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Stream: learning: questions

Topic: Set x Set is not monadic over Set

Martin Brandenburg (Apr 01 2026 at 22:19):

What is a good way to prove that there is no monadic functor Set x Set ---> Set? I am currently writing down a proof but I am not happy how it evolves, probably I miss something easy.

Kevin Carlson (Apr 01 2026 at 23:01):

Why isn't $(A,B)\mapsto A\times B$ such a functor? It preserves reflexive coequalizers, right, as in crude monadicity?

Jonas Frey (Apr 01 2026 at 23:11):

The product functor is not monadic since it's not conservative.

Aaron David Fairbanks (Apr 01 2026 at 23:12):

I think the category of algebras of that monad is equivalent to the full subcategory of $\mathbf{Set} \times \mathbf{Set}$ such that neither or both of the sets are empty.

Aaron David Fairbanks (Apr 01 2026 at 23:17):

It looks like $\mathbf{Set} \times \mathbf{Set}$ doesn't have a [[separator]], so it can't be monadic over $\mathbf{Set}$.

Jonas Frey (Apr 01 2026 at 23:19):

It has a separator, namely the pair $(1,0), (0,1)$.

Aaron David Fairbanks (Apr 01 2026 at 23:20):

I think that a category monadic over $\mathbf{Set}$ has a single object separator.

Jonas Frey (Apr 01 2026 at 23:20):

Ahh that makes sense! I was about to propose a more complicated proof using projective objects, but that's easier!

Jonas Frey (Apr 01 2026 at 23:28):

My argument would have been that the poset-reflection of the subcategory of projectives in a Set-monadic category can have at most two elements since every projective object is a retract of a free object. But in Set x Set, every object is projective, and the poset reflection of SetxSet has four elements.

Martin Brandenburg (Apr 01 2026 at 23:49):

Now this got me confused, I always thought that when a category has a separating set and coproducts, then it has a separating object. But I was wrong. It only works in special categories, for example those with zero morphisms.

Kevin Carlson (Apr 01 2026 at 23:53):

Jonas Frey said:

The product functor is not monadic since it's not conservative.

Oh, yeah, I think I've made that mistake before.

Martin Brandenburg (Apr 02 2026 at 00:25):

Ok since I now need to erase everything I was believing about generators ... I don't even know anymore if the category of simplicial sets has a generator :D (of course it has a generating set)

(generator = "single" separator, generating set = separating family)

Martin Brandenburg (Apr 02 2026 at 00:34):

ok Lemma: Let $S$ be a generating set in a category. Assume that between any two objects there is at least one morphism, and assume that the coproduct of the objects in $S$ exists. Then this coproduct is a generator.

In particular, $\coprod_n \Delta^n$ should be a generator in sSet.

Martin Brandenburg (Apr 02 2026 at 02:13):

Proof that $Set \times Set$ has no generator: Assume that $(A,B)$ is a generator. Of course, $A$ and $B$ cannot be both empty. Assume w.l.o.g. that $A$ is non-empty. Then there is no morphism $(A,B) \to (0,1)$, but there are two different morphisms $(0,1) \rightrightarrows (0,2)$.

Corollary: there is no monadic functor $Set \times Set \to Set$.

Martin Brandenburg (Apr 02 2026 at 02:14):

Can we classify the Grothendieck topoi that have a generator?

Morgan Rogers (he/him) (Apr 03 2026 at 10:07):

We can try! Based on Jonas' comment, we can observe that any generator must have support (the morphism to 1 must be epic). If its support (image of the terminal map) were some properly subterminal object $U$ then the generator couldn't distinguish the two pushout injections $1 \rightrightarrows 1+_U 1$.
The conclusion is that the topos must be 2-valued, or equivalent hyperconnected over Set: 0 and 1 are the only subterminal objects.

Morgan Rogers (he/him) (Apr 03 2026 at 10:12):

Conveniently, any non-initial object in a hyperconnected topos is well-supported. If the axiom of choice holds in our category of sets(!), the product of any generating set of objects will not be initial and all of the projections will be epimorphisms, so that object will be a generator. Two-valuedness is necessary and sufficient!