What is the way to see $(S^1\times S^1)/(S^1\vee S^1)\simeq S^2$? What is the way to see $S^1\times S^1/(S^1\vee S^1)\simeq S^2$? Even just an intuitive walkthrough. I can't visualize this quotient in my head.
 A: Alternatively, one can use the minimal CW complex structure of the sphere to show the following stronger result:
$$
(S^n \times S^m) / (S^n \vee S^m) = S^{n+m}.
$$
$S^n$ has one $0$-cell and one $n$-cell. $S^m$ has one $0$-cell and one $m$-cell. Hence, the product $S^n \times S^m$ has one cell at each of the dimensions $0$, $n$, $m$, $n + m$.
We can identify $S^n \vee S^m$ with the result of attaching the $n$ and $m$-cells to the $0$-cell. By collapsing $S^n \vee S^m$ to a point, we are left with one $0$-cell and one $(n+m)$-cell. This is $S^{n+m}$.
A: You can view $S^1\times S^1$ as a torus which can be represented by a rectangle with opposite edges identified with the same orientation (see below). 
$\hspace{4.5cm}$
Now $S^1\vee S^1$, the wedge sum of two circles, is precisely the boundary of this rectangle (the four corners are identified with a single point which is the point where the two circles are joined). This may be clearer to see on the torus itself (see below). 
$\hspace{5.8cm}$
So $(S^1\times S^1)/(S^1\vee S^1)$ is the rectangle where we identify the entire boundary with a single point; this is one way to view the two dimensional sphere, as the one-point compactification of $\mathbb{R}^2$.
