A sufficient and necessary condition of a continuous map 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?

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.
