I'm going through Hatcher's Algebraic Topology, and after reading the paragraphs on simplicial and singular homology, I was wondering about this :

  1. Singular homology is homotopy invariant, that is, two homotopically equivalent topological spaces $X, Y$ have isomorphic homology groups.
  2. Simplicial and singular homology are equivalent, meaning that for a $\Delta$-complex $X$, there are isomorphisms between $H_n^\Delta(X)$ and $H_n(X)$ for all $n$.

Now, would this mean that, given a $\Delta$-complex $X$, any other topological space $Y$ of the same homotopy equivalence class would inherit a $\Delta$-complex structure, and hence have the same simplicial homology groups as $X$?

  • 1
    $\begingroup$ I don't see how you could use a homotopy equivalence to transfer a $\Delta$-complex structure from one space to another. In fact, I think the results presented in these slides on products of CW complexes, show that this is impossible, because you can construct contractible CW complexes $X$ and $Y$ such that $X \times Y$ is not a CW complex. (CW complexes generalise $\Delta$-complexes, and if $X$ and $Y$ are both contractible then $X$ and $X \times Y$ are homotopy equivalent.) $\endgroup$
    – Rob Arthan
    Commented Jun 11, 2023 at 20:01
  • $\begingroup$ PS: I should stress "think" rather than "know for certain" in the above comment as I haven't checked all the details. The results definitely show that the natural CW structure on the product of CW complexes does not deliver the product topology in general. I see also that Hatcher's book covers the relevat result of Dowker in an appendix. $\endgroup$
    – Rob Arthan
    Commented Jun 11, 2023 at 20:59
  • $\begingroup$ @RobArthan I'm glad to find out, however, that the doubt I'm having is legitimate. Thank you for your comment. $\endgroup$ Commented Jun 11, 2023 at 21:09
  • $\begingroup$ I am not sure exactly what it is that you doubt: if spaces $X$ and $Y$ are homotopy equivalent, and if their topologies are induced by a $\Delta$-complex structure, then the simplicial and singular homology groups will all agree. $\endgroup$
    – Rob Arthan
    Commented Jun 11, 2023 at 21:18
  • 2
    $\begingroup$ That's right. For a specific example, a product of uncountably many copies of $\Bbb{R}$ is contractible (and hence homotopically equivalent to a point) but is not compactly generated (and hence cannot be the underlying space of a CW complex or $\Delta$-complex). $\endgroup$
    – Rob Arthan
    Commented Jun 11, 2023 at 21:39

1 Answer 1


Having the same homology and "inheriting" $\Delta$-structure are very vaguely related things.

A simplest counterexample would be a contractible space $Y$ which is not a simplicial/$\Delta$/CW-complex. For example the comb space. Because it is not locally connected.

Homotopy equivalences rarely preserve topological properties.


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