Applications of the Fundamental Group.

May 8, 2010

What if someone came up to you and asked you if there were any way to take the real line and keep bending it around until we got the real plane.  I mean, it doesn’t seem all that far-fetched: maybe we could just keep bending it in half until we got something that looked like the plane!  Can we show that this is impossible?  After the cut.

Formally, this question is stated as: is there any homeomorphism (bijective, continuous, with a continuous inverse) from {\mathbb R} to {\mathbb R}^2?

Well, we first note that

\pi_{1}({\mathbb R}) = 0


\pi_{1}({\mathbb R}^2) = 0

so that’s a good sign.  But remember, we didn’t prove that spaces with isomorphic fundamental groups are homeomorphic; instead, we proved that homeomorphic spaces have isomorphic fundamental groups.  It seems like kind of the same thing, but it’s NOT, damnit.  It’s the converse.  So these two equivalences above tell us NOTHING about the spaces being homeomorphic.  Okay.  Okay.  Let’s calm down.  We can still figure something out.

Okay, let’s be clever.  If these spaces were homeomorphic, then if we removed a point from each, they would still be homeomorphic.  Think about this for a second.  Good.  Okay, now, if we remove a point from {\mathbb R}, what do we get?  We get two disjoint lines.   But {\mathbb R}^2 minus a point is not disjoint.  Commonly, we’d use  \pi_0 at this point to say that these spaces are now different (since \pi_0 tells us how many components a space has — we’ll talk about this a little later.) but we can use \pi_1 as well. 

First, let’s note that {\mathbb R}^2 minus a point is homomotopy equivalent to the circle (for now, all we care about is that their fundamental groups are the same, so that’s the only thing we’ll state.  it’s relatively easy to see this: the “puncture” in the plane acts the same way as the hole does in a circle.) S^1 and so we have that

\pi_{1}({\mathbb R}-\{(0,0)\}) = \pi_{1}(S^1) = {\mathbb Z}

but we note that whatever the fundamental group of the real line minus a point is, it will not be  {\mathbb Z}.  In fact, we have not gotten to a point where we can say what this is (in fact, I should have really talked about pointed spaces first; that is, spaces which have a distinguished point where every one of our rubber band loops start and end.) but if we think about this a little, this space is just going to be two disjoint points.  If one of these points is our distinguished point, then every loop that starts and ends there is the same (just a point) and so we have that the group is trivial.  Certainly not {\mathbb Z}

So what have we done?  We’ve assumed that the spaces were homeomorphic.  Then we’ve removed a point, but the spaces should remain homeomorphic after this.  If two spaces are homeomorphic, this means their fundamental groups are the same.  But we’ve shown that these spaces minus a point do not have the same fundamental groups.  Therefore, these removed-point spaces cannot be homeomorphic.  Therefore, the original spaces are not homeomorphic.

Using this exact same process (or one which is nearly identical), we can prove a little big more.   I leave this as an exercise!

Exercise:  Prove the following theorem:

Thm: {\mathbb R}^n is homeomorphic to {\mathbb R}^m if and only if m = n.

(Hint: Suppose n < m, without loss of generality.  What can we remove from {\mathbb R}^n that will be like removing a point from {\mathbb R}^1?  How does this bring us a contradiction?  It’s much easier to use \pi_0 in this proof.)

edit:  Thx to brk, for pointing out that the plane minus a point is not homeomorphic to the circle.  As she points out, try removing a point from each of these: for the circle, this becomes a line segment which has  trivial fundy group; for the plane, this becomes something which is homotopy equivalent to the figure 8, and so it certainly doesn’t have a trivial fundy group.

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