I hope this doesn't turn out to be a silly question.
There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms. But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnected spaces.
My Question: Is there a connected topological space $X$ and a continuous bijection $X\to X$ that is not a homeomorphism?
For the record, my example of a continuous bijection $X\to X$ that is not a homeomorphism is the following. Roughly, the idea is to find an ordered family of topologies $\tau_i$ (
$i\in \mathbb Z$) on a set $S$ and use the shift map to create a continuous bijection from $\coprod_{i\in \mathbb Z} (S, \tau_i)$ to itself. Let $S = \mathbb{Z} \coprod \mathbb Z$. The topology $\tau_i$ is as follows: if $i<0$, then the left-hand copy of $\mathbb Z$ is topologized as the disjoint union of the discrete topology on $[-n, n]$ and the indiscrete topology on its complement, while the right-hand copy of $\mathbb Z$ is indiscrete. The space $(S, \tau_0)$ is then indiscrete. For $i>0$, the left-hand copy of $\mathbb Z$ is indiscrete, while the right-hand copy is the disjoint union of the indiscrete topology on $[-n, n]$ with the discrete topology on its complement. Now the map $\coprod_{i\in \mathbb Z} (S, \tau_i)\to \coprod_{i\in \mathbb Z} (S, \tau_i)$ sending $(S, \tau_i) \to (S, \tau_{i+1})$ by the identity map of $S$ is a continuous bijection, but not a homeomorphism.
Best Answer
Here's a nice geometric example. Let $X\subset\mathbb{R}^2$ be the union of the $x$-axis, the line segments $\{n\}\times[0,2\pi)$ for $n\in \{\ldots,-3,-2,-1,0\}$, and circles in the upper half plane of radius $1/3$ tangent to the $x$-axis at the points $(1,0),(2,0),\ldots$.
Note that $X$ is connected.
Define a map $f\colon X\to X$ by $$ f(x,y) \;=\; \begin{cases}(x+1,y) & \text{if }x\ne 0 \\ \left(1+\frac{\sin y}{3},\frac{1-\cos y}{3}\right) & \text{if }x=0\end{cases}. $$ That is, $f$ translates most points to the right by $1$, and maps the line segment $\{0\}\times[0,2\pi)$ onto the circle that's tangent to the $x$-axis at the point $(1,0)$. Then $f$ is continuous and bijective, but is not a homeomorphism.