TITLE: Singular cohomology and birational equivalence
QUESTION [8 upvotes]: Let us remenber that we have the following proposition of Artin and Mumford (in "Some elementary examples unirational varieties which are not rational" proposition 1.):
"The torsion subgroup $T_2\subset H^3(V,\mathbb{Z})$ is a birational invariant of a complete non-singular complex variety $V$ of any  dimension $n$."
Question : Are there other any other type of singular cohomological information (even homotopical) which is known to be a birational invariant of complete non-singular complex varieties?

REPLY [10 votes]: Basically, any invariant $T$ which satisfies the following purity property will turn into a birational invariant for proper smooth varieties (aka complex compact manifolds): let $X$ be a smooth variety and $U$ be an open subspace of $X$ such that $\mathop{\rm codim}_X(X\setminus U)\geq 2$; then the injection $U\hookrightarrow$ induces an isomorphism $T(X)\simeq T(U)$.
The proof is as follows. Let $X,Y$ be proper smooth varieties and let $f\colon X\dashrightarrow Y$ be a birational isomorphism. Because $X$ is smooth and $Y$ is proper, the valuative criterion applies and shows that $f$ is defined on an open subset $U\subset X$ such that $\mathop{\rm codim}_X(X\setminus U)\geq 2$. Similarly, the inverse isomorphism $f^{-1}$ is defined on an open subset $V\subset Y$ such that $\mathop{\rm codim}_Y(Y\setminus V)\geq 2$. Say $T$ is covariant, for definiteness.
One gets maps $f_*\colon T(X)\simeq T(U)\to T(Y)$ and $g_*\colon T(Y)\simeq T(V)\to T(X)$. One checks that their composite is the identity.
A loot of interesting examples are given in the answer of R. van Dobben de Bruyn.
Another class of examples is given by the unramified cohomology. They are important in the study of rationality of varieties.