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Stream: learning: questions

Topic: Creation of colimits by monadic functors

Aaron David Fairbanks (Jul 23 2025 at 02:14):

If $T$ is a monad on a category $C$, then the monadic forgetful functor $C^T \to C$ creates colimits that are preserved by $T$ and $T^2$. This is Proposition 4.3.2 in Borceux's Handbook of Categorical Algebra 2. Is there an example that shows preservation by $T$ is not sufficient?

Mike Shulman (Jul 23 2025 at 03:53):

This bit me the last time someone asked about it, because my first instinct is always to say "if $T$ preserves some colimits, then certainly so does $T^2$"! But I gather the point is that that's true for a particular shape of colimit that if $T$ preserves all such colimits then so does $T^2$, but it might happen that $T$ preserves a particular colimit diagram but $T^2$ doesn't. (That is, if $T$ preserves the colimit diagram $D$, then $TD$ is also a colimit diagram, but $T$ might not preserve the colimit diagram $TD$ so that $T^2D$ is not a colimit diagram.) So in order to find an example of the sort you want, you need a monad $T$ that preserves some particular colimit diagram but doesn't preserve all colimits of the same shape, and I would guess that's pretty tricky to come by.

John Baez (Jul 23 2025 at 08:28):

Very nice distinction! I'm sure our team here could find an example.

Mike Shulman (Jul 23 2025 at 16:35):

If I were going to try, the first place I would start looking is to take $C$ to be a poset, since colimits in posets and monads on posets are particularly simple to construct "by hand". Maybe there is some small concrete poset that can be drawn by hand on paper, and an idempotent inflationary endofunction (= monad) on it, such that there are two elements that have a join that's preserved by the endofunction but not by its square.

Kevin Carlson (Jul 23 2025 at 17:21):

But if the endofunction is idempotent, it’s equal to its square so they can’t do things differently! So posets won’t work…

Mike Shulman (Jul 23 2025 at 17:32):

headdesk

Mike Shulman (Jul 23 2025 at 17:33):

Okay, the second place I would look is... um, I'm not sure.

Mike Shulman (Jul 23 2025 at 18:26):

The trouble is I can't think of any good ways to prove that a functor preserves some particular colimit that don't also prove that it preserves all colimits of the same shape.

Amar Hadzihasanovic (Jul 23 2025 at 19:29):

I tried coming up with an example of a comonoid $A$ in a monoidal closed category with tensor $-\otimes-$ and internal hom $[-,-]$, as well as objects $X$, $Y$, such that $[A, X] + [A, Y]$ is isomorphic to $[A, X + Y]$ but $[A \otimes A, X] + [A \otimes A, Y]$ is not isomorphic to $[A \otimes A, X + Y]$, but I did not succeed.

Amar Hadzihasanovic (Jul 23 2025 at 19:32):

The category cannot be one where the coproduct is actually a biproduct, because otherwise it would be always preserved by $[A, -]$.

Amar Hadzihasanovic (Jul 23 2025 at 19:35):

The idea is to try to categorify something like a ring with elements $a$, $b$ and a number $k$ such that $(a + b)^k = a^k + b^k$ but $(a + b)^{k^2} \neq a^{k^2} + b^{k^2}$ (I haven't come up with a concrete example, sorry)...

Amar Hadzihasanovic (Jul 23 2025 at 19:40):

Of course if one could find this in a cartesian closed category it would further simplify things because then every object has a canonical comonoid structure.

Kevin Carlson (Jul 23 2025 at 20:07):

Screenshot 2025-07-23 at 1.06.46 PM.png
OK, here's a tiny category that, I think, supports a counterexample.

Kevin Carlson (Jul 23 2025 at 20:08):

The monad $T$ I have in mind is idempotent on $\{d,e,f,g\}$, being the reflection onto $\{d,f,g\}$. On $\{a,b,c\},$ $T$ just "rounds up" to $d,e,f$ respectively. Therefore $T$ is not idempotent but does satisfy $T^3=T^2.$

Kevin Carlson (Jul 23 2025 at 20:10):

With the further point that $e$ is meant to be a retract of $g,$ I think everything is now uniquely specified. The unit points northwest at $a,b,c,e$ and is the identity at $d,f,g$; the only nontrivial multiplication component is the retraction $g\to e$, which shows the unitality axiom for a monad; and the associativity axiom is vacuous since by the time you're in $T$ you're in the idempotent monad we started with on $\{d,e,f,g\}.$

Kevin Carlson (Jul 23 2025 at 20:14):

Now, I want to argue that $T$ preserves the coproduct $b=a+c$ but $T^2$ does not. The key question is why $g$ is not itself the coproduct, and the reason is that the cocone $d,f\to g$ factors through $e$, which means this cocone is fixed by $g$'s nontrivial idempotent, which disproves uniqueness in the coproduct property for $g.$

Kevin Carlson (Jul 23 2025 at 20:15):

$e$ doesn't have any parallel pairs of maps out, so it's easier to see that it actually is the coproduct, which should finish the proof.

Kevin Carlson (Jul 23 2025 at 20:16):

But wait, this isn't a counterexample to Aaron's actual claim, because $a,b,c$ don't support algebra structures! Maybe it can be spruced up to work...

Amar Hadzihasanovic (Jul 24 2025 at 08:03):

Amar Hadzihasanovic said:

The idea is to try to categorify something like a ring with elements $a$, $b$ and a number $k$ such that $(a + b)^k = a^k + b^k$ but $(a + b)^{k^2} \neq a^{k^2} + b^{k^2}$ (I haven't come up with a concrete example, sorry)...

The smallest example of this that I have found is $a = 1$, $b = 3$, and $k = 4$ in the ring $\mathbb{Z}/29\mathbb{Z}$.
Here one can check that

• $(a + b)^4 = 4^4$ and $a^4 + b^4 = 1 + 3^4$ are both $24\mod 29$,
• $(a + b)^{16} = 4^{16}$ is $16\mod 29$ but $a^{16} + b^{16} = 1 + 3^{16}$ is $21\mod 29$.

Perhaps someone has an idea how to categorify this?

Amar Hadzihasanovic (Jul 24 2025 at 08:07):

(For the exponents $k = 2, 3$ the vanishing of the cross-terms in $(a + b)^k$ always implies the vanishing of the cross-terms in $(a + b)^{k^2}$, so $k = 4$ is the smallest exponent for which this can work.)

Amar Hadzihasanovic (Jul 24 2025 at 08:51):

I guess I don't know about any monoidal closed category which has a family of objects $(n)$ indexed by integers mod m such that the internal hom works as exponentiation in $\mathbb{Z}/m\mathbb{Z}$, that is, $[(n), (k)]$ is isomorphic to $(k^n)$...