TITLE: Rational maps from $\mathbb CP^n$ to $\mathbb CP^{n-1}$, fixing $\mathbb CP^{n-1}$
QUESTION [6 upvotes]: Consider $\mathbb CP^n$ and let $H\subset \mathbb CP^n$ be a hyperplane. Suppose $\varphi: \mathbb CP^n\to H$ is a rational map that fixes $H$ pointwise. I believe that $\varphi$ must be a projection from a point $p\in \mathbb CP^n\setminus H$ to $H$. How to prove this?
Added. F_L gave a nice counterexample to the above question. But I still hope that the answer is positive if one has the following restriction on $\varphi$:
The intersection of the locus of indeterminacy of $\varphi$ with $H$ has codimension $\ge 2$ in $H$.
Does the statement hold under this additional condition? (In the counter-example of F_L, $p_1\in L$ is a point of indeterminancy of $\varphi$)

REPLY [6 votes]: With the additional restriction, $\varphi$ is indeed a linear projection.
Applying a linear change of coordinates, $H$ is given by $x_0=0$, so $\varphi$ is
$[x_0:\cdots:x_n]\mapsto [0:P_1(x_0,\ldots,x_n):\cdots: P_n(x_0,\ldots,x_n)]$
for some homogeneous polynomials $P_1,\ldots,P_n\in \mathbb{C}[x_0,\ldots,x_n]$ of the same degree $d$. You may choose them without common factor. Your assumption is that the restriction to $H$ is the identity. So this means that
$[0:x_1:\cdots:x_n]\mapsto [0:P_1(0,x_1,\ldots,x_n):\cdots: P_n(0,x_1,\ldots,x_n)]$
is the identity on $H$. Hence there is a polynomial $A\in \mathbb{C}[x_1,\ldots,x_n]$, homogeneous of degree $d-1$, such that $$P_i(0,x_1,\ldots,x_n)=A\cdot x_i$$ for $i=1,\ldots,n$.
The locus where $A=0$ and where $x_0=0$ is then contained in the intersection of the base-locus of $\varphi$ with $H$. As you assumed that this latter should be of codimension $2$ in $H$, it implies that $A$ is a non-zero constant. Hence, $d=1$, so $\varphi$ is of degree $1$ and is a linear projection.