I have been reading the section on the cohomology of a blow-up in Griffths-Harris "Principles of algebraic geometry", and I am a bit confused about a result that their recipe, which uses the Mayer-Vietoris sequence, gives in a very simple example. Namely the blow up of $\mathbb{P}^2$ at a point. All cohomology is with coefficients in $\mathbb{Q}$.
If I understand their idea it goes like this. Let $\pi : P \rightarrow \mathbb{P}^2$ be the blow up at a point $p \in \mathbb{P}^2$. Take a tubular neighbourhood $U$, in this case a disk, around the point $p$. Then note that the blow up $P = \pi^{-1}(U) \cup (P \setminus E)$, where $E$ is the exceptional divisor in the blow-up. Let $\tilde{U}^* = \pi^{-1}(U) \setminus E$ and by $U^* = U \setminus p$. Because the blow up is an isomorphism outside the origin we have that $\tilde{U}^* \cong U^*$ . We apply the Mayer-Vietoris to this situation and get: $$ ...\rightarrow H^{i-1}(U^*) \rightarrow H^i(P) \rightarrow H^i(P\setminus E) \oplus H^i(\pi^{-1}(U)) \rightarrow H^i(U^*) $$ Now, we have also the fact that $\pi^{-1}(U)$ deformation retracts to the exceptional divisor $E$, and that $P \setminus E \cong \mathbb{P}^2 \setminus p$ again by the fact that the blow-up is an isomorphism outside of the origin.
However, if I look at this for $i=2$, I would excpect to get $H^2(P) \cong H^2(\mathbb{P}^2) \oplus H^2(E)$, but in the exact sequence above I would have $H^1(U^*)$ on the left of $H^2(P)$, which is 1-dimensional as it is the first cohomology group of a puncutred disk, which spoils this.
Is there an error in my reasoning here? Or multiple of them?
Also, is there another way to show that $H^2(P) \cong H^2(\mathbb{P}^2) \oplus H^2(E)$, or is this a standard argument?
Note that $U$ is a complex two-dimensional disc (you're thinking of a complex one-dimensional disc).