Regarding a holomorphic function from the unit disc to any domain mapping given points Let $\Omega\subset\mathbb{C}^n$ be an open connected set and let $\mathbb{D}$ be the open unit disc in $\mathbb{C}$. Now given $\lambda_z$, $\lambda_w\in \mathbb{D}$ and $z,w\in \Omega$, does there exist a holomorphic function $f:\mathbb{D}\longrightarrow \Omega$ such that $f(\lambda_z)=z$ and $f(\lambda_w)=w$?
I know that as $\Omega$ is path connected, there exists a continuous function $\gamma:[0,1]\longrightarrow\Omega$ such that $\gamma(0)=z$ and $\gamma(1)=w$.
 A: The answer is no without any conditions on $\Omega, z,w$ as we can see in the simple case $n=1, \Omega=\mathbb D$ since any holomorphic $f:\mathbb D \to \mathbb D$ is a contraction in the hyperbolic metric $d(w,z)=\frac{|w-z|}{|1-\bar w z|}$ by the Schwarz (Pick) Lemma, so in particular, the existence of $f$ above implies immediately that we need to have $d(w,z)=\frac{|w-z|}{|1-\bar w z|} \le d(\lambda_z, \lambda_w)=\frac{|\lambda_w-\lambda_z|}{|1-\bar \lambda_w \lambda_z|}$ and that doesn't hold in general as we can see with $w=\lambda_w=0, 1> |z| > |\lambda_z| >0$. 
If the hyperbolic metric inequality is satisfied, we can find $f$ as we first can move $w$  to zero with a Mobius automorphism of the disc, $h(y)=\frac{y-w}{1-\bar w y}$ (which is an isometry in the hyperbolic metric) and then with $h(z)=z_1$ the inequality becomes $|z_1|=d(z_1,0)=d(h(z),h(w))=d(z,w) \le d(\lambda_z, \lambda_w)=\frac{|\lambda_w-\lambda_z|}{|1-\bar \lambda_w \lambda_z|}$ so using the disc automorphism $g(y)=\frac{\lambda_w-y}{1-\bar \lambda_w y}$ that sends $\lambda_w \to 0$, we get $|z_1| \le |g(\lambda_z)|$, so we can scale and get $f_{z_1}(y)=ag(y), |a| \le 1$ s.t $f_{z_1}(\lambda_z)=z_1, f_{z_1}(\lambda_w)=0$ and then $f(y)=h^{-1}(f_{z_1}(y))$ will satisfy $f(\lambda_{w})=w, f(\lambda_z)=z$
as required.
(edited as per comments to explicit the general case)
