We denote $Aut \left (\mathbb{D} \times \mathbb{D} \right) $ the set of automorphisms of $\mathbb{D} \times \mathbb{D}$, where $\mathbb{D}$ is the unit disk in $\mathbb{C}$. Let $d_X$ denote the Kobayashi distance on $X$.
1. Describe precisely $Aut \left (\mathbb{D} \times \mathbb{D} \right) $.
2. Let $p=\left(p_1,p_2\right),q=\left(q_1,q_2\right)\in \mathbb{D} \times \mathbb{D}$. Prove that $$d_{\mathbb{D} \times \mathbb{D}}\left(\left(p_1,p_2\right),\left(q_1,q_2\right)\right)=\sup \left(d_{\mathbb{D}}\left(p_1,q_1\right),d_{\mathbb{D}}\left(p_2,q_2\right)\right)$$
3. Let $M\subset \mathbb{C}^m$ and $N\subset\mathbb{C}^n$ be complex manifolds. Prove that $$\sup \left(d_{M}\left(p_1,q_1\right),d_{N}\left(p_2,q_2\right)\right)\le d_{M\times N}\left(p,q\right)\le d_{M}\left(p_1,q_1\right)+d_{N}\left(p_2,q_2\right)$$
- My (hinted) idea is that if $f=\left(f_1,f_2\right)\in Aut \left (\mathbb{D} \times \mathbb{D} \right)$ then $f_j$ depends on only $z_j$, i.e, $f_j\in Aut \left(\mathbb{D}\right)$. But I have a trouble to show it. I can show that for fixed $z\in \mathbb{D} \times \mathbb{D}$ there exists $f_z\in Aut \left (\mathbb{D} \times \mathbb{D} \right)$ such that $f_z\left(z\right) =0$.
Thank you very much for your helps.