TITLE: Creating surjective holomorphic map from unit disc to $\mathbb{C}$?
QUESTION [12 upvotes]: I'm trying to formulate a surjective holomorphic map from the unit disc ($\mathbb{D}$) to $\mathbb{C}$. I know that there exists $f: \mathbb{D} \rightarrow \mathbb{H}$, which is a biholomorphism from the unit disc to the upper half plane.
I know that I can define $g(z) = z -i$ to shift the upper half plane down by 1 unit, then I can define $h(z) = z^2$ to square the result.
Then, $h \circ g \circ f$ is a holomorphic function $\mathbb{U} \rightarrow \mathbb{C}$. My question is: how does $h(z)$, (i.e. squaring the result of the shifted upper half plane), extend the image to the entire lower half plane?

Maybe I'm missing something obvious, but I'm not seeing how $\mathbb{C}$ is hit in its entirety.

REPLY [7 votes]: You know that $h \colon ℂ → ℂ$ is surjective (by the fundamental theorem of algebra or using polar coordinates).
For any $w ∈ ℂ$, if $z^2 = w$, then $(-z)^2 = w$. Then $\Im (-z) = - \Im (z)$, so one of $z, -z$ is in $\mathbb H$, if not in $ℝ$. Therefore $h \colon \mathbb H - \mathrm i → ℂ$ is surjective as well.
And $g ∘ f \colon \mathbb D → \mathbb H-i$ is obviously surjective, as you said.
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