Continuous Maps and Products

  • Thread starter jmjlt88
  • Start date
  • #1
96
0
Let f: A -> B and g: C -> D be continuous functions.

Define h: A x C -> B x D by the equation

h(a,c)=(f(a),g(c)).​

Show h is continuous.


A few weeks ago I completed this exercise. Now, I am working on a problem that would be almost too easy if the converse of the above claim were true. I had trouble trying to construct a counterexample; so I tried to prove it.

Suppose h is continous. Let U be open in B. Then U x D is open in B x D, and by our assumption, h-1(U x D) is open in A x C. Since h-1(U x D) = f-1(U) x g-1(D), f-1(U) is open and f is continuous.

Does f-1(U) x g-1(D) being open in the product space imply f-1(U) is open in A?
 

Answers and Replies

  • #2
22,089
3,286
Are projections open maps?
 
  • #3
96
0
Yes.

Forgive my naivety, but how come the exercise isn't presented as an "if and only if" statement?

Perhaps I should examine these problems more closely. I originally completed this problem rather quickly; I then moved on without a second thought about if the converse is true. I should pay more attention.
 
  • #4
22,089
3,286
Yes.

Forgive my naivety, but how come the exercise isn't presented as an "if and only if" statement?

Perhaps I should examine these problems more closely. I originally completed this problem rather quickly; I then moved on without a second thought about if the converse is true. I should pay more attention.
The converse seems much less interesting, so I can understand why it is not an exercise. In either case, thinking about the converse is always a good practice.
 

Related Threads on Continuous Maps and Products

  • Last Post
2
Replies
25
Views
2K
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
942
  • Last Post
Replies
5
Views
802
Replies
2
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
5
Views
1K
Top