# Is $[0, 1) \times (0, 1] \to D \setminus \{ 0 \}$, $(\varphi, r) \mapsto r e^{2 \pi i \varphi}$ not a homeomorphism?

In our lecture, it was claimed that $$\Psi \colon [0, 1) \times (0, 1] \to D \setminus \{ 0 \}, \ (\varphi, r) \mapsto r e^{2 \pi i \varphi},$$ where $$D$$ denotes the unit disk in $$\mathbb R^2 \cong \mathbb C$$, is a homeomorphism. On Wikipedia it states that "The fundamental group is a homotopy invariant—topological spaces that are homotopy equivalent (or the stronger case of homeomorphic) have isomorphic fundamental groups.", which I understand to mean: "If $$f \colon X \to Y$$ is a homeomorphism, then $$\pi_1(X) \cong \pi_1(Y)$$", where $$X$$ and $$Y$$ are connected topological spaces, so the fundamental group $$\pi_1$$ is independent of its basepoint.

It is well known, that $$\pi_1(D \setminus \{ 0 \}) \cong \mathbb Z$$ and we should have $$\pi_1([0, 1) \times (0, 1]) \cong \{ e \}$$, where $$e$$ is the identity element, since $$[0, 1) \times (0, 1]$$ is a convex set. Is my reasoning correct?

Nevertheless, the map $$\Psi$$ seems to be continuous as composition of continuous maps, and I think the inverse is given by $$\Psi^{-1}(t) = \left( \frac{1}{2 \pi i} \log\left( \frac{t}{| t |}\right), | t |\right)$$ which looks continuous also.

• It's not a homeomorphism. The inverse map is not continuous. Commented Nov 17, 2020 at 17:49

Yes, your reasoning is correct. If you want a direct proof of the fact that $$\Psi$$ is not a homeomorphism, consider the sequence $$\left(1-\frac1n,\frac12\right)_{n\in\Bbb N}$$ of points of $$[0,1)\times(0,1]$$. It diverges in that space, but it is mapped by $$\Psi$$ into the sequence $$\left(\frac12e^{2\pi i(1-1/n)}\right)_{n\in\Bbb N}$$, which converges to $$\frac12$$.
• We can replace $\frac{1}{2}$ by $1$ in your example, right? Commented Nov 17, 2020 at 18:22
• Your construction of $\Psi^{-1}$ is correct. Your error lies in assuming that there is a continuous $\log$ function from $S^1$ into $\Bbb C$. There is no such function. Commented Nov 17, 2020 at 18:33
• So if I replace the interval $[0, 1)$ by $[0, 1] / (0 \sim 1)$, $\Psi$ becoems a homeomorphism? Commented Nov 17, 2020 at 19:03
• Yes. That's a way of saying that $D\setminus\{0\}$ is homeomorphic to $S^1\times(0,1]$. Commented Nov 17, 2020 at 19:25