Homeomorphic

 

Def 1

Topological spaces $X$ and $Y$ are homeomorphic if

there exists 1-1 onto function $f : X \to Y$  s.t.

$f$  and  $f^{-1}$ are continuous

 

We write $X \cong Y$

$f$ : homeomorphsim

 

 

 

Thm 1

Homeomorphic is equivalence relation

 

  Show 1 : $X \cong X$

  Define $f : X \to X$ by

  $f(x) = x$

  It is clear that $f$ is 1-1 onto.

 

  Let $V$ be open set in $X$.

  Since $f^{-1}(V) = V$,

  $f^{-1}(V)$ is open set in $X$

  $\therefore$  $f$ is continuous

 

  Since $f^{-1} = f$,

  $f^{-1}$ is continuous 

 

  $\therefore$  $X \cong X$

 

  end

 

 

  Show 2 : $X \cong Y$  $\Rightarrow$  $Y \cong X$

  Since $X \cong Y$, there exists 1-1 onto function $f : X \to Y$  s.t.

  $f$  and  $f^{-1}$ are continuous

 

  Since $f$ is 1-1 onto,

  $f^{-1}$ is 1-1 onto

  Also

  $f^{-1}$  and  $(f^{-1})^{-1} = f$ are continuous

 

  $\therefore$  $Y \cong X$

 

  end

 

 

  Show 3 : $X \cong Y$, $Y \cong Z$  $\Rightarrow$  $X \cong Z$

  Since $X \cong Y$, there exists 1-1 onto function $f : X \to Y$  s.t.

  $f$  and  $f^{-1}$ are continuous

  Since $Y \cong Z$, there exists 1-1 onto function $g : Y \to Z$  s.t.

  $g$  and  $g^{-1}$ are continuous

 

  Since $f$, $g$ is 1-1 onto,

  $g \circ f$ is 1-1 onto

  Also by Thm 1,

  $g \circ f$  and  $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ are continuous

 

  $\therefore$  $X \cong Z$

 

  end

 

 

 

Thm 2

Let $X$, $Y$ be topological space  and  $X \cong Y$. Then

1.  $X$ is Hausdorff  $\Leftrightarrow$  $Y$ is Hausdorff

2.  $X$ is metrizable  $\Leftrightarrow$  $Y$ is metrizable

 

pf)

$\Rightarrow)$

  $\forall a, b \; (a \neq b) \in Y$

 

  Since $X \cong Y$, there exists 1-1 onto $f : X \to Y$  s.t.

  $f$  and  $f^{-1}$ are continuous

  Since $f$ is 1-1 onto,

  $f^{-1}(a) \neq f^{-1}(b)$

 

  Since $X$ is Hausdorff, there exists open set $U, V$ in X  s.t.

  $f^{-1}(a) \in U$,  $f^{-1}(b) \in V$,  $U \cap V = \varnothing$

  $\therefore$  $a \in f(U)$,  $b \in f(V)$

 

  Since $f^{-1}$ is continuous,

  $(f^{-1})^{-1}(U) = f(U)$,  $(f^{-1})^{-1}(V) = f(V)$ are open set in $Y$

  Also

  $f(U) \cap f(V) = f(U \cap V) = \varnothing$

 

  $\therefore$  $Y$ is Hausdorff

 

$\Leftarrow)$

  You can prove it in a similar way as above.

 

$\Rightarrow)$

  Since $X$ is metrizable, there exists metric $d : X \times X \to \mathbb{R}$

  Since $X \cong Y$, there exists 1-1 onto $f : X \to Y$  s.t.

  $f$  and  $f^{-1}$ are continuous

 

  Define $D : Y \times Y \to \mathbb{R}$ by

  $D(a,b) = d(f^{-1}(a), f^{-1}(b))$

 

 

  Show 1 : $D(a, b) \geq 0$     &     $D(a, b) = 0$  $\Leftrightarrow$  $a = b$

  Since $d$ is metric,

  $D(a, b) = d(f^{-1}(a), f^{-1}(b)) \geq 0$

 

  Also

  $D(a, b) = 0$  $\Leftrightarrow$  $d(f^{-1}(a), f^{-1}(b)) = 0$

                            $\Leftrightarrow$  $f^{-1}(a) = f^{-1}(b)$

                            $\Leftrightarrow$  $a = b$

 

  end

 

 

  Show 2 : $D(a, b) = D(b ,a)$

  Since $d$ is metric,

  $D(a, b) = d(f^{-1}(a), f^{-1}(b))$

                  $= d(f^{-1}(b), f^{-1}(a))$

                  $= D(b, a)$

 

  end

 

 

  Show 3 : $D(a, b) + D(b, c) \geq D(a, c)$

  Since $d$ is metric,

  $D(a, b) + D(b, c) = d(f^{-1}(a), f^{-1}(b)) + d(f^{-1}(b), f^{-1}(c))$

                                       $\geq d(f^{-1}(a), f^{-1}(c))$

                                       $= D(a, c)$

 

  end

 

  $\therefore$  $D$ is metric on $Y$

  $\therefore$  $Y$ is metrizable

 

$\Leftarrow)$

  You can prove it in a similar way as above.