Impossible to lift the identity map on the circle

[PsychonautQQ]

Suppose that L: $S^1$ ---> $R$ is a lift of the identity map of $S^1$, where e is the covering map from $R$ to $S^1$, where $R$ is the real numbers and $S^1$ is the circle.

Then the equation e * L = $Id_{S^1}$ (where * is composition) means that 2*pi*L is a continuous choice of angle function on the circle. it is intuitively evident that this cannot exist, because any choice of angle function would have to change by 2*pi as one goes around the circle, and thus could not be continuous on the whole circle.How does the angle function changing by 2*pi as one goes around the circle imply that it could not be continuous on the whole circle?
P.S. Crossing fingers for LaTeX to work out...
P.S.S. Woot!

 

[WWGD]

Suppose that L: $S^1$ ---> $R$ is a lift of the identity map of $S^1$, where e is the covering map from $R$ to $S^1$, where $R$ is the real numbers and $S^1$ is the circle.

Then the equation e * L = $Id_{S^1}$ (where * is composition) means that 2*pi*L is a continuous choice of angle function on the circle. it is intuitively evident that this cannot exist, because any choice of angle function would have to change by 2*pi as one goes around the circle, and thus could not be continuous on the whole circle.How does the angle function changing by 2*pi as one goes around the circle imply that it could not be continuous on the whole circle?
P.S. Crossing fingers for LaTeX to work out...
P.S.S. Woot!

You need to have a topology defined on $S^1$ to be able to talk about continuity. Are you seeing $S^1$ as a metric space with the chordal? Subspace? etc. metric?

 

[andrewkirk]

Rather than using angles, I find it easier to approach the proof as follows:

Since $e\circ L$ is an injection, $L$ must be too. Therefore $L$ is a bijection from $S^1$ to I am $L\subset \mathbb R$.

Since $e\circ L=Id_{S^1}$, $e^\dagger\equiv e|_{{\mathrm Im}\ L}$ is an inverse map of $L$ and, as a submap of a covering map, is continuous. Hence $\mathbb R$ contains a homeomorphic image I am $L$ of $S^1$. But $S^1$ is compact and connected, so I am $L$ must be too (by continuity of $L$), from which we can deduce (omitting several steps) that I am $L$ is a closed interval.

But the fundamental group of a closed interval is the trivial group of one element, whereas the fundamental group of $S^1$ is isomorphic to the integers, so the image cannot be homeomorphic, which gives a contradiction, and we conclude that $L$ cannot be continuous.

On reflection, that last step uses some heavy machinery - fundamental groups - that may not be necessary. Let's try without that.

If L is continuous then we have a closed interval $C=[u,v]\subset \mathbb R$ that is homeomorphic to $S^1$ via $L$ and $e^\dagger$ that are each other's inverses. Can we get a contradiction from that?
Hint: think of how we can surgically 'disconnect' an interval and whether the same procedure works on a circle.

 

[WWGD]

Along what Andrew was saying, elaborating on both approaches, it if we use the induced map $f_{*}$ on the (first) fundamental group, we have $f_{*}\ [a] := \[f{a]]=0$ Then use the fact that if f is a covering map then the induced map of a covering map is injective, to get a contradiction. EDIT:

The map is given by $p(x):=(sin 2\pi x, cos2 \pi x )$. Pull back an open set in the right ( or, equiv. left) half of $S^1$ to get a union of open intervals, each homeomorphic to $S^1$ (by definition of covering map) and restrict it to a compact set. Then the restriction to a compact set is a homeomorphism , using the simple but powerful result that a continuous bijection from Compact into Hausdorff is a homeomorphism, which cannot happen by what Andrew said, the connectivity number. Much nicer idea than my much clunkier proof.