Download Solutions - www-math.utsc.utoronto.ca

Math 131: Final Exam Solutions
(1) Let X be a compact Hausdorff space and let ∼ be an equivalence relation
on X. Show that the quotient space X/ ∼ is Hausdorff if and only if the
subset {(x, y) ∈ X × X : x ∼ y} is closed in X × X.
Suppose first that the quotient Y = X/ ∼ is Hausdorff. Let ∆ ⊆ Y × Y
be the diagonal. Then ∆ is a closed subset of Y × Y . The natural map f :
X × X → Y × Y is continuous, so that f −1 ∆ = {(x, y) ∈ X × X : x ∼ y} is a
closed subset of X × X.
Conversely, suppose that S = {(x, y) ∈ X × X : x ∼ y} is closed. Let
p1 , p2 : X × X → X be projection onto the first and second factor, respectively,
and let q : X → Y be the quotient map. We first claim that the map q is
closed. Let K ⊆ X be a closed subset; we wish to prove that its image q(K)
is a closed subset of Y . Since q is a quotient map, it suffices to show that
q −1 q(K) = p2 (S ∩ p−1
1 K) is closed. Since X is compact, the map p2 is closed;
it therefore suffices to observe that S ∩ p−1
1 K is closed (since S is closed and
p−1
K
is
closed
by
virtue
of
the
continuity
of p1 ).
1
Now suppose that we are given distinct points x, y ∈ Y ; we wish to prove that
there exist disjoint open sets in Y containing x and y, respectively. Note that
q −1 {x} = {x0 ∈ X : (x, x0 ) ∈ S} is a closed subset of X, and therefore compact.
Similarly, q −1 {y} is a compact subset of X. Since X is Hausdorff, we can choose
disjoint open subsets U, V ⊆ X such that q −1 {x} ⊆ U and q −1 {y} ⊆ V . Let
U0 = {z ∈ Y : q −1 {z} ⊆ U } and let V0 = {z ∈ Y : q −1 {z} ⊆ V }. Then
U0 = Y − q(X − U ). Since q is a closed map, q(X − U ) ⊆ Y is closed and
therefore U0 is open. Similarly, V0 is open. Thus U0 and V0 are disjoint open
sets in Y containing x and y, respectively.
(2) Let S 2 = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 1} be the 2-sphere and S 3 =
{(w, x, y, z) ∈ R4 : w2 + x2 + y 2 + z 2 = 1} be the 3-sphere. Show that S 2
is not homeomorphic to S 3 .
Suppose there exists a homeomorphism h : S 2 → S 3 . Without loss of
generality (composing h with a rotation if necessary) we may assume that h
carries the North pole x = (0, 0, 1) ∈ S 2 to the North pole x0 = (0, 0, 0, 1) ∈ S 3 .
Then h induces a homeomorphism S 2 − {x} → S 3 − {x0 }. By stereographic
projection, we have homeomorphisms
S 2 − {x} ' R2
S 3 − {x} ' R3 .
1
We therefore obtain a homeomorphism h0 : R2 → R3 . Without loss of generality
(composing with a translation if necessary), we may assume that h0 carries the
origin 0 ∈ R2 to the origin 0 ∈ R3 . It follows that we obtain a homeomorphism
v
, log |v|) determines a homeoh00 : R2 − {0} → R3 − {0}. The map v 7→ ( |v|
3
2
morphism R − {0} ' S × R. We know from class that S 2 and R are simply
connected, so the product S 2 × R is also simply connected. Using the homeomorphism h00 , we deduce that R2 − {0} is simply connected. This contradicts
our calculation π1 (R2 − {0}) ' Z.
(3) Let G be a topological group, and let e ∈ G be the identity element. Show
that the fundamental group π1 (G, e) is abelian.
Let m : G × G → G be the multiplication on G. Since m is continuous, it
induces a group homomorphism
f : π1 (G, e) × π1 (G, e) → π1 (G × G, (e, e)) → π1 (G, e).
Let e : [0, 1] → G be the constant map taking the value e, and let [e] denote
its homotopy class (which is the identity element of π1 (G, e)). For any loop
p : [0, 1] → G, we have m(e(t), p(t)) = p(t) = m(p(t), e(t)), so that
f ([p], [e]) = [p] = f ([e], [p]).
Now let [p], [q] be arbitrary elements of π1 (G, e). Since f is a group homomorphism, we compute
[p][q] = f ([p], [e])f ([e], [q]) = f ([p][e], [e][q]) = f ([e][p], [q][e]) = f ([e], [q])f ([p], [e]) = [q][p],
so that the group π1 (G, e) is abelian.
(4) Let T 2 = S 1 × S 1 = {(z1 , z2 ) ∈ C × C : |z1 | = |z2 | = 1} be a torus. Define
an equivalence relation ∼ on T 2 so that (z1 , z2 ) ∼ (z10 , z20 ) if and only if
(z1 , z2 ) = (z10 , z20 ) or (z1 , z2 ) = (−z10 , z10 ), and let X be the quotient space
2
T 2 / ∼. Show that the fundamental group of X is not abelian.
Let p : T 2 / ∼→ X be the quotient map, and let q : R2 → T 2 be the map
given by (x, y) 7→ (e2πix , e2πiy ). The map q is also a quotient map, so that
(p ◦ q) : R2 → X is also a quotient map. Let G denote the group of all linear
transformations from R2 to itself having the form
(x, y) 7→ (x + m, y + n)
or
m, n ∈ Z
1
(x, y) 7→ (x + m + , −y + n)
m, n ∈ Z.
2
Note that G the action of G on R2 is topologically free: for any v ∈ R2 , the
open ball U = B 41 (v) does not intersect g(U ) for any g ∈ G. Note also that
two points v, w ∈ R2 have the same image in X if and only if v = g(w) for
2
some g ∈ G. It follows that we can identify X with the quotient space R2 /G.
Since R2 is simply connected, we deduce that there is a canonical isomorphism
π1 (X, x) ' G (for any base point x ∈ X). It now suffices to observe that the
linear transformations
1
(x, y) 7→ (x + , −y)
2
(x, y) 7→ (x, y + 1)
belong to G and do not commute.
3