Check whether the map is a homeomorphism.

Question

Let $\mathbb R \times [-\pi, \pi]$ be given the topology induced from $\mathbb R^2 \simeq \mathbb C.$ Consider the map $\mathcal E : \mathbb R \times [-\pi, \pi] \longrightarrow \mathbb C^{\times}$ defined by $(x,y) \mapsto e^{x + iy} = (e^x \cos y, e^x \sin y).$

$(a)$ Let $X$ be the quotient space obtained from $\mathbb R \times [-\pi, \pi]$ under the equivalence relation generated by $(x, \pi) \sim (x, -\pi).$ Show that there is an induced continuous map $\widetilde {\mathcal E} : X \longrightarrow \mathbb C^{\times}.$

$(b)$ Show that $\widetilde {\mathcal E}$ is a bijection.

$(c)$ Prove or disprove $:$ The map $\widetilde {\mathcal E}$ is a homeomorphism.

My Attempt $:$ $(a)$ is clear as for any $x \in \mathbb R$ we have $\mathcal E (x, \pi) = \mathcal E (x, -\pi),$ as can be easily checked. For $(b)$ we notice that if $e^x \cos y = e^u \cos v$ and $e^x \sin y = e^u \sin v$ then by squaring and adding we have $e^{2 x} = e^{2 u}$ and now taking natural logarithm in both sides we can conclude that $x = u.$ So $\cos y = \cos v$ and $\sin y = \sin v$ which is only possible if $y = v + 2 n \pi,$ for some integer $n$ and $y - v = 2 n \pi.$ So if we restrict $y, v \in [-\pi, \pi]$ then $n = 0, \pm 1$ and in either case $(x,y) \sim (u,v).$ This shows that $\widetilde {\mathcal E}$ is injective. For surjectivity of $\widetilde {\mathcal E}$ what I got is that given $(p,q) \in \mathbb C^{\times}$ we have $\widetilde {\mathcal E} \left (\ln \sqrt {p^2 + q^2}, \arcsin \frac {q} {\sqrt {p^2 + q^2}} \right ) = (p,q),$ if $p \geq 0$ and $\widetilde {\mathcal E} \left (\ln \sqrt {p^2 + q^2}, \pi - \arcsin \frac {q} {\sqrt {p^2 + q^2}} \right ) = (p,q),$ if $p \lt 0.$ So $\widetilde {\mathcal E}$ is surjective and hence a bijection. The above formula in fact gives rise to the inverse of $\widetilde {\mathcal E}$ which is given by $$\widetilde F : (p,q) \longmapsto \begin{cases} \left [\left (\ln \sqrt {p^2 + q^2}, \arcsin \frac {q} {\sqrt {p^2 + q^2}} \right ) \right ] & \text{if}\ p \geq 0 \\ \left [\left (\ln \sqrt {p^2 + q^2}, \pi - \arcsin \frac {q} {\sqrt {p^2 + q^2}} \right ) \right ] & \text{if}\ p \lt 0 \end{cases}$$ and in order to show it's continuity it is enough to show the continuity of the following function $$F : (p,q) \longmapsto \begin{cases} \left (\ln \sqrt {p^2 + q^2}, \arcsin \frac {q} {\sqrt {p^2 + q^2}} \right ) & \text{if}\ p \geq 0 \\ \left (\ln \sqrt {p^2 + q^2}, \pi - \arcsin \frac {q} {\sqrt {p^2 + q^2}} \right ) & \text{if}\ p \lt 0 \end{cases}$$

which can be clearly seen to be continuous by pasting lemma. But I am not quite sure about the last part as to whether $F$ actually serves the purpose of $\widetilde {\mathcal E}$ and for some reason I believe that $F$ fails to be onto in the second factor on $[-\pi, \pi]$ as $\arcsin$ function only takes values in $\left [-\frac {\pi} {2}, \frac {\pi} {2} \right ].$ Did I make any mistake?

Any help in this regard would be greatly appreciated. Thanks for investing your valuable time on my question.

Answer 1

Your proof that the continuous map $\widetilde{\mathcal E}$ is bijective (needs to be fixed, as noted in comments, and) can be simplified as follows: it suffices to check that $(x,y)\mapsto e^{x+iy}$ is a bijection from $\Bbb R\times(-\pi,\pi]$ to $\Bbb C^\times.$ For this, simply write: $\forall(x,y)\in\Bbb R\times(-\pi,\pi],\forall z\in\Bbb C^\times,$ $$z=e^{x+iy}\iff x=\ln|z|\quad\text{and}\quad y=\operatorname{Arg}(z).$$

The restriction/corestriction of $\widetilde {\mathcal E},$ from the open subset $\Bbb R\times(-\pi,\pi)$ of the quotient onto $\Bbb C\setminus(-\infty,0]$, is classically known to be a homeomorphism (even a smooth diffeomorphism).

