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Suppose that I have an arbitrary regular CW complex. By associating a topological space to each vertex of the CW complex, I can have a diagram of topological spaces, denoted by $D$, over the CW complex. The definition of “diagram of spaces” goes as follows: The following data constitute what is called a diagram of topological spaces $D$ over a cell complex $A$:

  • For each vertex $v$ of $A$ we have a topological space $D(v)$
  • For each edge $(v\rightarrow w)$ we have a continuous map $D(v\rightarrow w): D(v)\rightarrow D(w)$. Additionally, we require that these maps commute over each triangle in A.

Given this diagram, we can take the colimit to get the “total space”.

Also, a manifold can be considered as a colimit of its atlas. Or equivalently, an atlas is essentially a way of viewing a manifold as a colimit of Euclidean balls. It seems to me that the constructions of manifold and “colimit of diagram of spaces” have some similarities. So my first question is, can a “diagram of spaces over a CW complex” be a manifold? If so, under what conditions?

In addition, I also read about the notion of “homotopy colimit”. According to the “Homotopy Lemma”, when all the spaces D(v) in the diagram are contractible, the homotopy colimit of the diagram is homotopy equivalent to the base complex. Since any Euclidean space is contractible, if all the spaces D(v) in the diagram are Euclidean, can I construct a manifold with a “homotopy colimit”? (If yes, the manifold should be homotopy equivalent to the base CW complex by the “homotopy lemma”)

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  • $\begingroup$ I think it would be helpful if you could give some references for some of your statements, in particular the homotopy lemma. I think there might be a mistake in the way you state it, since the diagram you associate to a CW-complex only depends on its 1-skeleton, it should not know about the whole homotopy type - maybe I also understand something wrong. However, your question about the relation to constructing manifolds via gluing together charts is very interesting indeed, I will need to think about that. $\endgroup$
    – Markus Zetto
    Commented Apr 15, 2022 at 13:49
  • $\begingroup$ @MarkusZetto the homotopy lemma can be found in the book “Combinatorial Algebraic Topology” by Dmitry Kozlov. I couldn’t find a pdf online. The lemma corresponds to Theorem 15.25 in the book. $\endgroup$
    – chriswest
    Commented Apr 15, 2022 at 14:02

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So my first question is, can a “diagram of spaces over a CW complex” be a manifold? If so, under what conditions?

Any smooth manifold is homotopy equivalent to the homotopy colimit of a diagram of contractible spaces indexed by a 1-category. More precisely, take any manifold $M$, take its good cover $\{U_i\}_{i∈I}$, and consider the diagram indexed by finite collections $K$ of indices in $I$ such that the intersection $⋂_{k∈K}U_k$ is nonempty (and hence contractible). Send such an index $K$ to the intersection $⋂_{k∈K}U_k$. Morphisms $K→K'$ are given by reverse inclusions $K⊃K'$. The resulting indexing category corresponds to the CW-complex given by the first barycentric subdivision of the nerve of $\{U_i\}_{i∈I}$.

Since any Euclidean space is contractible, if all the spaces D(v) in the diagram are Euclidean, can I construct a manifold with a “homotopy colimit”? (If yes, the manifold should be homotopy equivalent to the base CW complex by the “homotopy lemma”)

Yes, see the construction above. This result is classical and is commonly known as the nerve theorem. The hyperlinked article has a reasonably comprehensive list of sources, including original references.

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  • $\begingroup$ That is helpful. But I think my question is a little bit different (or may be I still haven’t get the point). What I have now is a diagram of contractible spaces and its homotopy colimit, does that mean it is automatically homotopy equivalent to a smooth manifold? You mention that any smooth manifold is homotopy equivalent to the homotopy colimit of a diagram of contractible spaces indexed by a 1-category, my question is actually whether the inverse of this statement holds. $\endgroup$
    – chriswest
    Commented Apr 16, 2022 at 0:49
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    $\begingroup$ @chriswest: Any homotopy type can be obtained as the homotopy colimit of a diagram of contractible spaces, indexed by an ordinary category. Thus, the answer to your question is negative because not every homotopy type is homotopy equivalent to a manifold. $\endgroup$
    – Dmitri Pavlov
    Commented Apr 16, 2022 at 6:03
  • $\begingroup$ Ok. Then if what I really want to obtain is a manifold, is there any way I can do this from homotopy colimit? For example, under what conditions the homotopy colimit can indeed be a manifold? $\endgroup$
    – chriswest
    Commented Apr 16, 2022 at 7:18
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    $\begingroup$ @chriswest: The homotopy colimit of a diagram of contractible spaces is always weakly equivalent to the nerve of the indexing category. Your question is thus equivalent to asking under what conditions the nerve of a category is homotopy equivalent to a manifold. As far as I am aware, no specific criteria are known for nerves, other than the generic criteria for arbitrary homotopy types. See this answer for more information: mathoverflow.net/questions/278901/… $\endgroup$
    – Dmitri Pavlov
    Commented Apr 16, 2022 at 16:03
  • $\begingroup$ Thanks! I got it, for this question. But that really gets me back to where it all begins...Since, ultimately, I want to get a smooth manifold to do calculus. However, the topology of a network, which can be modeled as a directed graph, or an acyclic category, or simplicial complex often does not naturally satisfy the conditions required for it to be a manifold (for example, the “link of every vertex being equivalent to a sphere” condition for simplicial complexes). So is there any other way to construct a smooth manifold from an arbitrary topology and the spaces associated with the vertices? $\endgroup$
    – chriswest
    Commented Apr 17, 2022 at 2:51

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