Middle East Technical University Spring 1994 Math 471 Elementary Topology and Geometry

TOPOLOGICAL PROPERTIES AND CONSTRUCTIONS

Solutions

Homeomorphisms

Problem 1. (2 pts) Prove that the complement of a closed interval in ℝ^{2} is homeomorphic to a punctured ℝ^{2}.

Problem 2. (3 pts) Prove that two surfaces below are homeomorphic:

Pictures for solutions of Problems 1, 2 to be given in the class.

Product topology

Problem 3. (2 pts) Prove that any metric is continuous as a map X × X → ℝ.

Solution: Denote the metric by d. Continuity of d means that ∀(x, y) ∈ X × X, ∀ϵ > 0

there is a neiborhood U ∋ (x, y) in X × X such that d(x, y) − ϵ < d(U) < d(x, y) + ϵ. Consider U = B\frac{ϵ}{2} (x) × B\frac{ϵ}{2} (y).

Quotient topology

Problem 4. (4 pts) Prove that the set of continuous maps X/Y → Y is in a 1–1 correspondence with the set of continuous maps X → Y which are constant on the equivalence classes of S.

Solution: Given a map f: X/S → Y , define f̂: X → Y as a composition of the projection p: X → Y and f. In the cathegory of sets (without considering continuity problem) this map obviously yields the 1–1 correspondence in question. If f is continuous, then f̂ is continuous as a composition of continuous maps. The only problem is to prove that if f̂ is continuous then f is continuous too. Let U ⊂ Y is open and V = f^{−1}(U). Since p^{−1}(V) = f̂^{−1}(U) is open for continuous f̂, V must be open in X/S, by definition of the quotient topology.

Separation properties

Problem 5. (2 pts) Prove that X is a Hausdorf topological space if and only if the diagonal in X × X is a closed subset.

Solution: The diagonal Δ ⊂ X × X is closed iff any point (x, y) from its complement contains a neiborhood U such that U ∩ Δ = ∅. Without loss of generality we can assume that this neiborhood belongs to the basis of product topology, i.e. U = U_{x} × U_{y}, where U_{x}, U_{y} are neiborhood of x, y. Since U ∩ Δ = (U_{x} ∩ U_{y}) × (U_{x} ∩ U_{y}), the equivalence is obvious now.

Typeset by AMS-TEX

Compactness

Problem 6. (1 pt) Prove that a locally bounded (not necessarily continuous!) map f: X → ℝ is bounded for a compact X.

Solution: Consider covering of X by neiborhoods of its points s.t. f is bounded on each of them and choose a finite subcovering {U_{i}}_{i=1...n}. If |f| ≤ C_{i} on U_{i}, then |f| ≤ C =

max_{1,...,n}(C_{i}).

Problem 7. (2 pts) Is it true that if F_{1} ⊂ F_{2} are two topologies in X, then compactness of (X, F_{1}) implies the compactness of (X, F_{2}) ? And vice versa ?

Solution: Not true: consider antidiscreete topology F_{1} and discreete F_{2}. The opposite implication is true. To prove it, note, that the identity map (X, F_{2}) → (X, F_{1}) is continuous iff F_{1} ⊂ F_{2}, hence, compactness of (X, F_{2}) implies compactness of its image (X, F_{1}).

Connectedness

Problem 8. (1 pts) Prove that if A ⊂ X is a connected subset and A ⊂ B ⊂ Cl(A), then B is also connected.

Solution: Just to repeat the proof given for B = Cl(A).

Problem 9. (2 pts) Prove that S^{1} is not homeomorphic to a subset of ℝ.

Solution: S^{1} is connected and compact. The only compact and connected subsets of ℝ are known to be closed intervals, which are not homeomorphic to S^{1}. To see it, note, that puncturing of S^{1} at any point leaves it being connected, while intervals does not have this topological property.

Problem 10. (4 pts) Prove that if G is a topological group (a group with a topological structure such that the multiplication G × G → G is continuous), then the connected component of the unit is a normal subgroup.

Solution: First, note that ∀g ∈ G the map G → G, x ↦ gx is homeomorphism. It is continuous as the product of continuous G → G × G, x ↦ (g, x) and multiplication G × G → G; the inverse map x ↦ g^{−1}x is continuous for similar reasons. The map G → G, x ↦ xg^{−1} is homeomorphism as well. So, the image B = gAg^{−1} of the connected component A of the unit e ∈ G is connected. Since geg^{−1} = e, B ∋ e and, hence, B ⊂ A, becouse A is the maximal connected set, which contains e.

Problem 11. (4 pts) Prove that a locally constant map f: X → Y is constant for a connected X.

Solution: Note that f^{−1}(x) is an open set ∀x ∈ Y , if f is a local constant map, hence,

f^{−1}(A) is open ∀A ⊂ Y . If y ∈ f(X), and Y −f(X) ≠ ∅, then f^{−1}(y) and f^{−1}(Y −{y})

forms covering of X by disjoint nonempty open sets, to make contradiction with connectedness of X. Hence, f(X) = {y}.