Q:

Why is $S^1\times\mathbb{R}^{n-1}$ the topology of $AdS_n$?

Anti-de Sitter $AdS_n$ may be defined by the quadric
$$-(x^0)^2-(x^1)^2+\vec{x}^2=-\alpha^2\tag{1}$$
embedded in ${\mathbb{R}^{2,n-1}}$, where I write ${\vec{x}^2}$ as the squared norm ${|\vec{x}|^2}$ of ${\vec{x}=(x^2,\ldots,x^n)}$. Now, I don't quite understand how is it justified that the topology of this space is $S^1\times\mathbb{R}^{n-1}$. As I understand it informally, I could write $(1)$ as
$$(x^0)^2+(x^1)^2=\alpha^2+\vec{x}^2\tag{2}$$
and then fix the ${(n-1)}$ terms ${\vec{x}^2}$, each one on $\mathbb{R}$, such that ${(2)}$ defines a circle ${S^1}$.
This is actually a reasoning I came up to later, based on the case of ${dS_n}$ (in which one just fixes the time variable) and when I saw what the topology was meant to be, but actually I first wrote ${(1)}$ as
$$\vec{x}^2=(x^0)^2+(x^1)^2-\alpha^2$$
which for fixed ${x^0,x^1}$, both in $\mathbb{R}$, defines a sphere ${S^{n-2}}$, so the topology would be something like ${S^{n-2}\times\mathbb{R}^2}$, (which is indeed similar to that of ${dS_n}$) right? I even liked this one better, since I could relate it as the 2 temporal dimensions on ${\mathbb{R}^2}$ and the spatial ones on ${S^{n-2}}$.
I don't really know topology, so I would like to know what is going on even if it's pretty basic and how could I interpret topological differences physically.
Update: I originally used $\otimes$ instead of $\times$ in the question. My reference to do this is page 4 of Ingemar Bengtsson's notes on Anti-de Sitter space; so is that simply a typo in the notes?
Update 2: I'm trying to understand this thing in simpler terms. If I write Minkowski 4-dimensional space in spherical coordinates, could I say that it's topology is ${\mathbb{R}\times{S}^3}$? If so, how come?

A:

Sketched proof: One may define a homotopy via the constraint 
$$x_0^2+x_n^2~=~ \alpha^2 +\lambda \sum_{i=1}^{n-1}x_i^2,\quad \alpha >0,$$
where $\lambda\in[0,1]$ is the homotopy parameter. Then $\lambda=1$ corresponds to $AdS_n \subset \mathbb{R}^{n+1}$, while $\lambda=0$ corresponds to $S^1\times \mathbb{R}^{n-1} \subset \mathbb{R}^{n+1}$.

