algebraic-topology
What is an easy way to see that torus with two points removed is homotopy equivalent to wedge of three circles?
I am trying to see it by viewing the torus as a square with two sides identified, but the intuition isn't clear to me, unlike the case with one point removed.
I am trying to see it as each triangle being retracted to its boundary, but I have no idea how that results in wedge of three circles.
Thanks!
It's a bouquet of $n+1$ circles, so the free group on $n+1$ generators. Think of a rectangle with $n-1$ horizontal lines across it. Roll it up into a tube, so you have a line segment with $n+1$ circles attached to it. Then identify the top and bottom circles. So you have a circle with $n$ circles attached to it which is homotopic to a bouquet of $n+1$ circles.
Yes. Taking the wedge sum with a circle is the same as identifying two points (up to homotopy, with a nice space like the sphere which is homogeneous).
Best Answer
You can actually see the three circles in this picture! Look at the three edges incident to the bottom-left corner in your picture. These all become circles after we do the identifications, and they meet at a single point (all four vertices of the rectangle become the same point after identifications). And, after identifications, these three edges are the entire space!