Discussion
Let X, Y be metric spaces and \(f : X \to Y\) be a surjective continuous map. Then :
Answer: C
\(f\) maps dense subsets of X to dense subsets of Y.
Let D be dense in X. Continuity gives \(f(\overline{D}) \subseteq \overline{f(D)}\), so \(Y = f(X) = f(\overline{D}) \subseteq \overline{f(D)}\), using surjectivity. Hence \(f(D)\) is dense in Y.
The others fail. \(f(x) = x^{2}\) from R onto \([0, \infty)\) maps the open set \((-1, 1)\) to \([0, 1)\), which is not open (A). The projection \((x, y) \mapsto x\) from \(\mathbf{R}^{2}\) onto R maps the closed hyperbola \(xy = 1\) onto \(\mathbf{R} - \{0\}\), which is not closed (B), and it maps the nowhere dense x-axis onto all of R (D).
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