Page 79 - DMTH503_TOPOLOGY

Unit 7: Continuous Functions

Theorem 4: Let (X, T) and (Y, U) be topological spaces. Notes
-1
Let f : (X, T)  (Y, U) be a map. Then f is continuous iff f (B) is open for every B  ,  being a base
for U on Y.
or
f is continuous iff the inverse image of each basic open set is open.
Proof: Let (X, T) and (Y, U) be topological spaces.
Let  be a base for U on Y. Let f : X  Y be a continuous map.
To prove that f (B) is open in X for every B  
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B  B  B  U
( B  U  B is open in Y.)
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 f (B) is open in X. Then f is continuous.
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Conversely, suppose that f : X  Y is map such that f (B) is open in X for each B  ,  being a
base for the topology U on Y. Let G  U be arbitrary. Then, by definition of base,
    s.t. G   B : B  
1 1

1

1
 f (G)  f  B : B  1 

1
  f (B) : B B 1 
= An arbitrary union of open subsets of X
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[ f (B) is open in X, by assumption]
 An open subset of X.
 f (G) is open in X
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Starting from an arbitrary open subset G of Y we are able to show that f (G) is open in X,
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showing thereby f is continuous.
Theorem 5: To show that a one-one onto continuous map f : X  X is a homeomorphism if f is
either open or closed.
Proof: For the sake of convenience, we take X = Y.

Suppose f : (X, T)  (Y, V) is one-one onto and continuous map. Also suppose that f is either open
or closed.
To prove that f is a homeomorphism, it is enough to show that f is continuous. For this we have
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to show that.

1


1
f (B)  f (B). For any set B  Y.
1

B  Y  f (B)  X is closed set
Also f is a closed map.



1
1
 f f (B)      f f (B)    


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