I was flipping through May's Concise Course in Algebraic Topology and found the following question on page 82.

  • Think about proving from what we have done so far that $\chi(X)$ depends only on the homotopy type of $X$, not on its decomposition as a finite CW complex.

Here, "what we have done so far" includes a fair amount of homotopy theory, but no homology theory. The only proof I know of the homotopy invariance of $\chi$ is the one in Hatcher using homology groups, so I am intrigued by this question.

Recall that for a finite CW complex $X$, we define the Euler characteristic as $$\chi(X)=\sum (-1)^n c_n,$$ where $c_n$ is the number of $n$-cells in $X$.

  • $\begingroup$ A professor of mine once defined the Euler Characteristic of surfaces using triangularizations (which is essentially CW-complexes), then argued that one couild prove that the EC was not dependent on the triangularization by refining two triangularizations to a third one and showing that the EC "agreed" on the refining. Maybe this idea gives some clue on how to solve your question (I really don't know if it does, since I never tried to do it because homology is quite powerful and solves the issue beautifully enough) $\endgroup$
    – Aloizio Macedo
    Aug 27, 2015 at 3:14
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    $\begingroup$ @AloizioMacedo Unfortunately your professor is incorrect in general, as the Hauptvermutung is super-false. It works for $n$-manifolds, $n \leq 3$, but that's all we've got. $\endgroup$
    – user98602
    Aug 27, 2015 at 16:19
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    $\begingroup$ What you want to do is make a bigger space that you can prove has Euler characteristic $\chi(X)$ and also $\chi(Y)$. One idea I toyed with for a bit right now, based on May's previous problem, is the mapping cylinder of a cellular homotopy equivalence $f: Y \to X$. You can prove this has Euler characteristic $\chi(Y)$ fairly easily. Somehow you should make a 'bigger' construction that includes both $f$ and (a choice of cellular) homotopy inverse $g: X \to Y$. I haven't figured out the right space, but that's the right idea. $\endgroup$
    – user98602
    Aug 27, 2015 at 16:34
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    $\begingroup$ @MikeMiller How about the mapping telescope of $Y \xrightarrow{f} X \xrightarrow{g} Y$? $\endgroup$
    – rj7k8
    Aug 27, 2015 at 22:09
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    $\begingroup$ @rj7k8: I could buy it. Maybe you could write up the details in an answer. $\endgroup$
    – user98602
    Aug 28, 2015 at 22:29

1 Answer 1


We will prove this by induction: suppose, we know that if $X$ consists of $\,<n$ cells then for all other finite $CW$-complexes $X'$ such that $X\approx X'$ we have $\chi(X)=\chi(X')$.

Let $Y$ and $Y'$ be two finite $CW$-complexes, $Y$ consists of $n$ cells, and $f:Y\to Y'$ is homotopy equivalence. Consider one higher-dimensional cell of $Y$: let $Y=Z\cup_\alpha D^k$, where $Z$ consists of $n-1$ cells, $D^k$ is just a cell and $\alpha:\partial D^k\to Z$ is attaching map. We see that $\chi(Z)=\chi(Y)-(-1^k)$.

Then consider the space $CZ\cup_{f|_Z}Y'$, here $CZ$ is a cone. $f$ supposed to be cellular map, so $CZ\cup_{f|_Z}Y'$ is a $CW$ complex; it contains all cells of $Y'$, all cells of $Z$ times $I$, and the vertex of the cone, thus $\chi(CZ\cup_{f|_Z}Y')=\chi(Y')-\chi(Z)+1$. But we know that this space has homotopy type of $S^k$, so by inductive hypothesis $\chi(CZ\cup_{f|_Z}Y')=1+(-1^k)$, and $\chi(Y)=\chi(Y')$ as desired.

EDIT: for the induction we also need the statement be truth in case $X\approx pt$ and $X\approx S^m$. If $X\approx S^m$, we may glue an $m+1$-disc and obtain contractible space.

Now suppose $X\approx pt$, $m$ is maximal dimension of cells of $X$ and $X$ has $p$ $\,m$-cells. The space $sk_{m-1}(X)$ is $(m-2)$-connected, so it is homotopy equivalent to the bouquet of $q$ $\,(n-1)$-spheres. Gluing $m$-cells determines a homomorphism $\phi:\mathbb Z^p\to\mathbb Z^q$, and equalities $\mathrm{coim\,}\phi=\pi_{m-1}(X)$ and $\ker\phi\subseteq\pi_m(X)$ give us $p=q$. Then we may remove $p$ $\,m$-cells, $p$ $\,(m-1)$-cells, and repeat. (this reasoning sounds more like homological argument)

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    $\begingroup$ The induction in the first half of the argument can be eliminated by just considering the mapping cone of $f$ directly, which is contractible and has Euler characteristic $1+\chi(Y')-\chi(Y)$. But it's still not clear how to handle the contractible case without homology, or at very least the computation $\pi_n(S^n)=\mathbb{Z}$ (which has not been done in May's book prior to the exercise in question). $\endgroup$ Aug 30, 2015 at 21:19
  • $\begingroup$ @EricWofsey, i come to conclusion that without notion of homology it's impossible to do $\endgroup$ Aug 31, 2015 at 9:16
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    $\begingroup$ That seems like a rather drastic conclusion to come to. I would bet it's just hard. $\endgroup$
    – user98602
    Aug 31, 2015 at 21:26
  • $\begingroup$ @EricWofsey I suppose that if we cheat a little bit, we can give an entirely homotopy theoretic proof that $\pi_n(S^n)=\mathbb Z$ by using the Freudenthal suspension theorem in the next chapter. $\endgroup$
    – Potato
    Sep 1, 2015 at 23:30
  • $\begingroup$ I do not understand the induction in your argument: For your induction to work you would have to prove that if $X, X'$ are h.e. finite CW complexes, then they have subcomplexes $Y, Y'$ obtained by removing 1 cell from $X, X'$ such that $Y, Y'$ are h.e. How would you prove this? $\endgroup$ Jun 24, 2022 at 17:23

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