Download Topology Midterm 3 Solutions

Topology Midterm 3 Solutions
(1) Let Xi (i = 1, 2) be topological spaces and let xi ∈ Xi be an arbitrary point.
Let X1 × X2 be the product, with the product topology. Let πi : X1 × X2 →
Xi be the projections. Prove that
π1∗ × π2∗ : π1 (X1 × X2 , (x1 , x2 )) → π1 (X1 , x1 ) × π1 (X2 , x2 )
is an isomorphism of groups.
Solution: We saw in class that each πi∗ is a well-defined group homomorphism, so the map in question is certainly a well-defined group homomorphism (by the universal property of the product group, if you like). For
injectivity, suppose γ and γ 0 are loops in X1 × X2 based at (x1 , x2 ) with
([π1 γ], [π2 γ]) = ([π1 γ 0 ], [π2 γ 0 ]). Then we have a homotopy (of loops based at
xi ) Hi : I × I → Xi from πi γi to πi γ 0 (i = 1, 2). By the universal property
of the product X1 × X2 , the map H := H1 × H2 : I × I → X1 × X2 is
continuous and is clearly a homotopy (of loops based at (x1 , x2 )) from γ to
γ 0 . Surjectivity is similar: ([γ1 ], [γ2 ]) is the image of γ1 × γ2 .
(2) Prove that π1 (S 2 ) = {1}. You can use, without proof, the fact that any loop
in S 2 is homotopic (as a loop) to a loop which is not surjective. (This fact
is not so obvious! There are loops in S 2 which are surjective.)
Solution: By the hint, any loop is homotopic to a loop missing some point
x ∈ S 2 and S 2 \ {x} ∼
= R2 . Since every loop in R2 is contractible (by
straight-line homotopy), we conclude that every loop in S 2 is contractible.
(3) Prove that S 1 × S 1 is not homeomorphic to S 2 .
Solution: By the result of Problem 1 and the computation π1 (S 1 ) ∼
= Z in
1
1 ∼ 2
2
class, we have π1 (S × S ) = Z , while π1 (S ) = 0, so the spaces in question
have non-isomorphic fundamental groups.
Choose any two of these last four problems and do only those two!
P
2
(4) The group {±1} acts on S n = {x ∈ Rn+1 : n+1
i=1 xi = 1} by ±1 · x = ±x.
1
Prove that S /{±1} (with the quotient topology) is homeomorphic to S 1 . We
noted in class that the projection map S 2 → S 2 /{±1} =: RP2 is a covering
space. Use your knowledge of covering spaces and the fundamental group to
prove that π1 (RP2 ) = {±1} is the group with two elements. Conclude that
S 2 is not homeomorphic to RP2 .
Solution: Let q : S 1 → S 1 /{±1} be the quotient map. Making the standard
identification R2 = C, we can view S 1 as the set of complex numbers of
magnitude 1. Then f (z) := z 2 is a continuous map f : S 1 → S 1 constant on
the fibers of q, so, since q is a descent map, there is a unique continuous map
g : S 1 /{±1} → S 1 such that gq = f . This map g is bijective since z 2 = w2 iff
z = ±w, hence it is a homeomorphism because its domain is compact (it is
the image of the compact space S 1 under q) and its codomain is Hausdorff.
2
(5) For n ∈ Z>0 , let Sn1 ⊆ R2 be the circle of radius 1/(2n) centered at (1/(2n), 0).
Let X ⊆ R2 be the Hawaiian earring (the union of the Sn1 ). Prove that there
does not exist a covering space f : X̃ → X with X̃ simply connected.
Solution: No one did this right. Let f : X̃ → X be a covering space.
Take some neighborhood`
U of x0 := (0, 0) in X such that f −1 (U ) is a (nonempty!) disjoint union i Ui of open subsets of X̃ with f |Ui : Ui → U
a homeomorphism for each i. Pick n large enough that Sn1 ⊆ U . Define
r : U → Sn1 by setting r(x) := x if x ∈ Sn1 ⊆ U , r(x) := x0 otherwise. As we
noted in class, this r is a continuous retract of the inclusion Sn1 ,→ U .
