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Let $X,Y$ be 2 topological spaces. $f\colon X\to Y$ be a mapping. It is known that

$f$ is continuous iff

  • $f[\bar{A}]\subseteq\overline{f[A]}$
  • $f^{-1}[\bar{B}]\supseteq\overline{f^{-1}[B]}$
  • $f^{-1}[\mathring{B}]\subseteq\overbrace{f^{-1}[B]}^\circ$

It is normally to ask that whether $f[\mathring{A}]\supseteq \overbrace{f[A]}^\circ$ can be added in this list?

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    @BrianM.Scott I agree that the circle-on-the-head is heavy notation, and often ugly. It shows up on some old textbooks, which were printed when the authors did not have to typeset the book themselves :-)2012-09-26

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Great question: I really had no idea how it was going to go. The answer is that $f[A^\circ]\supseteq f[A]^\circ$ is neither sufficient nor necessary for $f$ to be continuous.


Counterexample to Sufficiency

Generalizing from an example given in the comments, just send $X$ discontinuously to a set with empty interior. Then every subset of $X$ goes to a set with empty interior as well, and the condition holds trivially. It's not hard to make the function discontinuous: as long as $X$ isn't discrete and its image intersects two different open sets (e.g. is at least two points of a Hausdorff space), we can do it.


Counterexample to Necessity

Let $f:(X,\tau)\rightarrow (X,\sigma)$ where $X=\{a,b,c\}, \tau=\{X,\{a\},\{b,c\}\}, \sigma=\{X,\{a\},\{b\},\{a,b\}\}.$ Set $f(b)=f(c)=b, f(a)=a$. Then $f$ is continuous, since $f^{-1}(\{a,b\})=X,$ $f^{-1}(\{b\})=\{b,c\}$, and $f^{-1}(\{a\})=\{a\}$. But on the other hand $\{a,b\}$ has empty interior and is mapped to an open set $\{a,b\}$.

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    Ok, that's all right.2012-09-26