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 That is, 

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(1)

9. Homeomorphisms(위상동형사상)

Homeomorphisms (위상동형사상)

Definition.  topological spaces 

  bijective일 때, 

  is a homeomorphism

⇔  is continuous and   is also continuous.

Remark : ① Every homeomorphism (and its inverse) is an open mapping :  is also open for any open  in .

 

    



 

That is,      (by def.) is open, since ⊇ ↦ , and   is continuous

② Similarly, every homeomorphism (and its inverse) is a closed mapping :  is closed for any closed set  in . That is,

     is closed, since   is continuous.

Definition:   injective, continuous;



  ->   is bijective. If  → happens to be homeomophism, then  is called a topological imbedding or just imbedding (embedding) of  onto 

(2)

e.g.

(1)   cos  

 sin  



     

  

ⓐ  is 1-1 and continuous ⊆  ->   is of the form

=  ∪  ∞∪ ∞  open  

ⓑ   equipped with the induced topology form ;   is also continuous by the reason as above.

ⓒ But,  → is not a homeomorphism with respect to the induced topology;

 

∩ 

 

  

 ∪  ∞∩ ∞ 

= as above "three connected components"

are connected component. In fact, when   ,

       : not open in the induced topology.

ⓓ Next, we put a different topology on  ;

(3)

 a sub-basis for a topology as above

~>     ∞ ∪

 

  

 ∪  ∞ not open ;

(2)      ∀ 

is continuous, 1-1, but not homeomorphism, since   is not continuous!

(3)

   cos sin 

1-1, continuous, onto

But   is not continuous ; e.g.

[///) )   

  

   

(4)

Constructing continuous functions Basic facts about continuous functions

 topological spaces ①   is continuous  ↦  = constant

② ⊆ ->   ↪ is continuous inclusion subspace

④   &  

 ⊇



 ->  

is continuous

⑦    is continuous of  can be written as union of open sets 

s.t.   is continuous for each ∝.

(5)

Proof: ① ∀  open of  in 

⇒    is open in  ∴  is continuous

②   open in  ⇒    ∩ ⇒

 def  is open in  ∴   ↪ is continuous.

③ ∀  open in 

~>   is open in  ->   

  conti   ∘   is open in 

④ 

↪ 

→ 

~>  

  ∘  &  and  are continuous.

~>  

is continuous

⑤  open in  ⇒   ∩   open in  ⇒      ∩

   ∩   

 ∩

 

  

  : open

∵  is continuous

⑥ ∼

   ∘    &  are continuous ~> ∼

 is continuous on 

⑦ ∀ open in    

 

 

 ∩

 

 

 

  : open in 

(6)

q.e.d.

Theorem (Pasting lemma):

∪  ⊆  closed (resp. open)  → 

 →

     ∈∩  or  

(7)

Then,  

   ∈   ∈ are continuous well-defined over ∪

Proof :  ⊂ closed subset of 

  ⇔  ∈ ⇔  ∈ or  ∈  

  ∈  ∈

⇔  ∈   or  , since  is continuous, from  to .

Thus,   is closed in 

     is closed in  Similarly, since  is continuous from  to ,

  is closed in     is closed in  ⇒  ∪   is closed in 

q.e.d.

As an application, let    → continuous. Here  is a totally ordered set with the order topology.

Let   min  be a function given by  

    ≥ 

    ≤ .

(i) well-defined, since  is defined to be    where

    

But the set ∈  ≤  is closed∗, and ∈  ≥  is also closed

(8)

is continuous

~>  is continuous on  by Pasting Lemma.

q.e.d.

In particular, in case of , we have

  min  

      

  max  

      

. Proof of

:

Claim :  ∈    is open in  (Assume    otherwise we're done)

Let ∈ Then    . We may assume without loss of generality that there should be some element ∈ s.t.

   .

(Otherwise,         to )

moreover,

 

  

∩⊆  ∀ ∈∩

   

⇒   

∴  ∈

Theorem  ×

 ↦  

Then,  is continuous ⇔  → and  → are both continuous.

(9)

Proof : ⇒ × 

are projections onto the 1st and 2nd

factors are continuous, and open mapping. Thus, ¡ ¡∘  is continuous

⇐ ∀×  basis element for the product ×  × is open ?

But,  ×   ∩   ⊆

 ∈  × ⇔    

∈× ⇔ ∈  ∩   ~>  × is open.

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