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If a finite CW complex $X$ is the union of sub complexes $A$ and $B$, show that $\chi(X)=\chi (A)+\chi (B)-\chi (A \cap B)$.

some how I can imagine what is happening,it is counting numbers of all kind of holes and calculate the sum of them with respect to the signs,and it subtracts the holes that counted twice one in $A$ and one in $B$.

I must make it rigid to connect my imagination to the theory,I don't know how should I start.please give me guidance or any note thank you very much.

kpax
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  • What is your definition of $\chi$? – Grigory M May 22 '14 at 08:36
  • it is with homology groups of $X$:$\chi(X)=\sum (-1)^n rank(H_n(X))$,which rank means its betti number (number of $\mathbb{Z}$). – kpax May 22 '14 at 08:48
  • Depending on your answer to Grigory's definition, it may be useful to use the Mayer-Vietoris sequence. – Aaron May 22 '14 at 08:48
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    Great. Do you know any theorem linking together H(X), H(A), H(B) and H(intersection)?.. – Grigory M May 22 '14 at 08:51
  • because Mayer-Vietoris sequence is exact,can I replace $H_n(X)=\frac{H_n(A)\oplus H_n(B)}{H_n(A \cap B)}$? – kpax May 22 '14 at 08:54
  • Well, can you indeed? If unsure, consider some examples first... – Grigory M May 22 '14 at 09:27
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    For what it's worth, I think this proof is easier if you use cell counts in the definition instead. – Ayman Hourieh May 22 '14 at 20:04
  • All you need is the Mayer-Vietoris sequence, and that the rank is an additive function (i.e for every short exact sequence the rank of the middle term is the sum of the ranks of the outer terms). If you have now a finite long exact sequence (i.e at most finitely many terms are not zero) you get that the alternated sum of the ranks is zero. – Agustín Moreno May 26 '14 at 21:17

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