homeomorphism non-example 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)!
 A: A generic way to obtain examples of bijective continuous functions whose inverse is not continuous is to fix some set $X$ and put two different topologies $T_1$ and $T_2$ on $X$, such that $T_2\subset T_1$. Now the identity function from $X$ to itself (with the topology being $T_1$ on the domain and $T_2$ on the codomain) will be continuous and bijective, but unless $T_1 = T_2$ it will not be a homeomorphism.
A: $$
(\theta\in [0,1))\mapsto((\cos\theta,\sin\theta)\in\text{circle}).
$$
This is continuous.
But as you move along the circle, $\theta$ has a jump discontinuity where it jumps from $1$ to $0$ or from $0$ to $1$ (depending on whether you're going clockwise or counterclockwise).
A: 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).
A: Constant functions are continuous.  So consider some space that has more than one point, such the circle $S = \{(\cos \theta, \sin\theta) \mid \theta\in\Bbb R\}$, considered as a subspace of $\Bbb R^2$.  A mapping $f:S\to\{x\}$ is continuous, but is not a homeomorphism, because it is not a bijection. And indeed the circle is not homeomorphic to a single point.
Now consider the mapping from the circle $S$ to the closed interval $[-1,1]$ where each point $(x, y)$ is mapped onto the value $y$.  This mapping is continuous but is not a homeomorphism (it is not invertible) and indeed the circle is not homeomorphic to the closed interval $[-1,1]$.
For an example which is a bijection, consider the mapping $f$ from the space $X = [0, 1) \cup [2,3]$, considered as a subset of the real line, defined as follows:  $$f(x) = \begin{cases} x & \text{if $x< 1$} \\
 x-1 & \text{if $x>1$}\end{cases}$$
This is a continuous bijection from $X$ to the interval $[0,2]$, but its inverse is not continuous and indeed $X$ is not homeomorphic to the interval $[0,2]$, because $[0,2]$ is connected and $X$ isn't.
