0
$\begingroup$

Let $X$ be a topological space. Assume that $X$ admits a finite decomposition of the form $X=\bigsqcup\limits_{i=1}^n V_i$ where each $V_i$ is homeomorphic (in the subspace topology of $X$) to an open cell of dimension $d_i$. By convention, the open cell of dimension $0$ is a point. We are not assuming that the disjoint union is a stratification of $X$, i.e., we allow the closure of $V_i$ to intersect cells $V_j$'s of dimension larger or equal to $d_i$.

Q: Does it necessarily follow that $H_k(X,\mathbb{Z})$ is a finitely generated $\mathbb{Z}$-module?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

This can fail badly even if the closure of each cell only intersects lower-dimensional cells. For instance, take an open disk together with some subset of the boundary that has infinitely many connected components, and glue it to an open interval by identifying the boundary set with a homeomorphic subset of the interval. The resulting space will not have finitely generated $H_1$ (in fact, I think its $H_1$ is uncountable).

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks Eric for the instructive exemple. I can now better appreciate the constraints that come with the definition of a (finite) CW complex where the gluing does not affect the open cells but only their boundaries. $\endgroup$
    – Hugo Chapdelaine
    Commented Jan 11, 2015 at 23:27
  • $\begingroup$ Eric, now that I'm thinking about your construction, why do you get a finite disjoint union of open cells? $\endgroup$
    – Hugo Chapdelaine
    Commented Jan 11, 2015 at 23:55
  • $\begingroup$ It's a disjoint union of the open disk and the open interval. $\endgroup$
    – Eric Wofsey
    Commented Jan 12, 2015 at 4:09
  • $\begingroup$ Ok, I got it. As a special case of your construction, you may take the open ends sine curve which you glue to the x axis along the intersection points. $\endgroup$
    – Hugo Chapdelaine
    Commented Jan 12, 2015 at 11:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .