Let $X$ be a (possibly singular) compact complex manifold of complex dimension $n$ and let $Y\subset X$ be a submanifold along we blow-up $X$. So we get a blow-down map $P:Bl_Y X\to X$.
Is $P^*: H^{2n}(X,\mathbb R)\to H^{2n}(Bl_YX,\mathbb R)$ an isomorphism?
So far I looked at the Mayer Vietoris sequence for $Bl_YX = (Bl_YX\setminus E) \cup E$, where $E$ is the exceptional divisor. We also call the corresponding intersection $Z$. If $X$ is non-singular, then $Z=S^{2n-1}$.
$Bl_YX\setminus E \simeq X\setminus Y$ and as $\dim Y \le 2n-2$, it follows that $H^{2n}(X) = H^{2n}(X\setminus Y)$ and $H^{2n-1}(X) = H^{2n-1}(X\setminus Y)$. Also, $\dim E = 2n-2$, so $H^{2n}(E)=H^{2n-1}(E) =0$.
Thus $$\to H^3(X) \stackrel j \to H^3 (Z)\to H^4(Bl_YX)\stackrel {i^*}\to H^4(X) \to 0$$ is exact.
Thus in order for $H^4(Bl_YX)$ to be isomorphic to $H^4(X)$, we need $j$ to be surjective.
But even so, this would only tell me that $i^*$ is an isomorphism. I am interested in $P^*$.
In the situation where $Y\subset X$ are both complex manifolds, this is true. As both $X$ and $Bl_YX$ are complex manifolds and all complex manifolds are orientable, both have top cohomology $\Bbb R$. So it suffices to show that $P^*$ is either injective or surjective on top cohomology in order to prove it is an isomorphism. We first note that $P_*: H_*(Bl_YX)\to H_*(X)$ is surjective, as $P$ is an isomorphism away from $Y$ and $Y$ is small. By duality, this gives us the injectivity required on the top piece of cohomology, and so $P^*$ is surjective and thus an isomorphism.