# Showing that $f^{-1}$ is not continuous.

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.

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 $$[0,2\pi)$$