Prove that a map \(f:X\to Y\) between topological spaces is continuous if and only if \[f(\overline A)\subseteq\overline{f(A)}\] for all \(A\subseteq X\).
Prove that if \(f\) is continuous and \(f(\overline A)\) is closed for some \(A\subseteq X\) then \[f(\overline A)=\overline{f(A)}.\]
Solution
1. If \(f:X\to Y\) is continuous, then for every \(A\subseteq X\),\[ f(\overline A)\subseteq\overline{f(A)}. \]
Let \(x\in\overline A\). To show \(f(x)\in\overline{f(A)}\), let \(V\) be any open neighborhood of \(f(x)\) in \(Y\). Continuity of \(f\) makes \[ f^{-1}(V) \] an open neighborhood of \(x\). Since \(x\in\overline A\), this neighborhood meets \(A\); choose \[ a\in f^{-1}(V)\cap A. \] Then \[ f(a)\in V\cap f(A), \] so every open neighborhood of \(f(x)\) meets \(f(A)\). Thus \(f(x)\in\overline{f(A)}\).
2. Conversely, suppose that\[ f(\overline A)\subseteq\overline{f(A)} \]for every \(A\subseteq X\).Then \(f\) is continuous.
It is enough to show that the inverse image of every closed subset of \(Y\) is closed. Let \(C\subseteq Y\) be closed and put \[ A=f^{-1}(C). \] By the assumed closure property, \[ f(\overline A) \subseteq \overline{f(A)}. \] Since \(f(A)\subseteq C\) and \(C\) is closed, \[ \overline{f(A)}\subseteq C. \] Hence \[ f(\overline A)\subseteq C, \] which says \[ \overline A\subseteq f^{-1}(C)=A. \] The reverse inclusion \(A\subseteq\overline A\) always holds, so \(A=\overline A\) is closed. Therefore inverse images of closed sets under \(f\) are closed, and \(f\) is continuous.
3. This proves part (a):\[ f\text{ is continuous} \quad\Longleftrightarrow\quad f(\overline A)\subseteq\overline{f(A)} \text{ for all }A\subseteq X. \]
The forward implication is 1 and the converse is 2.
4. Suppose now that \(f\) is continuous and that \(f(\overline A)\) is closed for a fixed \(A\subseteq X\).Then\[ f(\overline A)=\overline{f(A)}. \]
By part (a), continuity gives \[ f(\overline A)\subseteq\overline{f(A)}. \] For the reverse inclusion, \(A\subseteq\overline A\) implies \[ f(A)\subseteq f(\overline A). \] The set \(f(\overline A)\) is closed by hypothesis, so it contains the closure of its subset \(f(A)\): \[ \overline{f(A)}\subseteq f(\overline A). \] Combining the two inclusions yields the desired equality.