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0 Let $X$ be the half open interval $[0,2pi)$ . Let $Y$ denote the unit circle in the plane. Let $f$ be the map defined by $f(t)=(cos(t),sin(t))$ . I checked that $f$ is continuous and bijective. Since it is bijective, inverse exist. I want to show that $f^{-1}$ is not continuous at $(1,0)$ using multiple ways(just to check whether I know concepts well or not) Using compactness. If $f^{-1}$ was continuous then it contradicts the fact that continuous image of a compact set is compact. $epsilon-delta$ proof. (Idea: If we approach $(1,0)$ from below and above there is a jump from $0$ and $2pi$ .) How to use the most general definition of continuity(that inverse image of open set is open) to show that $f^{-1}$ is not continuous. Thanks in advance. real-a...