Is the Converse of a Continuous Function on Product Spaces Also Continuous?

[jmjlt88]

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?

 

[micromass]

Are projections open maps?

 

[jmjlt88]

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.

 

[micromass]

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.