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!



The Betti numbers of of $CP^n\sharp CP^n$
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](https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence) 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](https://math.stackexchange.com/a/187417).)


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 $n > 1$.)


