Suppose that $f$ is a continuous function, $K$ is a compact set, and $K=f^{-1} (\cap_n U_n)$, where each $U_n$ is open. Can I conclude that $K=\cap_n f^{-1}(U_n)$? In other words, is it true that $f^{-1} (\cap_n U_n)=\cap_n f^{-1}(U_n)$? If not, is there an additional condition (maybe that $f^{-1}$ has to be one to one) that would make this true somehow?
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It is true. http://math.stackexchange.com/questions/359693/overview-of-basic-results-about-images-and-preimages – Math.StackExchange Mar 19 '15 at 03:53