Let $X$ be a smooth projective variety and $W \subset X$ a smooth, projective subvariety. Let $\pi:\tilde{X} \to X$ be the blow-up of $X$ along $W$. Let $E$ be the exceptional divisor of $\pi$ and $i:E \hookrightarrow \tilde{X}$ the natural closed immersion. Let $\alpha \in H^k(\tilde{X})$ such that $i^*\alpha=0$ in $H^k(E)$. Does this imply that $\pi^*\pi_*\alpha=\alpha$?

## 1 Answer

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Yes. Put $U:=\widetilde{X}\smallsetminus E=X\smallsetminus W$. By the open-closed exact sequence, $\alpha $ comes from a class in $H^k_c(U)$. But the map $H^k_c(U)\rightarrow H^k(\widetilde{X})$ factors through $H^k(X)$, hence $\alpha =\pi ^*\beta $ for some class $\beta $ in $H^k(X)$. Applying $\pi _*$ gives $\beta =\pi _*\alpha $, hence your formula.