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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?

enter image description here

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

r123454321
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2 Answers2

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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.

k.stm
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    I don’t understand, what is the point of shifting H down by -i then? – lightnesscaster Dec 12 '21 at 20:28
  • Oh, I think to map onto when i = 0. If that’s right then shifting by .000000000000000001 i would also work. – lightnesscaster Dec 12 '21 at 20:31
  • I still don't understand why shifting H down by $-i$ will give $\mathbb{C}$ – fieke_2000 Dec 29 '21 at 14:04
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    @fieke_2000 $A = { z ∈ ℂ;~\Im z ≥ 0} ⊆ \mathbb H - \mathrm i$, and $∀ w ∈ ℂ ~ ∃ z ∈ A \colon w = z^2$, so $ℂ ⊆ h(A)$, so $ℂ ⊆ h (\mathbb H - \mathrm i) = \operatorname{img} h$. – k.stm Dec 31 '21 at 15:00
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I know this is an old post, but here's for future viewers.

Shifting $\mathbb{H}$ down by 1 is performed in order to ensure each complex number $z=re^{i\theta}$ has a root within our domain.

Let $f$, $h$ be the functions you mentioned. Notice that $z=1$ doesn't have a square root in $f(\mathbb{D})=\mathbb{H}$, so $h\circ f$ isn't surjective.

Now let's "shift $\mathbb{H}$ down by $i$" with $g$.

Mark $g(\mathbb{H})="\mathbb{H}-i$"

Let $z=re^{i\theta}$, $z$ has $2$ square roots in the Complex plane, $\sqrt{r}e^{i\theta/2}$ and $\sqrt{r}e^{i(\theta/2+\pi)}$.

Because the difference between their angles with the real axis is $\pi$ we are guaranteed at least one of them (mark it as $z_1$) will be in $\mathbb{H}\cup\mathbb{R}$ (draw it down to see why). Now because $\mathbb{H}\cup\mathbb{R}$ is contained in $\mathbb{H}-i$ we know $z_1\in\mathbb{H}-i$.

We know $g\circ f$ is surjective from $\mathbb{D}\to\mathbb{H}-i$ then there exists $z_2$ so that $g\circ f(z_2)=z_1$ and that $z_1^2=z$ so $h\circ g\circ f(z_2)=z$.

As this is true for any $z\in\mathbb{C}$ we get $h\circ g\circ f$ is surjective.

Indeed we could have chosen a different "shift" of $\mathbb{H}$, so long as its a shift down in the imaginary axis.

Danifa10
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