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Does there exist a manifold with finitely generated homology groups that is not homotopy equivalent to a compact manifold with boundary?

I am also interested in several variations of this question. Does it exist..

  1. ..a smooth manifold without boundary
  2. .. a smooth manifold (possibly with boundary)
  3. ..a manifold without boundary
  4. .. a manifold (possibly with boundary)

is not homotopy equivalent to

A)..a compact manifold without boundary

B)..a compact manifold (possibly with boundary)

The minimal example that I know for all type A questions of is four-dimensional: the configuration space of two-element subsets of the plane. Its first homology group is isomorphic to Z (abelinization of the braid group), and the third cohomology group is trivial (proved by Vasiliev see "Topology of complements to discriminants" § Cohomology of braid groups with constant coefficients, Proposition 1), which contradicts Poincare duality.

P.S. I wrote earlier if the finitely generated of the homology groups implies the finitely generated of the fundamental group (noticed in the comments that this is not true), then for 3-manifolds, by Scott's compact kernel theorem, there are no such manifolds: a three-dimensional manifold admits a smooth structure, then it is triangulable, hence homeomorphic to a CW-complex, and CW-complexes with isomorphic homotopy groups are homotopy equivalent by Whithead's theorem.

Improve this question
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    $\begingroup$ Sure, take an infinite connected sum of integer homology 3-spheres which are not $S^3$. The result will have infinitely generated fundamental group but trivial homology. $\endgroup$
    – mme
    Commented Oct 14, 2021 at 11:21
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    $\begingroup$ So you need to add the assumption "$\pi_1$ is finitely presented". After this, there are two more obstructions. First, you could find a finite-dimensional $\pi_1$-module $A$ so that the local system homology $H_*(X; A)$ is infinite-dimensional (impossible if $X$ is equivalent to a compact manifold). If these are always finite-dimensional then iirc $X$ is a retract of a finite CW complex and there is one final obstruction: the Wall finiteness obstruction. $\endgroup$
    – mme
    Commented Oct 14, 2021 at 12:07
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    $\begingroup$ According to Wall there exist spaces which are dominated by finite CW complexes but not homotopy equivalent to a finite complex. Such a space necessarily has the homotopy type of a countable, finite-dimensional complex with finitely-presented fundamental group and homology groups. Take one such space $X$ and find a homotopy equivalent locally-finite, finite-dimensional simplicial complex. This simplicial complex embeds in some Euclidean space, and a regular neighbourhood $M$ of it here is the manifold you are looking for. Since $M\simeq X$, it has all the requried (non-)finiteness conditions. $\endgroup$
    – Tyrone
    Commented Oct 25, 2021 at 12:21

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