Functions which are Continuous, but not Bicontinuous

Solution 1:

A bijective map that is continuous but with non-continuous inverse is the following parametrization of the unit circle $\mathbb{S}^1$:

$$f: \colon [0, 2\pi) \to \mathbb{S}^1, \qquad f(\theta)=e^{i \theta}.$$

This map cannot have continuous inverse, because $\mathbb{S}^1$ is compact, while $[0, 2\pi)$ is not. Indeed, $f^{-1}$ jumps abruptly from $2\pi$ to $0$ when we travel round the unit circle.

Another example, somewhat similar in nature, is the map $g\colon [0,1] \cup (2, 3] \to [0, 2]$ defined by

$$g(x)=\begin{cases} x & 0 \le x \le 1 \\ x-1 & 2 < x \le 3 \end{cases}$$

The inverse map is $$g^{-1}(y)=\begin{cases} y & 0 \le y \le 1 \\ y+1 & 1 < y \le 2\end{cases}$$

and it is not continuous because of a jump at $y=1$. Note that, again, the range of $g$ is compact while the domain is not.

More generally, every bijective map $h\colon X \to K$ with $X$ non-compact and $K$ compact cannot have a continuous inverse.

Solution 2:

Define $f: [0,1) \cup [2,3] \rightarrow [0,2]$ by

$$f(x)=\begin{cases} x & x \in [0,1) \\ x-1 & x \in [2,3] \end{cases}$$

Solution 3:

Let $X$ be a set and $\tau_1,\tau_2$ two topologies on $X$ with $\tau_2\subsetneq\tau_1$. Then the identity function from the topological space $(X,\tau_1)$ to $(X,\tau_2)$ is a continuous bijection but the inverse function (the identity function from $(X,\tau_2)$ to $(X,\tau_1)$) is not continuous.