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

Question

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.

Answer 1

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 $\frac12$.