Exceptional locus of rational map has codimension two


Let $f: X\dashrightarrow\mathbb{CP}^n$ be a birational map where $X$ is a smooth, projective variety. Then there are some closed subvarieties $Z'\subset X$ and $Z \subset \mathbb{CP}^n$ such that $f$ induces an isomorphism $X \setminus Z' \cong \mathbb{CP}^n \setminus Z.$ Can we assume $Z \subset \mathbb{CP}^n$ has codimension at least two? My intuition is that no codimension one things in $\mathbb{CP}^n$ can be "contracted."

Best Answer

This not true. The simplest counterexample is the linear projection $Q \dashrightarrow \mathbb{P}^2$ from a smooth quadric surface $Q \subset \mathbb{P}^3$ from a point. The inverse map blows up two points on $\mathbb{P}^2$ and contracts the line connecting them. So, to obtain an isomorphism one has to remove this line.

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