examples-counterexamplesgeneral-topology
We know that continuous bijections between compact and Hausdorff spaces are homeomorphisms. I.e. if $(X,\tau_{X})$ is a compact topological space, $(Y,\tau_{Y})$ is a Hausdorff space and $f:(X,\tau_{X})\rightarrow(Y,\tau_{Y})$ is a continuous bijective function, then $f$ is a homeomorphism.
However, does this hold if we relax the assumption that $(Y,\tau_{Y})$ is a Hausdorff space, i.e. only a topological space? If not, provide a counterexample.
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.
There's an error in your statement: you need the domain to be compact, and the range to be connected.
Easy counterexample: restrict the exponential map $\mathbb{R} \rightarrow S^1$ to some interval like (0, 1.5) so that the fibers have different cardinalities.
Best Answer
Take $X$ to be a finite set ($|X|>1$) with discrete topology and $X'$ be the same set with indiscrete topology.
Consider $i : X \to X'$, the identity function.
Clearly, $X$ is compact (as it is finite). $i$ is clearly bijective and continuous (you cannot demand more open sets in the left space).
But, obviously, this is not a homeomorphism.