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I am trying to prove this using the definition of continuity (and complements) where $f$ is continuous iff for every open set $U$, $f^{-1}(U)$ is also open. But I am uncertain about my proof when I try to write it formally. If somebody can present a formal proof, it will be really appreciated.

Thanks!

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A function is continuous if and only if the pre-image of an open set is open. Let $f:X\to Y$ be such that for every closed set $C$ in $Y$ the set $f^{-1}(C)$ is closed. Now, consider an open set $U$. Then $Y-U$ is closed by definition, so $f^{-1}(Y-U)$ is closed. But $f^{-1}(Y-U)=f^{-1}(Y)-f^{-1}(U)=X-f^{-1}(U).$ Since $X-f^{-1}(U)=(f^{-1}(U))^c$ is closed, its complement, i.e., $f^{-1}(U)$ is open, thus $f$ is continuous.

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    Suppose $x\in f^{-1}(Y-U)$. Then $f(x)\in Y-U$, i.e., $f(x)\notin U$, so $x\notin f^{-1}(U)$, hence $x\in f^{-1}(Y)-f^{-1}(U)$. Conversely, let $x\in f^{-1}(Y)-f^{-1}(U)$. Then $x\notin f^{-1}(U)$, so $f(x)\notin U$. Therefore $f(x)\in Y-U$ so we have $x\in f^{-1}(Y-U)$. This gives us containment in both directions, so we have equality.2012-12-07
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The two key facts are: (a) a set is open iff its complement is closed; and (b) inverse image preserves complements, i.e. if $f:X\to Y$, and $U \subseteq Y$, then $f^{-1}(U^c) = (f^{-1}(U))^c$.

Given this: suppose $U \subseteq Y$ is open. We need to show that $f^{-1}(U)$ is open; equivalently, that its complement is closed. But its complement is $(f^{-1}(U))^c = f^{-1}(U^c)$, the inverse image of a closed set, so is closed by the assumption on $f$.

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Suppose that $f:X\to Y$ is such that $f^{-1}[F]$ is closed in $X$ whenever $F$ is closed in $Y$. Let $U$ be an arbitrary open set in $Y$. Then $Y\setminus U$ is closed in $Y$, so $f^{-1}[Y\setminus U]$ is closed in $X$, and therefore $X\setminus f^{-1}[Y\setminus U]$ is open in $X$. Now show that $X\setminus f^{-1}[Y\setminus U]=f^{-1}[U]$, and you’re done.

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    @UH1: Excellent!2012-12-07
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Suppose $f:X \to Y$ satisfies the condition, and let $U \subset Y$ be an open set. Then $U^c$ is closed subset of $Y$, and so we have that $f^{-1}(U^c)$ is closed in $X$. But $f^{-1}(U^c) = f^{-1}(U)^c$, and so $f^{-1}(U)$ is open since its complement is closed.