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Stream: learning: questions

Topic: When does F ~ F U F?

fosco (Jan 06 2021 at 14:49):

This is probably a very naive question, but I feel I am confused by something.

Assuming enough choice, an infinite set $A$ has the property that $A\cong A\sqcup A$ (the disjoint union of two copies of A). However, no such bijection can be natural, in the sense that there can't be a natural transformation, restricted to the subcategory of infinite sets, whose components are said isomorphisms $\alpha_A : A \cong A\sqcup A$. This for a very elementary reason, the identity functor preserves some (co)limits that the functor ${\sf Set} \to {\sf Set} : X \mapsto X\sqcup X$ does not preserve.

Now, does this obstruction mean that given a small category $C$, there can't be a natural isomorphism $F\cong F\sqcup F$ for $F : C \to Set$, but only some unnatural bijections $FA\cong FA\sqcup FA$ as soon as $FA$ is infinite?

Jacques Carette (Jan 06 2021 at 15:00):

I am confused too: while $A$ and $A \sqcup A$ are equi-inhabited, there is no bijection. Sure, you can use choice to pick a particular $f : A \rightarrow A\sqcup A$, and there's an obvious map $g : A\sqcup A \rightarrow A$, but I don't see any amount of choice that would let you get $f ; g : A\sqcup A \rightarrow A\sqcup A$ be the identity. The map $g$ forgets all the choices that $f$ made.

Nathanael Arkor (Jan 06 2021 at 15:13):

@Jacques Carette: you can replace the set $A$ with the cardinal $|A|$; the addition of two infinite cardinals is their maximum (assuming choice), so that $|A| + |A| = |A|$. The explicit bijections are then given by the definition of cardinality, and the bijections constructed by cardinal arithmetic.

Morgan Rogers (he/him) (Jan 06 2021 at 15:15):

Jacques Carette said:

I am confused too: while $A$ and $A \sqcup A$ are equi-inhabited, there is no bijection. Sure, you can use choice to pick a particular $f : A \rightarrow A\sqcup A$, and there's an obvious map $g : A\sqcup A \rightarrow A$, but I don't see any amount of choice that would let you get $f ; g : A \rightarrow A\sqcup A \rightarrow A \rightarrow A\sqcup A$ be the identity. The map $g$ forgets all the choices that $f$ made.

Of course the bijection isn't going to involve the canonical codiagonal map $A \sqcup A \to A$, since that map is not injective...

fosco (Jan 06 2021 at 15:19):

Nathanael Arkor said:

Jacques Carette: you can replace the set $A$ with the cardinal $|A|$; the addition of two infinite cardinals is their maximum (assuming choice), so that $|A| + |A| = |A|$. The explicit bijections are then given by the definition of cardinality, and the bijections constructed by cardinal arithmetic.

This is exactly what I meant, sorry for the misunderstanding :smile:

Nathanael Arkor (Jan 06 2021 at 15:23):

Now, does this obstruction mean that given a small category $C$, there can't be a natural isomorphism $F\cong F\sqcup F$, but only some unnatural bijections $FA\cong FA\sqcup FA$ as soon as $FA$ is infinite?

If you take the unique endofunctor $F : 1 \to 1$ on the terminal category (or the initial category), then $F \cong F \sqcup F$.

Nathanael Arkor (Jan 06 2021 at 15:26):

(Or is $F$ supposed to be a functor into $\mathrm{Set}$?)

Morgan Rogers (he/him) (Jan 06 2021 at 15:27):

I think the question was going to be asked more generally, but for the latter half to make sense it needs to be a setting where infinite objects are a meaningful concept.

fosco (Jan 06 2021 at 15:31):

whoops, a part is missing; $F : C \to {\sf Set}$

Jacques Carette (Jan 06 2021 at 15:33):

Ah, cardinals, that makes more sense. I did say I was confused (by the phrasing of the question).

Nathanael Arkor (Jan 06 2021 at 15:39):

If $C$ is the empty category, then there is a natural isomorphism between the two functors.

Nathanael Arkor (Jan 06 2021 at 15:41):

Which really only means you will at least need some nontriviality condition on $C$ or $F$.

Nathanael Arkor (Jan 06 2021 at 15:42):

(Same with any constant functor to an infinite set.)

fosco (Jan 06 2021 at 15:52):

@Nathanael Arkor exactly what I expected (also, if $C$ is discrete, or something? In that case, naturality in an empty condition on a family of maps)

Nathanael Arkor (Jan 06 2021 at 15:56):

(These conditions are all described by "$F$ is constant on connected components of $C$".)

fosco (Jan 06 2021 at 15:58):

Ah. Very nice.

Reid Barton (Jan 06 2021 at 16:21):

$FA$ can even be $FA = A \times S$ for some fixed infinite set $S$

fosco (Jan 06 2021 at 16:44):

hmmm

fosco (Jan 06 2021 at 16:44):

yes, true