Let $X$ be a finite cell-complex and $A\subset X$ be a sub-cell complex.
The post has been answered here: Euler characteristic for CW complexes asked about the proof for $$\chi(A)-\chi(X)+\chi(X/A)=1.$$ This question induces a similar question about what $\chi(A)-\chi(X)+\chi(X,A)$ is.
I know that Euler characteristic is the alternating sum of the number of cells in each dimension. That is $$\chi(A)=c_{0}(A)-c_{1}(A)+c_{2}(A)-\cdots+(-1)^nc_{n}(A)$$ $$\chi(X)=c_{0}(X)-c_{1}(X)+c_{2}(X)-\cdots+(-1)^nc_{n}(X).$$
Also, we know that the cellular chain complex of $(X,A)$ is given by the quotient groups $$C_{k}(X,A)=C_{k}(X)/C_{k}(A)=\mathbb{Z}^{\{\#\ \text{of}\ k-\text{cells in}\ X\ \text{but not in}\ A\}}$$ Does this imply that $c_{k}(X,A)=c_{k}(X)-c_{k}(A)$?
If so, then $\chi(A)-\chi(X)+\chi(X,A)=0$… but this seems not correct… I actually expect it to equal $1$.
Is my proof correct or I am missing something? Thank you!
Best Answer
The confusion you're having ultimately comes from the fact that $H_*(X,A;\Bbb Z)$ is not isomorphic to $H_*(X/A;\Bbb Z)$, but rather to the reduced homology $\tilde H_*(X/A)$. The former is larger by exactly a copy of $\Bbb Z$ in degree zero, so $$\chi(X/A) = \chi(X,A) + 1.$$