Example Sheet 7

MA3F1 Introduction to Topology November 11, 2019

Example Sheet 7

Please let me (Martin) know if any of the problems are unclear or have typos. Please turn in a solution to one of Exercises (7.1) or (7.4) by 12:00 on 21/11/2019, to the dropoff box in front of the undergraduate ofﬁce. If you collaborate with other students, please include their names.

For the next three problems we need the following deﬁnition. Suppose X is a topological space. We deﬁne CX to be the cone on X: that is,

CX = X × I/(x, 1) ∼ (y, 1) for all x, y ∈ X.

The point a = [(x, 1)] is called the apex of the cone.

(7.1) Equip the integers Z with the discrete topology. Show that CZ is homeomorphic to the wedge sum of a countable collection of unit intervals at 1.

Solution (7.1) The disjoint union of a countable collection of unit intervals is deﬁned to be S {n} × I = × I, the latter with the product topology. The wedge sum n∈Z Z ∨n∈ZI at 1 is then deﬁned in the same way as the cone CZ.

2 (7.2) Let In ⊂ R to be the line segment connecting (0, 1) to (n, 0), for n ∈ Z. Set D = ∪n∈ZIn and equip D with the subspace topology. Show that CZ is not homeomorphic to D.

Solution (7.2) Suppose that they are homeomorphic via q : D → CZ. Note that q sends the apex to the apex (removing the apex gives a space with a countable number of connected components in both cases, and the apex is the only point with this property). Also, q|In is a homeomorphism to its image. Deﬁne the set U ⊂ D where U ∩ In is the half-open interval containing the apex and having length 1/|n|. Note that U is not open in D. However, q(U) is open in CZ. Could one simply argue that U ∩ In is not open in the subspace topology, but its image is open in the cone topology?

(7.3) For any space X, show that CX is contractible. Deduce that π1(CX, a) is trivial.

Hint: You may use the following fact: if X is a topological spaces, and Y = X/ ∼ a quotient space, then Y × I =∼ (X × I)/ ∼0, where (x, t) ∼0 (y, s) if x ∼ y and t = s. That is, the product topology of the quotient and the interval is the same as the quotient topology of the product (you may try to prove this!)

1 MA3F1 Introduction to Topology November 11, 2019

Solution (7.3) Construct a homotopy CX×I → CX by ﬁrst constructing a homotopy H :(X × I) × I → CX and then identifying,

(X × I) × I H q×IdI CX × I G CX

By the Hint, the resulting map G is continuous. As a map, take H((x, s), t) = [(x, st)]. This map induces a homotopy G from IdCX to the apex.

(7.4) Suppose G and H are nontrivial groups. Show that the free product G ∗ H is not isomorphic to Z2.

Solution (7.4) The free product of two non-trivial groups is not abelian.

(7.5) Suppose that {Gα} is a countable collection of countable groups. Show that ∗αGα is countable.

Solution (7.5) The disjoint union of the Gα, less their identity elements, is countable. Thus the set of reduced words is countable. Since this is the underlying set for ∗αGα, we are done.

2 For the next two problems we need the following deﬁnition. Let Cn ⊂ R be the circle of radius 1/n centered at (1/n, 0) ∈ R2. We deﬁne H ⊂ R2, the Hawaiian ∞ earring, to be the union H = ∪n=1Cn, equipped with the subspace topology. We take H to be a pointed space, with basepoint at h = (0, 0). Let Γ = π1(H, h).

(7.6) For all n > 0 give a retraction rn : H → Cn. Explain why rn is continuous. Show that Γ = π1(H, h) is uncountable. Brieﬂy explain why Γ is not isomorphic to

1 ∼ π1 ∨n∈NS = ∗n∈NZ.

Solution (7.6) The retraction rn is defined by sending Cm to h, for all m 6= n, and fixing Cn pointwise. The only thing to check is that this is continuous. Fix U ⊂ Cn an open interval. If h 6∈ U then the preimage of U is exactly U. Since U is open in Cn it is open in H and we are done. If h lies in U then the preimage is H − (Cn − U). This is again relatively open. Consider two infinite binary strings {an} and {bn}. We use these strings to code elements of π1. These elements are equal in π1 if and only if the strings are identical - this is because all retractions must have the same image. (Actually, see Hatcher Example 1.25) One is countable, the other is not. Note that the spaces are not homeomorphic for a more basic reason - H is compact, ∨S1 is not.