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Say $f:X\to X$ is continuous and bijective. Then is the pre-image of a set $I\subset X$ under the function $f$ the same as the image of $I$ under the inverse function $f^{-1}$?

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  • $\begingroup$ yes, obviously. continuity is irrelevant here though. bijection is enough. $\endgroup$ – Karolis Juodelė Nov 8 '14 at 6:49
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Yes, this uses bijectivity and the definitions. By $g$ I will denote the inverse of $f$ (to avoid confusion with the general $f^{-1}[I]$ definition). So $f(x) = y$ iff $g(y) = x$, for all $x,y \in X$.

Then $x \in f^{-1}[I]$ iff $f(x) \in I$ iff $\exists y \in I: f(x) = y$ iff $\exists y \in I: g(y) = x$ iff $x \in g[I]$.

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