunch of new elements" that you wouldn't if you were taking the coproduct in $Set^T$. That's all they're saying here.

John Baez (May 26 2020 at 20:38):

Okay, Nathaniel said everything I said. This increases the probability that it's true.

Todd Trimble proved for me that for any multisorted Lawvere theory, its category of models has colimits - it's Theorem 3.1 here. Moreover he proved it for models, not just in Set, but in any "cartesian monoidally cocomplete" category, or CMC. This is a cartesian category with colimits such that the product distributes over colimits.

Mike Stay (May 26 2020 at 20:47):

Jens Hemelaer said:

In this case, the original guess for the coproduct is correct, right?

I think so.

John Baez (May 26 2020 at 20:50):

For what it's worth it looks right to me. In general there's an inductive construction of coproducts of algebras of a Lawvere theory, which is quite tiresome to state. You're lucky because your operation obeys no equations, and it never generates new elements of L, just M.