# Math Help - quotient space

1. ## quotient space

Let X be a topological space and A a subset of X . On X × {0, 1} deﬁne the
partition composed of the pairs {(a, 0), (a, 1)} for a ∈ A , and of the singletons {(x, i)} if x ∈ X \ A and i ∈ {0, 1}.
Let R be the equivalence relation deﬁned by this partition, let Y be the quotient space
[X × {0, 1}] /R and let p : X × {0, 1} → Y be the quotient map.

(1) Prove that there exists a continuous map f : Y → X such that f ◦ p(x, i) = x for every x ∈ X and i ∈ {0, 1} .
Prove that Y is Hausdorﬀ if and only if X is Hausdorﬀ and A is a closed subset of X .

(2) Consider the above construction for X = [0, 1] and A an arbitrary subset of [0, 1].
Prove that Y is compact. Prove that K = p(X × {0}) and L = p(X × {1}) are compact, and that K ∩ L is homeomorphic to A .

Any input is appreciated!

2. Originally Posted by raulk
Let X be a topological space and A a subset of X . On X × {0, 1} deﬁne the
partition composed of the pairs {(a, 0), (a, 1)} for a ∈ A , and of the singletons {(x, i)} if x ∈ X \ A and i ∈ {0, 1}.
Let R be the equivalence relation deﬁned by this partition, let Y be the quotient space
[X × {0, 1}] /R and let p : X × {0, 1} → Y be the quotient map.

(1) Prove that there exists a continuous map f : Y → X such that f ◦ p(x, i) = x for every x ∈ X and i ∈ {0, 1} .
What do you think?

Prove that Y is Hausdorﬀ if and only if X is Hausdorﬀ and A is a closed subset of X .
c
The key point to notice is that the restriction $p^{*}X-A)\times\{0,1\}\to Y" alt="p^{*}X-A)\times\{0,1\}\to Y" /> is an injection, and thus if $U\subseteq (X-A)\times\{0,1\}$ is open then so is $p(U)$ since $U=p^{-1}(p(U))$

(2) Consider the above construction for X = [0, 1] and A an arbitrary subset of [0, 1].
Prove that Y is compact.
Uhh... $X$ is compact and evidently $X\times\{0,1\}$ is compact and $Y=p\left(X\times\{0,1\}\right)$ which is continuous?

Prove that K = p(X × {0}) and L = p(X × {1}) are compact, and that K ∩ L is homeomorphic to A .
$X\times\{0,1\}$ is compact and both $x\times\{0\}$ and $X\times\{1\}$ are closed subspaces and so automatically compact. And since $K,L$ are the continuous images of compact spaces they are trivially compact.

Think about the other part. If $x\in X-A$ then $p((x,1))=\{(x,1)\}\notin p\left(X-A\times\{0\}\right)$ and so you can think of $K\cap L$ as being $A\times\{0,1\}$ but treating $(a,0)$ and $(a,1)$ as the same point. Doesn't this sound eerily familiar to $A\times\{0\}$ which is naturally homeomorphic to $A$?