The restriction/corestriction of $\widetilde {\mathcal E},$ from the open subset $[(\Bbb R\times[-\pi,0))\cup(\Bbb R\times(0,\pi])]/\sim$ onto $\Bbb C\setminus[0,+\infty)$, may be naturally identified (translating by $(0,2\pi)$ the first piece $\Bbb R\times[-\pi,0)$) with the map $\Bbb R\times(0,2\pi)\to\Bbb C\setminus[0,+\infty),(x,y) \mapsto e^{x + iy} ,$ which, similarly, is also clearly a homeomorphism. (The only new ingredient here is the identification of $\left([-\pi,0)\cup(0,\pi]\right)/(\pi\sim-\pi)$ with $(0,2\pi),$ via the homeomorphism $\widetilde e$ defined by $e(x)=x+2\pi$ if $x\in[-\pi,0)$ and $e(x)=x$ if $x\in(0,\pi]$.)

Since the two open subsets $\Bbb C\setminus(-\infty,0]$ and $\Bbb C\setminus[0,+\infty)$ cover $\Bbb C^\times$, this proves that $\widetilde {\mathcal E}^{-1}$ is continuous.

Alternatively, you can use the following theorem: every continuous bijection from a compact Hausdorff space onto a Hausdorff space is a homeomorphism, applying it, for every $M>0,$ to the restriction/corestriction of $\widetilde {\mathcal E},$ from $([-M,M]\times[-\pi,\pi])/\sim$ onto $V_M:=\{z\in\Bbb C\mid e^{-M}\le|z|\le e^M\},$ and noticing that the $V_M$'s contain an open cover $\left(\{z\in\Bbb C\mid e^{-M}<|z|<e^M\}\right)$ of $\Bbb C^\times.$

Answer 2

You have the right ideas, but the proof of surjectivity is not correct. Also the proof of injectivity can be simplified a bit.

1. Injectivity.

All we have to know is that $e^x > 0$ and $\lvert e^{iy} \rvert = 1$.
If $e^{x +i y} = e^{u +i v}$, then $e^x = \lvert e^{x +i y} \rvert = \lvert e^{u +i v} \rvert = e^{iu}$ which implies $x = u$. But then we get $e^{iy} = e^{iv}$ and as you have shown this means $y = v$ or $y, v \in \{-\pi, \pi\}$.

2. Surjectivity.

All we need to know is that $e^x : \mathbb R \to (0,\infty)$ is a homeomorphism and $e^{iy} : [-\pi,\pi] \to S^1$ is surjective.

Given $z \in \mathbb C^\times$, let $x \in \mathbb R$ satisfy $e^x = \lvert z \rvert$ and $y \in [-\pi,\pi]$ satisfy $e^{iy} = z/\lvert z \rvert$. Then $e^{x+iy} = e^x \cdot e^{iy} = \lvert z \rvert \cdot z/\lvert z \rvert = z$.

3. $\bar{\mathcal E}$ is a homeomorphism.

We shall show that $\mathcal E$ is a closed map. This implies that it is a quotient map. Since $\bar{\mathcal E}^{-1} \circ \mathcal E$ is nothing else than the quotient map $q : \mathbb R \times [-\pi,\pi] \to X$ (which is continuous), we see that $\bar{\mathcal E}^{-1}$ is continuous (universal property of quotient maps!) which proves 3.

So let $A \subset \mathbb R \times [-\pi,\pi]$ be closed. We have to prove that $\mathcal E(A)$ is closed in $\mathbb C^\times$. Let $(z_n)$ be a sequence in $\mathcal E(A)$ which converges to some $z \in \mathbb C^\times$. We show that $z \in \mathcal E(A)$ which proves that $\mathcal E(A)$ is closed in $\mathbb C^\times$. Pick $(x_n,y_n) \in A$ with $\mathcal E(x_n,y_n) = z_n$. We have $e^{x_n} = \lvert \mathcal E(x_n,y_n) \rvert = \lvert z_n\rvert \to \lvert z \rvert$. Thus $x_n \to x := \ln \lvert z \rvert$. Since $[-\pi,\pi]$ is compact, $(y_n)$ has a covergent subsequence $(y_{n_k})$ with limit $y \in [-\pi,\pi]$. Hence $(x_{n_k},y_{n_k}) \to (x,y)$. Since $A$ is closed, we have $(x,y) \in A$. Since $\mathcal E$ is continuous, we get $\mathcal E(x_{n_k},y_{n_k}) \to \mathcal E(x,y) \in \mathcal E(A)$. But clearly $\mathcal E(x_{n_k},y_{n_k}) \to z$, thus $z \in \mathcal E(A)$.