Understanding reduced suspension of $S^1$ I know this is just $S^2$. To see it, I use the CW structure of $S^1$ x $S^1$ , consisting of one 0-cell, two 1-cells and a 2-cell. Then since the reduced suspension is the cartesian product identifying the wedge (or smash product) , what just remains is the 0.cell and the 2-cell...This is very theoretical and I don't visualize what is going on.
I imagine a torus with two circular threads being pulled to a point and then?
If anyone has thought of this and is convinced by some visualization, I'll thank his idea....
 A: Here is a way to see this. After you collapse one circle, you end up with a sphere with two points attached. Then, the remaining circle (1-cell) is the image of the path connecting this two points. If you first glue the two points and then collapse the whole segmnent between them, is the same as collapsing the segment to a point from the very beginning. But collapsing an arc on a sphere does nothing, so you get a sphere. Alternatively, instead of attaching the two points, connect them with an external path. Then make the two points collide, and you get a bucket of a 2-sphere and a cirle. Then you have to collapse the circle to the point, since this is the second 1 cell, and you end up with a 2-sphere. 
A: I like to think about it this way: Start with a circle and cross with an interval rather than another circle, to get a cylinder. You're next going to identify the two ends of the cylinder and a path connecting them. If you can't picture that immediately, start by pinching off each end of the cylinder, to get a sphere, with a line connecting the two poles you just created. (This is the unreduced suspension.) Now shrink that line to a single point, which still gives you a sphere.
