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## Main Question or Discussion Point

Over a map f that is continuous, and I believe it also has to be onto. I keep trying to come at this at different angles and can't seem to get anywhere.

So more formally...

If f:X→Y, and f is continuous and onto. Let A be a dense set in Y. Then f^1(A) is dense in X.

I'm not entirely sure how to do this. My approach was to let U be any open set in X such that U intersected with any neighborhood of f^1(a) such that a is in A. But I keep getting all mixed up.

Using the fact (from continuity) that every open neighborhood of V(f(x)) for x in X, there exists a U(x) such that f(U(x)) is contained in V(f(x)). Since A is dense in Y, there exists and a in A such that a is in V(f(x)). BUT this doesn't mean that f^1(a) is in U(x), as V(f(x)) could contain elements that are not mapped from f(U(x)).

However, if this map is continuous, does that mean I can get the f^1(a) in U(x), and since x is arbitrary f^1(A) intersect all open sets in X are nonempty, and this f^1(A) is dense? Or am I just way off in all regards?

So more formally...

If f:X→Y, and f is continuous and onto. Let A be a dense set in Y. Then f^1(A) is dense in X.

I'm not entirely sure how to do this. My approach was to let U be any open set in X such that U intersected with any neighborhood of f^1(a) such that a is in A. But I keep getting all mixed up.

Using the fact (from continuity) that every open neighborhood of V(f(x)) for x in X, there exists a U(x) such that f(U(x)) is contained in V(f(x)). Since A is dense in Y, there exists and a in A such that a is in V(f(x)). BUT this doesn't mean that f^1(a) is in U(x), as V(f(x)) could contain elements that are not mapped from f(U(x)).

However, if this map is continuous, does that mean I can get the f^1(a) in U(x), and since x is arbitrary f^1(A) intersect all open sets in X are nonempty, and this f^1(A) is dense? Or am I just way off in all regards?