Constructing a homeomorphism $S^1\times S^1/S^1\vee S^1\to S^2$ I want to construct a homeomorphism  $(S^1\times S^1)/(S^1\vee S^1)\to S^2$, by giving a quotient map that is constant on $S^1\vee S^1$, and "when cutting it out" is a bijection on $S^2$.
I tried the following.
We have an easy quotient map $[0,1]\to S^1$ given by $t\mapsto (\sin(2\pi t), \cos(2\pi t))$.
I tried to give a quotient map $S^1\times S^1\to S^2$ using sphere coordinates.
Sending $(\sin(2\pi t),\cos(2\pi t)), (\sin(2\pi s),\cos(2\pi s))\mapsto (\sin(2\pi t)\cdot \cos(2\pi s), \sin(2\pi t)\cdot\sin(2\pi s),\cos(2\pi\cdot t))$, but this should not yield a bijection.
Has someone an idea for a quotient map?
Thanks in advance.
 A: Let us first understand what $(S^1 \times S^1)/(S^1 \vee S^1)$ looks like.
Define $p : [-1,1] \to S^1, p(t) = e^{\pi it}$ and
$$P =  p\times p :  S = [-1,1] \times  [-1,1] \to S^1 \times S^1 .$$
It is a quotient map (because it is a closed map, its domain  being compact and its range Hausdorff).
$P$ identifies all pairs of points $(t,\pm 1)$ to the point $(p(t), -1)$ and all pairs of points $(\pm 1,s)$ to the point $(-1,p(s))$. Thus, letting $\partial S  = [-1,1] \times \{\pm 1\} \cup \{\pm 1\} \times [-1,1]$ denote the boundary of the square $S$, we see that $P(\partial S) = S^1 \times \{-1\} \cup  \{-1\} \times S^1 = S^1 \vee S^1 \subset S^1 \times S^1$.
Let $q : S^1 \times S^1 \to (S^1 \times S^1)/(S^1 \vee S^1)$ denote the quotient map. Then $Q = q \circ P$ is a quotient map which identifies  $\partial S$ to a single point. In other words, $(S^1 \times S^1)/(S^1 \vee S^1)$ is nothing else than $S/\partial S$.
Thus we have to find a homeomorphism $S/\partial S \to S^2$. This is not too hard.

*

*There exists a homeomorphism $g : S \to D^2 = \{\xi = (x,y) \in \mathbb R^2 \mid \lVert \xi \rVert = \sqrt{x^2+y^2} = 1\}$. It maps $\partial S$ to $S^1$ and induces a homeomorphism $\bar g : S/\partial S \to D^2/S^1$.
This is intuitively clear. For a formal proof see Showing that $\gamma$ is a homeomorphism betrween $[-1,1]\times [-1,1]$ and $D^2$.


*There exists a homeomorphism $h : D^2/S^1 \to S^2$.
This is also intuitively clear, but technically complicated. For a formal proof see my answer to Equivalence between two homotopy groups definition.
Remark:
What about using spherical coordinates as you suggested in your question?
It suffices to consider the square $S' = [0,2\pi] \times [0,\pi]$ and to find a homeomorphism $S'/ \partial S' \to S^2$.
Working with spherical coordinates leads us to define
$$H : S' \to S^2, H(\theta,\phi) = (\cos \theta \sin \phi,\sin \theta \sin \phi,\cos \phi) .$$
It is well-known (though not completely trivial) that $H$ is surjective. We have $H([0,2\pi] \times \{0\}) = \{(0,0,1)\}$ and $H([0,2\pi] \times \{\pi\}) = \{(0,0,-1)\}$. Moreover, $H$ maps the line segments $\{0\} \times [0,\pi]$ and $\{2\pi\} \times [0,\pi]$ to the semicircle $C \subset S^2$ in the half-plane $x \ge 0, y = 0$ connecting $(0,0,1)$ and $(0,0,-1)$.
Thus $H$ does not identify $\partial S'$ to a point. However, one can show that $S^2/C \approx S^2$ which would gives another proof.
