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 $X$.

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  • 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) – Aloizio Macedo Aug 27 '15 at 03:14
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    @AloizioMacedo Unfortunately your professor is incorrect in general, as the [Hauptvermutung](https://en.wikipedia.org/wiki/Hauptvermutung) is super-false. It works for $n$-manifolds, $n \leq 3$, but that's all we've got. –  Aug 27 '15 at 16:19
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    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. –  Aug 27 '15 at 16:34
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    @MikeMiller How about the mapping telescope of $Y \xrightarrow{f} X \xrightarrow{g} Y$? – rj7k8 Aug 27 '15 at 22:09
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    @rj7k8: I could buy it. Maybe you could write up the details in an answer. –  Aug 28 '15 at 22:29
  • @rj7k8 I would love to see your mapping telescope solution. – Potato Aug 30 '15 at 19:13
  • @Potato It didn't work out like I thought it might! I will keep thinking though. – rj7k8 Aug 30 '15 at 20:16
  • @rj7k8 That seems like a very natural idea -- hopefully it works out eventually! – Potato Aug 30 '15 at 20:20

1 Answers1


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)

Andrey Ryabichev
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    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). – Eric Wofsey Aug 30 '15 at 21:19
  • @EricWofsey, i come to conclusion that without notion of homology it's impossible to do – Andrey Ryabichev Aug 31 '15 at 09:16
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    That seems like a rather drastic conclusion to come to. I would bet it's just hard. –  Aug 31 '15 at 21:26
  • @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. – Potato Sep 01 '15 at 23:30