Rational map having a birational restriction. Let $\sigma\subset |\mathcal{O} _{\mathbb{P} _{\mathbb{C}}^N}(2)|$ be 
an $N-$dimensional linear systems of quadrics on $\mathbb{P} _{\mathbb{C}}^N$ ($N\geq 4$), 
$F:\mathbb{P} _{\mathbb{C}} ^N\dashrightarrow \mathbb{P} _ {\mathbb{C}}^ N$ the rational map associated to $\sigma$,
and $Q\in \sigma$ a fixed smooth quadric
such that $F|_{Q}:Q\dashrightarrow \mathbb{P} _{\mathbb{C}}^{N-1}$ is birational.
Is $F$ birational?
 A: This is a bit too long for a comment, but since it was requested, below a positive answer in the case $N=2$ is given. This might also help to understand the question in the simplest case.
So we start with a smooth quadric $Q=0$ in $\mathbb P^2$ and two more quadrics $Q_1=0$, $Q_2=0$ such that the non-fixed locus of intersection of $Q$ with the linear system  $Q_1+tQ_2=0$ is a single point (so the corresponding map $Q\to \mathbb P^1\supset t$ is degree one, i.e. birational). Such situation can happen only if both $Q_1$ and $Q_2$ intersect $Q$ at the same set consisting of  three points, or two points of which one is with multiplicity 2, or at one point with multiplicity $3$. Let us denote this set by $x$. Then it is clear that for any generic pencil in the family generated by $Q,Q_1,Q_2$ the fixed locus is $x$ plus one point. If you translate this into the notations of the original problem you see that the map from $\mathbb P^2$ is birational. 
A: Nice question. I think that the answer is yes and more general. I have never seen this but seems natural.
${\bf Lemma}$ Let $f\colon \mathbb{P}^n_{\mathbb{C}}\to \mathbb{P}^n_{\mathbb{C}}$ be a rational map of degree $d$, given by $(x_0:\dots:x_n)\to (f_0:\dots:f_n)$, where the $f_i$ are homogeneous of degree $d$.
Suppose that the hypersurface $H\subset  \mathbb{P}^n_{\mathbb{C}}$ given  $f_0=0$ is irreducible and that the map from $H$ to $\mathbb{P}^{n-1}_{\mathbb{C}}$ given by the restriction (i.e. $(x_0:\dots:x_n)\to (f_1:\dots:f_n)$ on $H$) is birational.
Then, the map $f$ is birational.
${\bf Proof}$ Let $\sigma$ be the linear system associated, which corresponds to hypersurfaces of $\mathbb{P}^n$ of equation $\sum_{i=0}^n a_i f_i=0$. The restriction being birational, the intersection of $n-1$ general elements of $\sigma$ with $H$ give one mobile point, and a non-mobile part that we call $R$. Since every member of $\sigma$ contains $R$ and because elements have all the same degree, the intersection of $n$ general elements of $\sigma$ gives $R$, plus exactly one mobile point (the last part follows from Bézout). This shows that $f$ is birational.
${\bf Remark:}$ We could also view $R$ as a set of points, curves,... with some multiplicities and use intersection form on the blow-up.
