I have known that $b_2(CP^n\sharp CP^n)=2$, however I have no idear how to prove this fact ! I appreciate any help for this simple question! Thank you!
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$\begingroup$ When you say you know this, where did you see this claimed? And how much algebraic topology have you studied? $\endgroup$– Yemon ChoiCommented Oct 16, 2017 at 2:51
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$\begingroup$ I have know this fact from the paper I am reading. In fact I have studied little about the algebraic topology. I am very thankful, if you offer me the reference $\endgroup$– FaithCommented Oct 16, 2017 at 2:58
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This is one of those calculations which is easier with the general formalism of homology, rather than just Betti numbers. Using the Mayer-Vietoris theorem a few times, one can show that if $M$ and $N$ are connected $n$-manifolds and $k\ne n-1,n$, then
$$H_k(M\mathbin{\#}N)\cong H_k(M)\oplus H_k(N).$$
(For a proof and explanation, see this Math.Stackexchange answer.)
Hence, when $n > 1$,
$$H_2(\mathbb{CP}^n\mathbin\#\mathbb{CP}^n)\cong H_2(\mathbb{CP}^n)\oplus H_2(\mathbb{CP}^n)\cong \mathbb Z\oplus\mathbb Z,$$
so its Betti number is
$$b_2(\mathbb{CP}^n\mathbin\#\mathbb{CP}^n) = \dim H_2(\mathbb{CP}^n\mathbin\#\mathbb{CP}^n) = 2.$$
(When $n = 1$, $\mathbb{CP}^1\cong S^2$, so $\mathbb{CP}^1\mathbin\#\mathbb{CP}^1\cong S^2\mathbin\# S^2\cong S^2$, and $b_2(S^2) = 1$, so to get $b_2 = 2$ one must assume $n > 1$.)
answered Oct 16, 2017 at 3:21