No bijective continuous function from upper half plane to $\mathbb{C}$

Question

If we take $H := \{z \in \mathbb{C}\ : im(z) \geq 0\}$. Is there a continuous bijective function from $H$ to $\mathbb{C}$?

$H$ is not isomorphic to $\mathbb{C}$, because we include the real line in $H$. So there can't be a bijective continuous map with an inverse that is also continuous. But I can't seem to find a map $H \longrightarrow \mathbb{C}$ that would be continuous and bijective at all.

I'm wondering if it's the same situation as in the real numbers, where two connected $X,Y \subset \mathbb{R}$ are already homeomorphic if there's a continuous bijection from $X \longrightarrow Y$, and there's no necessity to check that the inverse is continuous.

But to go back to the example with $H$ and $\mathbb{C}$. It could be that they're not isomorphic, but that there is a continuous bijection from $H$ to $\mathbb{C}$, but the inverse wouldn't be continuous. So that's why I'm wondering if there is a continuous bijection from $H$ to $\mathbb{C}$ at all.

I think this was asked few years ago and the answer was negative. The reason was that every closed subset of the plane which is the image of a continuous bijection from a line, has to separate the plane. — Moishe Kohan, Jan 18 '20 at 21:03
Just to be clear, by "answer is negative" you mean that there is no bijective continuous function $H \longrightarrow \mathbb{C}$, right? — M.G., Jan 18 '20 at 21:35
Your statement about subsets of $\mathbb {R}$ is wrong. For example, let $X = [0,1) \cup {2}$, $Y=[0,1]$ and let $f$ be the identity on $[0,1)$, $f(2) = 1$. — David Hartley, Jan 19 '20 at 01:04
I should have been more specific, what I meant were connected subsets $X$ and $Y$. Would that change anything? — M.G., Jan 19 '20 at 01:10

score 3 · Answer 1 · answered Jan 19 '20 at 19:41

Yes, indeed, this closely related question was asked and answered earlier.

Theorem. There is no continuous bijection $H\to {\mathbb C}$.

Proof. Suppose that there exists $f: H\to {\mathbb C}$, a continuous bijection. The restriction of $f$ to the open half-plane $$ U= \{z: Im(z)>0\} $$ is still a continuous injective map. Hence, by the invariance of domain theorem, $f(U)$ is open in ${\mathbb C}$. Since $f(H)={\mathbb C}$, the image $f(L)$ is closed in ${\mathbb C}$, where $L$ is the boundary line of $H$, i.e. the real line in the complex plane.

Now, we have a continuous injective map $f: L\to f(L)\subset {\mathbb C}$ with closed image. By my answer to the linked question, $f(L)$ separates ${\mathbb C}$ into at least two components. But ${\mathbb C}- f(L)=f(U)$ is connected. A contradiction. qed

No bijective continuous function from upper half plane to $\mathbb{C}$

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