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pivoxa15
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Homework Statement
THe quotient map f is open but is it also closed?
The Attempt at a Solution
I think it is. Consider f: X->Y
FOr every open set V in Y there exists by definition an open set f^-1(V) in X. There is a one to one correspondence between open sets in X and open sets in Y by definition.
So for every closed set V complement in Y there exists a closed set f^-1(V) complement in X. So f is both closed and open.
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