$X:=\{\cos\theta(0,...,0,e^{2\pi it/m})+\sin\theta(z_1,...,z_{n-1},0):t \in [0,1], \theta \in [0,\pi/2], \sum |z_k|^2=1 \}\cong D^{2n-1}$ Let $m>1$ be a fixed positive integer, and consider the subset $X:=\{\cos\theta(0,...,0,e^{2\pi it/m})+\sin\theta(z_1,...,z_{n-1},0) \in \Bbb C^n : t \in [0,1], \theta \in [0,\pi/2], |z_1|^2+...+|z_{n-1}|^2=1 \}$
of $S^{2n-1}$.
How can I show that this space is homeomorphic to the $2n-1$ dimensional closed unit disk $D^{2n-1}$?
This is from an example in Hatcher's Algebraic Topology, giving a CW complex structure on a lens space. That $X$ is homeomorphic to $D^{2n-1}$ is written without proof, but I can't verify it.
 A: The set $S = \{(z_1,\dots,z_{n-1})\in \mathbb C^{n-1} \mid \sum_{i=1}^{n-1} \lvert z_i \rvert^2 = 1 \}$ is nothing else than the standard unit sphere in $\mathbb C^{n-1} = \mathbb R^{2n-2}$, i.e. we have $S = S^{2n-3}$.
Consider the map
$$f : [0,1] \times [0,\pi/2] \times S^{2n-3} \to X \subset \mathbb C^n, \\f(t, \theta,(z_1,\dots,z_{n-1})) = \cos\theta(0,...,0,e^{2\pi it/m})+\sin\theta(z_1,...,z_{n-1},0) .$$
This is a continuous surjection with compact domain and Hausdorff range, thus it is a closed map and therefore a quotient map. Note that if $t \ne t'$, then $e^{2\pi it/m} \ne e^{2\pi it'/m}$. Let us write $\mathbf z = (z_1,...,z_{n-1})$.
We claim that $f(t, \theta,\mathbf z) = f(t', \theta',\mathbf z')$ if and only if $(t, \theta,\mathbf z) = (t', \theta',\mathbf z')$ or [$\theta = \theta' = 0$ and $t = t'$] or [$\theta = \theta' = \pi/2$ and $\mathbf z = \mathbf z'$].
The "if" part is trivial. For the "only if" part observe that


*

*$f$ is obviously injective on $[0,1] \times (0,\pi/2) \times S^{2n-3}$.

*If exactly one of $\theta, \theta'$ is $0$, then $f(t, \theta,\mathbf z) \ne f(t', \theta',\mathbf z')$ because on one side of the equation the first $(n-1)$ coordinates are $0$, but on the other side they are not all $0$.

*If exactly one of $\theta, \theta'$ is $\pi/2$, then $f(t, \theta,\mathbf z) \ne f(t', \theta',\mathbf z')$ because on one side of the equation the $n$-th coordinate is $0$, but on the other side it is not $0$.

*If $\theta = \theta' = 0$ and $t \ne t'$, then $f(t, \theta,\mathbf z) \ne f(t', \theta',\mathbf z')$ because the $n$-th coordinate is not the same on both sides.

*If $\theta = \theta' = \pi/2$ and $\mathbf z \ne \mathbf z'$, then $f(t, \theta,\mathbf z) \ne f(t', \theta',\mathbf z')$ because the first $(n-1)$ coordinates are not the same on both sides.
Thus $X$ is homeomorphic to the quotient space $[0,1] \times [0,\pi/2] \times S^{2n-3} / \sim$, where $(t, \theta,\mathbf z) \sim (t', \theta',\mathbf z')$ iff $(t, \theta,\mathbf z) = (t', \theta',\mathbf z')$ or [$\theta = \theta' = 0$ and $t = t'$] or [$\theta = \theta' = \pi/2$ and $\mathbf z = \mathbf z'$]. This space is nothing else than the join $[0,1] * S^{2n-3}$ as defined in Hatcher p.9.
But we have $[0,1] \approx P * P$, where $P$ is single point space. Thus
$$[0,1] * S^{2n-3} \approx (P * P) * S^{2n-3} \approx P * (P * S^{2n-3}) = P * C S^{2n-3} \approx P * D^{2n-2} = CD^{2n-2} \\ \approx D^{2n-1} .$$
