$\DeclareMathOperator\PSL{PSL}$Let $U\subset\mathbb C^2$ be an open set, $f:U\to \PSL(2,\mathbb C)$ a holomorphic map. If the image of $f$ is contained in $\operatorname{PSU}(2,\mathbb C)$, I guess that it can only be a constant map. Is it correct ?



Holomorphic map to Möbius group
Condition $U\subseteq \mathbb{C}^2$ can be replaced by $U\subseteq \mathbb{C}^1$ (just restrict your map
on a little disk in a complex line in $U$. Moreover, the restricted map lifts to a map into $SU(2)$. Then existence of a map to $SU(2)$
essentially means that you have two analytic functions satisfying
$|f|^2+|g|^2=1$. Such functions must be constant. To see this write
$f=u+iv,\; g=u\_1+iv\_1$, differentiate $u^2+v^2+u\_1^2+v\_1^2=1$ twice wrt $x$ and $y$, then add, and you you obtain that sum of squares of all first derivatives is $0$.


