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  "body": "\n## Proof\n\nWe first show that preimages commute with complements.\n\n1. Let $X$ and $Y$ be [[topological spaces|0CNQXVBHCX]] and let $f \\colon X \\to Y$.\n2. Let $A \\subseteq Y$ and $x \\in X$. Then $x$ lies in the [[preimage|0FGCFY1AZX]] $f^{-1}(Y \\setminus A)$ if and only if $f(x) \\notin A$.\n3. This holds if and only if $x \\notin f^{-1}(A)$. So $f^{-1}(Y \\setminus A) = X \\setminus f^{-1}(A)$, with $X \\setminus f^{-1}(A)$ the [[complement|0B1FSS1VZW]].\n4. Suppose $f$ is [[continuous|0SY9ECQ19Q]], and let $F$ be a [[closed set|0G5AV9Q4RR]] of $Y$.\n5. By the definition of a closed set, $Y \\setminus F$ is open. By continuity, $f^{-1}(Y \\setminus F)$ is open in $X$.\n6. By step 3, $X \\setminus f^{-1}(F)$ is open. By the definition of a closed set, $f^{-1}(F)$ is closed.\n7. Conversely, suppose that each closed $F$ of $Y$ has a closed preimage. Let $V$ be open in $Y$.\n8. By [[complement of an open set is closed|0K67Q60Z8P]], $Y \\setminus V$ is closed in $Y$. By hypothesis, $f^{-1}(Y \\setminus V)$ is closed in $X$.\n9. By step 3, $X \\setminus f^{-1}(V)$ is closed. By the same result, $f^{-1}(V)$ is open.\n10. As $V$ was arbitrary, $f$ is continuous.\n\n## Idea\n\nClosed sets are the complements of open sets. Taking preimages commutes with taking complements. So the condition on open sets and the condition on closed sets say the same thing.\n",
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