Let $X$ be the half open interval $[0,2\pi)$. 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)

  1. Using compactness. If $f^{-1}$ was continuous then it contradicts the fact that continuous image of a compact set is compact.

  2. $\epsilon-\delta$ proof. (Idea: If we approach $(1,0)$ from below and above there is a jump from $0$ and $2\pi$.)

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.


Look at a neighbourhood of $0$ in $X=[0,2\pi)$, say $U=[0,1)$. Then $f(U)$ is not open in $Y$. As $(1,0)\in f(U)$ and every neighbourhood of $(1,0)$ in $Y$ contains points $(x,y)$ with $y<0$, but $f(U)$ has no such points then $U$ is not open. You need $f$ to be an open map in order for $f^{-1}$ to be continuous.


For $0<\epsilon <2\pi$ not that $(f^{-1}) ^{-1} [0,\epsilon)=f([0,\epsilon)$ is not open even though $[0,\epsilon)$ is open on $[0,2\pi)$


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