A homeomorphism is a continuous function between topological spaces that has a continuous inverse function.
Can someone provide examples of a continuous function between topological spaces that does not have a continuous inverse function?
It would help me grasp the intuition for what a homeomorphism is.
Feel free to use different topological spaces for your examples (but at least one with $\mathbb{R}$ would be nice)!
Best Answer
The standard example is an interval and a circle. Consider the interval $[0,2\pi)$ and the circle $S^1 = \{(x,y) \in \mathbb{R}^2 \, | \, x^2 + y^2 = 1 \}$ (with the induced topology as a subset of $\mathbb{R}^2$). You have a continuous map $\varphi \colon [0,2\pi) \rightarrow S^1$ given by $\varphi(\theta) = (\cos(\theta), \sin(\theta))$. This map is one to one and onto, but it doesn't have a continuous inverse.
The inverse map $\varphi^{-1} \colon S^1 \rightarrow [0,2\pi)$ takes a point $(x,y)$ on the circle and returns the angle the vector $(x,y)$ makes with the $x$-axis, in radians, in the range $[0,2\pi)$. This map is not continuous because if you look approach the point $(0,1)$ on the circle using points that have a positive $y$-coordinate, the angle $\varphi^{-1}$ will approach $0$ while if you approach the point $(0,1)$ with points having a negative $y$-coordinate, the angle $\varphi^{-1}$ will "approach $2\pi$" so the "left limit" and the "right limit" at $(0,1)$ are not equal for $\varphi^{-1}$.
(This is not entirely rigorous because we consider the map $\varphi^{-1}$ as a map with codomain $[0,2\pi)$ so $2\pi$ is not in the codomain and in fact when we approach $(0,1)$ with points having a negative $y$-coordinate, the angle $\varphi^{-1}$ won't have a limit in $[0,2\pi)$ at all! However, this is enough to show that $\varphi^{-1}$ can't be continuous).