Is it true that if you have a presentable category, there is a small set of \kappa-compact objects for any cardinal \kappa?

@IshanLevy Yes, the full subcategory of kappa-compact objects of an accessible category is essentially small. A reference is HTT.5.4.2.11.

Great, thanks a lot!

To nitpick, \kappa must be regular.

I'm sure you could cook up some weird counterexample for $\kappa = \aleph_\omega$, the first (infinite) regular cardinal.

I think it might be true even if kappa isn't regular. You can probably replace kappa with a sufficiently large regular cardinal

I'm not really an expert on set theory, but is false that for any cardinal, there is a regular cardinal larger than it?

No, that's true; maybe if you replace \kappa with cof(\kappa) it works. All $\kappa$-filtered colimits are $\mathsf{cof}(\kappa)$-directed. So I retract the assuredness of my previous statement :P I am now 50/50 on whether or not i think it's true for all \kappa

@ReubenStern As it happens I discussed with Marc Hoyois a couple of weeks ago. You can always replace $\kappa$ with $\kappa^+$, which is regular, that is if $\kappa$ is not regular every $\kappa$-filtered poset is $\kappa^+$-filtered

I have a very silly question. I read on wikipedia that Whitehead brackets vanish for H-spaces, but it is known that [id,id] generates the 6-th homotopy group of S^3. What is wrong ?

[id,id] in pi_5(S^3)

Where did you see this?

Beware of Wikipedia.

Link please @elidiot

@DenisNardin oh cool, good to know!

@elidiot I think you are confused about the Whitehead bracket notation and are thinking of the commutator map. If $X$ is a topological group then the commutator $c(x,y) = xyx^{-1} y^{-1}$ defines a map $c: H \wedge H \to H$. For $H = S^3$ James (IIRC) proved that this is the generator of $\pi_6 S^3$. As far as I know this construction has no relationship to the Whitehead bracket.

Ok thanks to you all, I was indeed confused. It seems that there can be a relation between the Whitehead bracket though, namely, this element in the \pi_6 seems to be the whitehead product [\alpha, \alpha] where \alpha generates \pi_4(HP^\infty), under the isomorphism \pi_7(HP^\infty) \simeq \pi_6(S^3), I just read that in Samelson Groups and spaces of loops.

That's a nice relation I didn't know about, thanks for the reference

The Samelson product in \Omega X is adjoint to the (generalised) Whitehead product in X.