A 'projective' property of the Haar U(n) measure Let $U(n)$ be the compact manifold of unitary $(n \times n)$-matrices and let $\mu_n$ denote the Haar-probability measure on $U(n)$. For $m < n$ does there exists a measurable (maybe even continuous or smooth) map
$$
F: \ U(n) \rightarrow U(m)
$$
with the property, that
$$
\mu_m(A) = \mu_n(F^{-1}(A))
$$
for every $A \in \mathcal{B}(U(n))$?
Are there maybe certain necessary conditions on $(m,n)$?
 A: For any $m<n$ the $n\times n$ unitary matrix $\Omega$ has the block decomposition
$$\Omega=\begin{pmatrix} A&B\\ C&D\end{pmatrix},$$
where $A$ has dimensions $m\times m$, $D$ has dimensions $(n-m)\times(n-m)$, $B$ has dimensions $m\times(n-m)$ and $C$ has dimensions $(n-m)\times m$. Up to a set of measure zero, the matrix $D$ will not have a unit eigenvalue, so $I-D$ is invertible. We then define the continuous map $F$ from $U(n)$ to $U(m)$ by
$$F(\Omega)=A+B(I-D)^{-1}C.$$
One readily checks that $F(\Omega)$ is unitary$^\ast$ and as David Speyer points out $F(\Omega)$ inherits$^{\ast\ast}$ the Haar measure from $\Omega$.


$^\ast$ More generally, for any $\Omega\in U(n)$ and $V\in U(n-m)$ the matrix $U=A+B(I-VD)^{-1}VC$ is unitary. One can think of $V$ as the reflection matrix of a barrier that closes off $n-m$ scattering channels. Then the remaining $m$ channels have scattering matrix $U=A+\sum_{k=0}^\infty B(VD)^kVC$, where $k+1$ counts the reflections off the barrier. As a further check for the unitarity of $U$, I can offer a Mathematica Notebook.
$^{\ast\ast}$ Since for any $g\in U(m)$, $G={{g\;\;0}\choose{0\;\;I}}\in U(n)$ one has $F(G\Omega)=gF(\Omega )$, the measure remains left-invariant.

A: You can get a map which is continuous outside a set of lower dimension.
Let $K\subset L$ be compact Lie groups and let $s:K\backslash L\to L$ be a section to the projection $L\to K\backslash L$. Now this section can be chosen continuous outside a set of lower dimension, since the projection is a fibre bundle. For $x\in L$ we write $s(x)$ for $s(Kx)$.
We define a map $f:L\to K$ by $f(x)=xs(x)^{-1}$. Since $f(kx)=kf(x)$ for $k\in K$, the measure $f_*($Haar$)$ is invariant, hence Haar.
