Problem Set #5
Math 453 – Differentiable Manifolds
Assignment: Chapter 7 #1, 5, 7, 8, 9, 11
Clayton J. Lungstrum
February 20, 2013
Exercise 7.1
Let f : X → Y be a map of sets, and let B ⊆ Y . Prove that f (f −1 (B)) = B ∩ f (X).
Therefore, if f is surjective, then f (f −1 (B)) = B.
Solution.
Let y ∈ f (f −1 (B)). Then there is an x ∈ f −1 (B) such that f (x) = y. By definition, this
means f (x) = y ∈ B. Since x ∈ X, then we have f (x) ∈ B ∩ f (X), as desired. The reverse
containment follows analogously.
Suppose f is surjective; then f (X) = Y , so for any B ⊆ Y , B ∩ Y = B. Therefore,
f (f −1 (B)) = B, as was to be shown.
Q.E.D.
1
Exercise 7.3
Deduce Theorem 7.7 from Corollary 7.8.
Solution.
We want to show that a topological space S is Hausdorff if and only if the diagonal ∆ in
S × S is closed implies that for an open equivalence relation ∼, the quotient space S/ ∼ is
Hausdorff if and only if the graph R of ∼ is closed in S × S.
First, suppose that S/ ∼ is Hausdorff. Let [x] 6= [y] be elements of S/ ∼. Then there
exist disjoint open subsets U and V of S/ ∼ such that [x] ∈ U and [y] ∈ V . Then π −1 (U )
is open by the continuity of π, and similarly for π −1 (V ). Also note that they are disjoint,
hence (x, y) ∈ π −1 (U ) × π −1 (V ) ⊆ S × S is an open set disjoint from R. It follows that since
x and y were arbitrary, that R is closed.
Conversely, suppose R is closed. Then, for a point (x, y) ∈
/ R, there is an open set, say
U , containing it that does not intersect R. This implies there are open sets V and W such
that V × W ⊆ U and the point is in V × W . Thus, we get that x ∈ V and y ∈ W . Now, the
openness of π implies that π(x) ∈ π(V ) and π(y) ∈ π(W ) where π(V ) and π(W ) are open
disjoint sets, hence S/ ∼ is Hausdorff.
Q.E.D.
2
Exercise 7.5
Suppose a right action of a topological group G on topological space S is continuous; this
simply means that the map S ×G → S describing the action is continuous. Define two points
x, y of S to be equivalent if they are in the same orbit; i.e., there is an element g ∈ G such
that y = xg. Let S/G be the quotient space; it is call the orbit space of the action. Prove
that the projection map π : S → §/G is an open map. (This problem generalizes Proposition
7.14, in which G = R× and S = Rn+1 − {0}. Because R× is commutative, a left R× -action
becomes a right R× -action if scalar multiplication is written on the right.)
Solution.
Let U be an open subset of S. For each g ∈ G, since right
S multiplication by g is a
−1
homeomorphism S → S, the set U g is open. But π (π(U )) = g∈G U g, which is a union of
open sets, hence is open. By the definition of the quotient topology, π(U ) is open.
Q.E.D.
3
Exercise 7.7
(a) Let {(Uα , ϕα )}2α=1 be the atlas of the circle S 1 in Example 5.7, and let ϕα be the map
ϕα followed by the projection R → R/2πZ. On U1 ∩ U2 = A t B, since ϕ1 and ϕ2 differ
by an integer multiple of 2π, ϕ1 = ϕ2 . Therefore, ϕ1 and ϕ2 piece together to give a
well-defined map ϕ : S 1 → R/2πZ. Prove that ϕ is C ∞ .
(b) The complex exponential R → S 1 , t 7→ eit , is constant on each orbit of the action of
2πZ on R. Therefore, there is an induced map F : R/2πZ → S 1 , F ([t]) = eit . Prove
that F is C ∞ .
(c) Prove that F : R/2πZ → S 1 is a diffeomorphism.
Solution.
Q.E.D.
4
Exercise 7.8
Exercise 7.9
Show that the real projective space RP n is compact.
Solution.
By Exercise 11, we know it is homeomorphic to S n / ∼, which is the continuous image of
a compact set, hence RP n is compact.
Q.E.D.
5
Exercise 7.11
pP
i 2
For x = (x1 , . . . , xn ) ∈ Rn , let kxk =
i (x ) be the modulus of x. Prove that the map
n+1
n
f :R
− {0} → S given by
x
f (x) =
kxk
induces a homeomorphism f : RP n → S n / ∼.
Solution.
Define f : RP n → S n / ∼ by
x
f [x] =
.
kxk
It’s clear that this map is well-defined as multiplying it by any nonzero scalar would give
the positive or negative of the equivalence class, but since we’re in a linear subspace, it’s the
same. Now, by Proposition 7.1, the map f is continuous.
Now we can define g : S n → Rn+1 − {0} by g(x) = x and we see that it induces a map
g : S n → RP n which, by the argument above, is well-defined and continuous. Moreover,
g ◦ f [x] = [x]
and
f ◦ g [x] = [x].
Thus, we have that f and g are inverses to each other.
Q.E.D.
6