0

Show that $X \simeq Y$ implies that $\chi (X)=\chi (Y)$ for $X$ and $Y$ finite CW-complexes

Since $X$ and $Y$ are homotopy equivalent I know we have the maps $f: X \to Y$ and $g:Y \to X$ such that $g \circ f =id_X$ and $f \circ g =id_Y$. How can I apply this to the Euler characteristic to show they are equal?

  • 1
    What is your definition of the Euler characteristic? There are many definitions. – J126 May 22 '23 at 14:19
  • Euler characteristic being $ \Sigma^{n}_{i=0} (-1)^i k_i$ where $k_n$ is the number of n-cells –  May 22 '23 at 14:25
  • It will be covered in any algebraic topology textbook. Consider reading one. – Moishe Kohan May 22 '23 at 14:39
  • Have you learned about homology or de Rham cohomology? Without something like that, it might be hard to prove: https://math.stackexchange.com/q/1410974/2838 – J126 May 24 '23 at 02:36

0 Answers0