MATH 234, SPRING 2008HOMEWORK 5 SOLUTIONS

1. Prove that any two lifts of a continuous circle map (not necessarily a homeomorphism) f

must differ by an integer. That is, show that if F1, F2 are any two lifts of f , then thereexists an integer m such that F2(x) = F1(x) + m, for all x ∈ R.

Proof: Suppose that F1 and F2 are two lifts of the same circle map f . Then for all x ∈ R,

exp{2πiF1(x)} = f(e2πix) = exp{2πiF2(x)}.

It follows that, for every x, N(x) = F2(x)−F1(x) is an integer. Since F1 and F2 are continu-ous maps, so is N : R → Z. The image of a connected set by a continuous map is connected,so N(R) is a non-empty connected subset of Z. The only such subsets are singletons, so N

is constant. Therefore, F1 and F2 differ by an integer. �

2. Which of the following maps g : [0, 1] → R can be used to construct a lift F : R → R of anorientation preserving homeomorphism f : S1 → S1 of the circle?(a) g(x) = x2

(b) g(x) = x2 − 2x

(c) g(x) = 2x2 − x.

Proof: For a continuous map F : R → R to be a lift of a continuous circle map f : S1 → S1,F has to satisfy

F (x + m)− F (x) ∈ Z, (∗)

for every integer m and every x ∈ R. For F to be a lift of a homeomorphism, it needs to bestrictly monotonic and onto (and satisfy F (x+ 1)−F (x) ∈ {−1, 1}). Finally, in order for itto be a lift of an orientation preserving homeomorphism, it needs to be strictly increasingand satisfy F (x + 1) = F (x) + 1, for all x ∈ R.

In (a)–(c), the only x for which both x ∈ [0, 1] and x + 1 ∈ [0, 1] is x = 0, so it is enoughto check that g(1) − g(0) is an integer. This is the case in (a)–(c), so all the given mapscan be extended to lifts of circle maps. This can be done in the following way: define anextension G : R → R of g by

G(x) = g(x− n) + σn,

where n is the integer part of x (i.e., n ≤ x < n + 1) and σ = g(1)− g(0). Then G satisfies(∗), so it is a lift of a circle map.

Maps in (a) and (b) are both 1–1, whereas the map in (c) is not (g has a minimum at1/4), so it cannot be a lift of a homeomorphism. In (b), g(1) = g(0)− 1, so g is a lift of anorientation reversing homeomorphism. In (a), g(1) = g(0)+1, so g is a lift of an orientationpreserving homeomorphism. �

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3. Suppose that f : S1 → S1 is an orientation reversing homeomorphism and let F : R → Rbe a lift of f .(a) Show that F is strictly decreasing.(b) Show that F (x + 1) = F (x)− 1, for all x ∈ R.(c) Show that f has two fixed points.

Proof: Part (b) holds by definition.For part (a), suppose x < y and |x− y| < 1. We will show that F (x) > F (y). Clearly,

F (x) 6= F (y), since F is 1–1. Assume F (x) < F (y). Recall that F (x + 1) = F (x) − 1, soF (x + 1) < F (x). By the intermediate value theorem, there exists z ∈ (y, x + 1) such thatF (z) = F (x), violating injectivity of F .

If x < y, but y − x > 1, let n be the greatest integer ≤ y − x. Then x < y − n < x + 1,so by the previous discussion, F (x) > F (y− n) = F (y) + n > F (y). Therefore, F is strictlydecreasing everywhere.

(c) If p is a fixed point of f , then for any lift x of p there exists an integer n such thatF (x) = x + n. Therefore, fixed points of f are the projections of solutions to the family ofequations

F (x) = x + n,

for n ∈ Z. Without loss of generality, we can fix a lift F of f for which F (0) ∈ [0, 1). (Ifthis is not the case for some lift F , replace it by F −n, where n is the integer part of F (0).)

Since F is strictly decreasing, for every n ∈ Z its graph intersects the line y = x + n in asingle point (xn, xn + n). If F (0) = 0 (i.e., x0 = 0), then p1 = 1 is a fixed point of f . SinceF (1) = F (0) − 1 = −1, it is not hard to see that x−1 is forced to be in the interval (0, 1).Thus p2 = exp(2πix−1) is another fixed point of f , different from p1.

If 0 < F (0) < 1, then similarly x0, x−1 ∈ (0, 1) give rise to two different fixed points p1, p2

of f .Observe also that

F (x0 + n) = F (x0)− n = x0 − n = (x0 + n)− 2n

andF (x−1 + n) = F (x−1)− n = x−1 − 1− n = (x−1 + n)− (2n + 1),

for every n ∈ Z. Thus by definition

x2n = x0 + n and x2n+1 = x−1 + n. (1)

The lift of any fixed point of f has to equal xn, for some n ∈ Z. By (1), the set {xn : n ∈ Z}projects to exactly two distinct points on the circle, namely, p1, p2. This implies that f hasexactly two fixed points. �