In $\mathbb{C}^n,\ n\geq 2$, there is no bijection between unit disk $B^n(0,1)$ and unit polydisk $P^n(0,1)$. But if we wish to find injective holomorphic mapping from unit disk to polydisk(whose image contains origin), inclusion is the obvious mapping or suitable automorphisms of unit disk(which is sort of inclusion, after applying automorphism). But can we go beyond unit disk in polydisk, by means of injective holomorphic mapping. Are there results in this direction available?

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    $\begingroup$ I'm not exactly sure what your question is - to find an injective map from the unit ball to the polydisk with the image not in the unit ball? You can map $B^n(0,1)$ to some small in the "corner" of the polydisk. $\endgroup$ – Wojowu Jun 3 '20 at 15:59
  • $\begingroup$ I want to see, if we can find a map, whose image contains origin, but also consists points out of $B^n(0,1)$. Edited my question a bit. $\endgroup$ – 123N Jun 3 '20 at 16:08
  • $\begingroup$ How about $z\mapsto 2z$? $\endgroup$ – Henri Jun 4 '20 at 6:24
  • $\begingroup$ @Henri Its image not be within polydisk $P^n(0,1)$ $\endgroup$ – 123N Jun 4 '20 at 9:02
  • $\begingroup$ Ok, I hadn't seen that you wanted the image to be contained in the polydisk. $\endgroup$ – Henri Jun 4 '20 at 12:23

Take $f(z_1,z_2):=(z_1,sz_2+(1-s)z_1)$, $0<s<1$. For example, $s=1/2$. Then $f$ is an injective holomorphic mapping from $B(0,1)$ to $P(0,1)$, ($n=2$), sending $0$ to $0$. Moreover, $\|f(s,0)\|>1$ for $s<1$, near $1$, i.e. $f(s,0)\notin B(0,1)$..


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