I read from somewhere the following proposition (maybe more condition needed):
Let $X$ be a $\mathbb C$-scheme, $Z$ its closed subscheme. Let $B=Bl_Z X$ be the blow-up along $Z$ and $E$ be the exceptional divisor. Then we have $$H^*(B)\oplus H^*(Z)=H^*(X)\oplus H^*(E)$$ where $H^*$ is the cohomology ring.
I want to know its proof.
My Attampt: Since we have $B\backslash E \cong X\backslash Z$, so it should be natural to have $$H^*(B)\ominus H^*(E)=H^*(X)\ominus H^*(Z)$$ and then follows the result. However, I don't know how to formulate the "$\ominus$" in a good way. But I believe this should make sense.