@D.ZackGarza Yes, that's what it means. Having the weak homotopy type of a CW-complex is a non-assumption, but having the homotopy type of a CW-complex is quite strong (although it is a condition satisfied by most "reasonable" spaces like smooth manifolds or complex varieties)
Typically the easiest way to prove that something has the homotopy type of a CW complex is to embed it in R^N somehow and find a tubular neighborhood (since all open subsets of R^n are homotopy equivalent to a CW complex)
@DenisNardin i'm recalling that a while back you told me something which i'm now thinking is not true. (so i'm probably misremembering.) namely, for a finite group G, the H-fixedpoints of the G-category of G-spaces carries a residual action of Weyl(H), and i recall you saying that that action is trivial when G is abelian. is that correct? i've convinced myself that already for the G-category of G-sets, the action is always nontrivial (whether or not G is abelian).
@DenisNardin WRT a space having the weak homotopy type of a CW complex, does this just come from the existence of its Postnikov tower? And do you happen to know of any references that mention any necessary properties on a space for it to admit a strong equivalence?
(Or maybe even other sufficient properties. E.g. I was looking spaces admitting Morse functions, where its gradient flow produces a CW decomposition, although I haven't sorted out if this is a strong/weak equivalence yet.)
@D.ZackGarza No, it's just an easy exercise to prove: if $X$ is your space, first you pick a point on each path connected component and get a map $Y^0→X$ from a discrete set which is a bijection on π_0, then you choose for each connected component a set of generators of π_1 and use them to attach loops to Y^0 getting Y^0' (which is a disjoint union of boquets of S¹'s), but then you need to add the relations and you add S² etc... It's the usual proof and it just works without any assumptions
may I ask @DenisNardin If a user changes their screen name here, is there a way to find the history of names? Or at least the most recent name before the change?
@DenisNardin Ah right! So this gives you a space $X'$ with $\pi_k(X') \cong \pi_k(X)$ for every $k$. Probably minor, but any idea how the actual equivalence arises out of that? IIRC you end up needing a map $f: X \to X'$ that induces all of those isomorphisms.
By the way, the Morse function decomposition is actually a strong homotopy equivalence, although you can prove a stronger statement by more elementary means (i.e. that every smooth manifold has a PL-structure, of course every PL manifold has a CW decomposition)
@DenisNardin thanks, i appreciate it. the argument is just the following. i may be botching the notation, but let me write $Fin_G$ for the ordinary category of finite G-sets and $\underline{Fin_G}$ for the G-category of finite G-sets. then, we have $(\underline{Fin_G})^H := (Fin_G)_{/(G/H)}$, and the action of $W(H) = Aut(G/H)$ is via postcomposition with automorphisms of $G/H$.
now, the general claims is that for any category $C$ and any object $x \in C$, the $Aut(x)$-action on $C_{/x}$ by postcomposition is always nontrivial (unless $Aut(x)$ is trivial). namely, a trivialization of the action of $x \xrightarrow{a} x$ gives a trivialization of $a$ itself (i.e. a homotopy $a \simeq id_x$). [i haven't made the argument homotopy-coherent yet, but surely this is true "in families" as well.]
@AaronMazel-Gee Hrmm... I thought that the action you describe was equivalent to a trivial one when G is abelian. But maybe I'm mistaken. I'll try to think about it tomorrow though, it's too late here now for me to think clearly :)
actually, i think what i said is false. (for instance, if C=BG then C_{/x}=pt.) but in any case, i'd love to see a proof of whatever's true, one way or the other.
