Ripping Points Out of the Plane: A Homology Adventure.
February 26, 2011
[This homology adventure will, unfortunately, have no pictures. I want to make some, but my bamboo pad is being irritable.]
Here’s the game: Take the plane. I remove, say, points from the plane. You tell me the homology groups of the resulting space.
Well, is this all that hard? If you think in terms of cells, this might be a bit difficult: how do I make the plane with cells? If you think in terms of singular maps, this is even more complicated — but, isn’t it always? This is one of those times (and there are many of these times!) where thinking about homotopy equivalences is going to go far.
We note the following: we can deform retract the plane minus points to the wedge of circles. Think about this for and from there you’ll see the pattern. Now, at this point, the problem becomes easy: the cellular homology states that for , and the abelianization of the fundamental group will just be , or you can just compute this directly from the CW structure resulting in the same solution.
Yes, that is interesting and all, but what about if instead of points, we removed every rational coordinate on the x-axis. What do we get then? Do we get a wedge of infinitely many circles? Is this still a CW-complex (consider the definition of a CW complex, and think about how many cells a compact set can intersect…)? How do we take the homology of this?
Further, remove every point which is in . What is the homology of this space?
What if we were to remove the set of points where both coordinates are irrational. Is the homology of this space the same as the one just mentioned in the last paragraph? If not, how is it different?
If this last question got you bummed out, what about just removing ? Does this give you the same homology as removing ? Make sense of this intuitively.