I am trying to read and understand the paper:



And I am having trouble with a statement. First of all, definitions:

First remark The definition of $k$-simplex should be with $t_i \in (0,1]$.

Second remark The above definition differs from the common one.

Now we continue and we introduce my problem:

Remark It is clear it is not an homotopic invariant.


  • Is there another easier (which needs less machinery) way to prove the topological invariance of this Euler Characteristic?
  • I mean, I have seen that Borel-Moore Homology is defined using sheafs or sheaves. I am still a masters student and I have never heard about that. Is there a proof of the statement using machinery from a first course in Algebraic Topology (Some homotopy theory, a bit of homology and cohomology)?
  • In the case some heavy machinery is needed could you provide me some references to look at?


  • I have tried using Cohomology with compact supports since in Massey's book titled Singular Homology Theory he uses it to deal with non compact manifolds in order to prove Poincarè Duality. He refers to

    H. Cartan, Seminaire Henri Cartan 1948/49: Topologie Algebrique

    which I have partially read. However, since the spaces we are dealing with are not locally-compact, I can't use that stuff.

  • I have read in the paper: T. Beke, “Topological invariance of the combinatorial euler characteristic of tame spaces,” an idea about using one-point compactifications ... but I didn't achieve my goal yet.
  • I have had a look at Stack Exchange. There is a related this question. However, I haven't found an answer there. I quote the part I find most related.

    The definition of combinatorial Euler characteristic is great for "finite polyhedral complexes", I think. By a "finite polyhedral complexes" I mean glue together finitely many polyhedra, but you're allowed to leave some faces open, so that unlike a CW complex not every cell must have complex closure. Then you can calculate Euler characteristic with the usual formula: (number of cells of even dimension) - (number of cells of odd dimension). I think this is a topological (but not homotopy!) invariant.

So thanks in advance and any help will be appreciated.

  • 1
    $\begingroup$ The Borel-Moore homology is not that bad, if you already understand singular homology. Instead of the complex of finite chains you take the complex of locally finite chains (this is the second definition on wikipedia). You can prove that the combinatorial Euler characteristic coincides with the one defined using B-M homology exactly in the same way as you do for the ordinary one. $\endgroup$ – Denis Nardin Oct 25 '16 at 21:23
  • $\begingroup$ You can use any homology theory, e.g., singular. (Any homology theory gives the same result on finite CW-complexes.) I doubt that it can be avoided altogether, if you want a really rigorous proof. However, this is textbook stuff, not research level. Accidentally, $\chi$ is a homotopy invariant. $\endgroup$ – Alex Degtyarev Oct 25 '16 at 21:24
  • $\begingroup$ @DenisNardin you mean I can prove the homology axioms and then create a kind of "cellular Borel-Moore homology" so I can prove that the combinatorial Euler characteristic coincides with the one defined using B-M homology? $\endgroup$ – D1811994 Oct 25 '16 at 21:30
  • $\begingroup$ @AlexDegtyarev we are dealing with spaces which are not CW-complexes since the closure of the cells doesn't need to be in the complex. So "(Any homology theory gives the same result on finite CW-complexes.)" is not useful here. And the statement : "Accidentally, $\chi$ is a homotopy invariant" is not true since with the definitions obove $\chi((0,1)) \neq \chi([0,1])$. $\endgroup$ – D1811994 Oct 25 '16 at 21:35
  • $\begingroup$ @AlexDegtyarev I am not so sure that the combinatorial Euler characteristic coincides with the topological one. This question seems to indicate that it doesn't. What I meant was simply a sketch of the proof that it is an homeomorphism invariant (since Borel-Moore homology obviously is). $\endgroup$ – Denis Nardin Oct 25 '16 at 21:36

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.