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Prove that a map $f:X\to Y$ is a continuous iff $$\text{cl}f^{-1}(A)\subseteq f^{-1}(\text{cl}A)$$ for each $A\subset Y$, where $X$ and $Y$ are both topological spaces and $\text{cl}$ denotes the closure.

In fact,for any $A\subset Y$,we have $A\subset \text{cl}A$ where $\text{cl}A$ is closed in $Y$, thus $\text{cl} f^{-1}(A)\subseteq f^{-1}(\text{cl}A)$. But how about the converse?

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Let $A$ be closed in $Y$. This can be expressed by $A=\text{cl}A$. By hypothesis $f^{-1}(A)=f^{-1}(\text{cl}A)\supseteq \text{cl}f^{-1}(A)$. Can you go on from here?

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  • $\begingroup$ Since $f^{-1}(A)\subset clf^{-1}(A)$,thus $f^{-1}(A)$ is closed in $X$,hence $f$ is continuous.But $A$ is selected in $Y$ arbitrarily,how is that? $\endgroup$ – mathon Jun 27 '13 at 15:08
  • $\begingroup$ $A$ was an arbitrary closed set in $Y$ and we have shown that the preimage is closed in $X$. $\endgroup$ – Stefan Hamcke Jun 27 '13 at 15:21
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Suppose $f$ is continuous. Then the preimage of any closed set is closed, and since $f^{-1}(A) \subseteq f^{-1}(\overline{A})$, where $f^{-1}(\overline{A})$ is a closed set, it follows that

$$\overline{f^{-1}(A)} \subseteq f^{-1}(\overline{A})$$

For the converse, take $A \subseteq Y$ to be a closed set. Since $A = \overline{A}$, we have

$$f^{-1}(A) = f^{-1}(\overline{A}) \supseteq \overline{f^{-1}(A)}$$ and clearly, $f^{-1}(A) \subseteq \overline{f^{-1}(A)}$ so $f^{-1}(A)$ is closed.

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  • $\begingroup$ how can you guarantee that there are closed sets oin $Y$? in every topological space there are closed sets? $\endgroup$ – user286485 Aug 25 '16 at 18:43

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