Now pick any i and let x̃0 ∈ Ui be the unique lift of x0 in Ui . Let S̃n1 be
the lift (preimage) of Sn1 in Ui . Let r̃ : Ui → S̃n1 be the retract of S̃n1 ,→ Ui
corresponding to r under the homeomorphisms Ui ∼
= U , S̃n1 ∼
= Sn1 . The
compact space S̃n1 ∼
= S 1 is closed in X̃ since X̃ is Hausdorff (it is easy to see
that any covering space of a Hausdorff space is Hausdorff). Let g : X̃ \ S̃n1 →
S̃n1 denote the constant map onto the point r̃(x̃0 ) ∈ S 1 . The maps g and r̃ are
defined on open subspaces of X̃ covering X̃ and they agree on the overlap, so
they glue to a continuous map s : X̃ → S̃n1 ∼
= S 1 which is clearly a retract of
the inclusion S 1 ∼
= S̃n1 ,→ X̃, hence the fundamental group of X̃ must surject
onto π1 (S 1 ) ∼
= Z, so X̃ can be simply connected.
(6) A topological group is a set G equipped with a group structure and a topology
which are compatible in the sense that the multiplication map G × G → G
and the inverse map G → G are continuous. Given two loops γ, ν based at
1 ∈ G, define a new loop γ ⊗ ν based at 1 by (γ ⊗ ν)(t) := γ(t) · ν(t). (This
is continuous because multiplication is continuous.)
(a) Show that ⊗ is well-defined on homotopy classes (of loops based at 1)
and that (π1 (G, 1), ⊗) is a group.
(b) Show that, in fact, the group operation ⊗ is the same as the usual
“concatenation of loops” group operation on π1 (G, 1).
(c) Show that π1 (G, 1) is abelian.
Solution: Let 1 : I → G denote the constant loop based at 1 ∈ G.
(a) If H is a homotopy (of loops in G based at 1) from γ to γ 0 and
K is a homotopy from ν to ν 0 , then H · K is a homotopy from γ ⊗ ν to
γ 0 ⊗ ν 0 . Evidenlty 1 is the identity for ⊗. Associativity of ⊗ follows from
associativity of · and the existence of inverses for ⊗ follows from the existence
of a continuous inverse map for G (γ −1 (t) := γ(t)−1 ).
(b) It is clear from the definitions that
(γ ∗ 1) ⊗ (1 ∗ ν) = γ ∗ ν
(on the nose, not just up to homotopy). Since γ ∼ γ ∗ 1 and ν ∼ 1 ∗ ν, the
equality [γ ∗ ν] = [γ ⊗ ν] now follows from the invariance established in (a).
3
(c) Just compute
[ν] ∗ [γ] = [1 ∗ ν] ∗ [γ ∗ 1]
= [1 ∗ ν] ⊗ [γ ∗ 1]
= [γ] ∗ [ν].
(7) Suppose f : X̃ → X is a covering space with X̃ simply connected. Fix a
base point x0 ∈ X and a point x̃0 ∈ f −1 (x0 ). Prove that [γ] 7→ γ̃(1) (c.f. the
Path Lifting Lemma) is a well-defined bijection from the set of homotopy
classes of paths in X based at x0 to X̃. (This is supposed to motivate the
general “universal cover” construction we discussed in class.)
Solution: Well-definedness follows from the Homotopy Lifting Lemma (using uniqueness in the Path Lifting Lemma) and the fact that each fiber f −1 (x)
is totally disconnected, so there are no non-constant paths in any such fiber.
Injectivity follows from simple-connectedness of X̃: If γ̃(1) = ν̃(1), then
there must be a homotopy (of paths) H from γ̃ to ν̃, else γ ∗ ν would be a
nontrivial loop in X̃, then f H is a homotopy of path from γ to ν. Surjectivity
follows from path-connectedness of X̃.