Homeomorphism and project space

[rainwyz0706]

1. (1) (a) Let X be a topological space. Prove that the set Homeo(X) of home-
omorphisms f : X → X becomes a group when endowed with the binary operation f ◦ g.
(b) Let G be a subgroup of Homeo(X). Prove that the relation ‘xRG y ⇔ ∃g ∈ G such
that g(x) = y ’ is an equivalence relation.
(c) Let G and RG be as in (b), and let p : X → X/RG be the quotient map. Prove that for every U open in X , p(U ) is open in X/RG .
2.(i) Let X = Rn \ {O} , n ≥ 2 , and let G be the subgroup of Homeo(X) composed of the
maps gλ (x) = λx. Prove that X/RG is the real projective space P Rn−1 .
(ii) Prove that the graph of the relation RG described in (i) is closed.
(iii) Prove that P Rn−1 is Hausdorff.

The setup was pretty straightforward. Could anyone give me some hints how to deal with later ones? Any input is appreciated!

 

[lippyka]

1. (a) To show that Homeo(X) is a group, we need to show that it is closed under composition and has an identity element. To show closure under composition, let f and g be two homeomorphisms in Homeo(X). Then, their composition, f ◦ g, is also a homeomorphism since it is the composition of two continuous functions. To show that there is an identity element, consider the identity map, id : X → X, which is a homeomorphism and is the identity element of Homeo(X). (b) To show that the relation 'xRG y ⇔ ∃g ∈ G such that g(x) = y' is an equivalence relation, we need to show that it is reflexive, symmetric, and transitive. To show that it is reflexive, let x be an arbitrary element of X. Then there exists the identity element id ∈ G such that id(x) = x, so xRGx. To show that it is symmetric, let x and y be elements of X such that xRGy. Then there exists a g ∈ G such that g(x) = y. Since G is a subgroup of Homeo(X), then g-1 ∈ G. Thus, g-1(y) = x, so yRGx. To show that it is transitive, let x, y, and z be elements of X such that xRGy and yRGz. Then there exist g, h ∈ G such that g(x) = y and h(y) = z. Thus, g ◦ h ∈ G, and (g ◦ h)(x) = z, so xRGz. Therefore, the relation is an equivalence relation. (c) Let U be an open subset of X. Then for any x ∈ U, there exists an open ball Bx around x such that Bx ⊂ U. Since p is a continuous map, then p(Bx) is an open ball around p(x) in X/RG. Thus, p(U) = {p(x) | x ∈ U} is open in X/RG. 2. (i) To show that X/RG is the real projective space PRn−