- #1
jmjlt88
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Let f: A -> B and g: C -> D be continuous functions.
Define h: A x C -> B x D by the equation
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?
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?