In this answer, mme shows that for any compact Lie group $G$, there is a model for its classifying space $BG$ which is the direct limit of closed manifolds. If $G$ is discrete (i.e. $\dim G = 0$), then a model for $BG$ is also a model for $K(G, 1)$ since $\pi_i(BG) \cong \pi_{i-1}(G)$.

Let $G$ be a countable abelian group and $n > 1$. Does there exist a model for $K(G, n)$ which is a direct limit of closed manifolds?

Recall, a space $X$ is a model for $K(G, n)$ if $\pi_n(X) \cong G$ and $\pi_i(X) = 0$ for $i \neq n$.

There is such a model for $K(\mathbb{Z}, 2)$, namely $\mathbb{CP}^{\infty}$ which arises as the direct limit of finite-dimensional complex projective spaces $\{\mathbb{CP}^m\}_{m\geq 1}$. This can also be viewed in the context of the linked answer as a model for $BU(1)$ is a model for $K(\mathbb{Z}, 2)$.

The above question may be too difficult to answer in general. I would be interested to know what happens for the next simplest cases, namely $K(\mathbb{Z}/m\mathbb{Z}, 2)$ and $K(\mathbb{Z}, 3)$.

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    $\begingroup$ I believe that any homotopy type that is represented by a countable CW complex is also represented by an increasing union of closed manifolds. $\endgroup$ Commented May 22 at 1:29
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    $\begingroup$ @Tom Goodwillie: just out of curiosity, how would you represent a bouquet of circles? $\endgroup$
    – Nandor
    Commented May 22 at 8:55
  • $\begingroup$ @Nandor: one way would be to take a connect-sum of $k$ copies of $S^1 \times S^n$, and then take the natural inclusions as $n$ varies. $\endgroup$ Commented May 22 at 16:43
  • $\begingroup$ @Ryan Budney: is the connected sum a manifold? $\endgroup$
    – Nandor
    Commented May 22 at 20:43
  • $\begingroup$ @Nandor: yes it is. I don't think I know any category other than connected, oriented manifolds where it is defined. $\endgroup$ Commented May 23 at 3:07

1 Answer 1


Any homotopy type that is represented by a countable CW complex is also represented by an increasing union of closed manifolds:

First consider a finite CW complex $X$. Let $M\sim X$ be a compact manifold (with boundary), let $M_n$ be $M\times D^n$ and use the inclusions $$\dots \subset \partial M_n\subset M_n\subset \partial M_{n+1}\subset M_{n+1}\subset \dots $$ where we are viewing $\partial M_{n+1}$ as the double of $M_n$. The union as $n\to\infty$ is both homotopy equivalent to $X$ and an increasing union of closed manifolds.

For a countably infinite complex, do something similar: Write it as an increasing sequence of finite complexes $X(k)$. For each $k$ choose a compact manifold $M(k)\sim X(k)$, and arrange for $M(k)$ to be embedded in $M(k+1)$ (by a map in the homotopy class of the inclusion $X_k\to X_{k+1}$). We can arrange for $M(k)$ to be in the boundary of $M(k+1)$. Then $X$ is equivalent to the union of $$ \dots \subset M(k)\subset M(k+1)\subset \dots, $$ which is also the union of the closed manifolds $$ \dots \subset \partial M(k)\subset \partial M(k+1)\subset \dots, $$

  • $\begingroup$ Given a finite CW complex, I know you can embed it into Euclidean space of large enough dimension, and then take a thickening to obtain a manifold with boundary with the same homotopy type. From this perspective, I can see how we can arrange for $M(k)$ to be embedded in $M(k+1)$, but I don't see how we can arrange for $M(k)$ to be in the boundary of $M(k+1)$. Can this be seen from this thickening idea or should I be constructing these manifolds differently? $\endgroup$ Commented May 22 at 13:16
  • $\begingroup$ @MichaelAlbanese $M(k+1)=M(k+1)\times 0$ is contained in the boundary of $M(k+1)\times [0,1]$, so replace $M(k+1)$ by $M(k+1)\times [0,1]$. $\endgroup$ Commented May 22 at 16:11
  • $\begingroup$ Thanks. Is it clear that there is a countable CW complex which is a model for $K(G, n)$? I guess it follows from here, but I was hoping for something more direct. My idea was to take a presentation for $G$, build the analogue of a presentation complex using $n$ and $(n+1)$-cells, then kill higher homotopy groups by attaching more cells. However, its not clear to me if this will only require countably many cells. It does if $G$ is finitely presented (the first CW complex will be finite and hence have finitely generated homotopy groups by Serre). $\endgroup$ Commented May 22 at 20:39
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    $\begingroup$ Here's a simplicial construction. Let $S^n$ be a simplicial model for a based $n$-sphere, such as $\Delta[n]/\partial\Delta^n$. Let $S^n[G]$ be the simplicial abelian group obtained from $S^n$ by applying the functor from based sets to abelian groups (left adjoint to the forgetful functor). By Dold-Kan, the homotopy groups of (the realization of) $S^n[G]$ are the reduced homology groups of $S^n$ with coefficients in $G$; this CW complex is a $K(G,n)$. If $G$ is countable, then in each simplicial degree this simplicial abelian group is countable, so that the CW complex has countably many cells. $\endgroup$ Commented May 22 at 21:16

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