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A pointed, $n$-connected space $X$ admits a CW structure using just $n$-connected spheres, so $X$ is an iterated homotopy colimit of $n$-connected spheres (i.e. you start with spheres and keep taking colimits, and eventually get whatever you want). On the other hand, if we take a pointed simplicial set $X_\bullet$ representing a given homotopy type $X$, the identity $X \simeq |X_\bullet|$ expresses $X$ as a bona fide, 1-step homotopy colimit of wedge powers of $S^0$ (i.e. it gives a single diagram, whose values are wedge sums of $S^0$'s, whose homotopy colimit is $X$).

Question: Let $X$ be an $n$-connected pointed space. Then is $X$ a homotopy colimit of wedges of $n$-connected spheres?

I think it's reasonable to hope the answer is yes -- possibly after assuming everything is simply-connected.

Notes:

  • Any homotopy colimit can be consolidated into a geometric realization at the cost of introducing some more wedges. So it's equivalent to ask whether $X$ is the geometric realization of a simplicial object whose values are wedges of $n$-connected spheres.

  • One approach I've considered is to take an $n$-reduced simplicial model of $X$, and maybe use this to express $X$ as a colimit of $\Delta[m] / sk_n \Delta[m]$'s (these skeleta are always wedges of spheres). But I'm not sure how to ensure that the resulting geometric realization is really a homotopy colimit.

  • Clearly if $X$ is an $n$-fold suspension space the answer is yes.

  • I'd be happy to know the answer "up to retract" -- i.e. to know whether every $n$-connected $X$ is a retract of a homotopy colimit of $n$-connected spheres.

  • I'm also happy to assume that $X$ is finite.

  • Everything here is pointed -- all maps / homotopies should be pointed, and all homotopy colimits are meant in the pointed sense.

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    $\begingroup$ Note that by iterated hocolim you can build any $n$-connected space out of copies of $S^{n+1}$ (because a point is an empty hocolim and suspension of $X$ is a hocolim of $X$ and points). So you might even ask whether every $n$-connected space is a hocolim of wedges of $S^{n+1}$. This is the case for $n=-1$, as you observe, and I believe it is also the case for $n=0$ by letting $X$ be $BG$ for a simplicial group $G$ that is levelwise free: $B$ commutes with hocolim and $B$ of a free group is a wedge of circles. $\endgroup$
    – Tom Goodwillie
    Commented Aug 26, 2021 at 0:06
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    $\begingroup$ There is a canonical map $\text{Bar}(S^n,\Omega^n\Sigma^n,\Omega^nX)\to X$ which is a weak equivalence if $X$ is $(n-1)$-connected. This exhibits $X$ as the geometric realisation of a simplicial space in which each layer has the form $S^n\wedge T$ for some $T$, but $T$ is not discrete so this does not do exactly what you ask for. Perhaps you can replace $T$ by its singular complex and then diagonalise? $\endgroup$
    – Neil Strickland
    Commented Aug 26, 2021 at 11:08
  • $\begingroup$ @NeilStrickland That sounds perfect -- I think the diagonalization should work. And the convergence of this bar complex to $X$ is related to the May recognition theorem for $n$-fold loop spaces, right? $\endgroup$
    – Tim Campion
    Commented Aug 26, 2021 at 11:13
  • $\begingroup$ @NeilStrickland Could you clarify how $\Omega^n \Sigma^n$ acts on $S^n$? I thought this requires $S^n$ to be framed. $\endgroup$
    – Connor Malin
    Commented Aug 26, 2021 at 21:37
  • $\begingroup$ @TimCampion I think the result Neil is using is called "Nonabelian Poincare duality". $\endgroup$
    – Connor Malin
    Commented Aug 26, 2021 at 21:40

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