Prove Euler characteristic is a homotopy invariant without using homology theory

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 $X$.

Solution 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)