It's known that $\mathbb{R}$ is a CDH space. Does anyone have any idea why $\omega^{\omega}$ is a CDH space? Then, Does anyone have any suggestion why $\mathbb{R}\times\omega^{\omega}$ is not a CDH space?
Also, the Cantor set $K$ in $[0,1]$ is CDH, and any compact metrizable space is continuous image of $K$. Then why this implies that continuous images of CDH-spaces need not be CDH?
Finally, note that $\mathbb{Q}$ is not CDH, so the property of being CDH is not hereditary.
Thanks
Theorem $5.2$ of this PDF says that every topologically complete separable metric space that is strongly locally homogeneous is CDH. A space $X$ is strongly locally homogeneous if it has a base $\mathscr{B}$ such that if $x,y\in B\in\mathscr{B}$, there is an autohomeomorphism $h$ of $X$ such that $h(x)=y$, and $h(z)=z$ for all $z\in X\setminus B$. Since $\omega^\omega$ is strongly locally homogeneous, it is also CDH.
To see why $\Bbb R\times\omega^\omega$ is not CDH, note that $\Bbb R$ is connected, $\omega^\omega$ is zero-dimensional, and autohomeomorphisms permute the connected components. The components of $\Bbb R\times\omega^\omega$ are the sets $\Bbb R\times\{x\}$ for $x\in\omega^\omega$. If $D$ is a dense subset of $\Bbb R\times\omega^\omega$ that has at most one point in each component, and $E$ is one that has more than one point in some component, there is no autohomeomorphism of $\Bbb R\times\omega^\omega$ that sends $D$ to $E$.
Finally, $[0,1]$ is the continuous image of $K$ but is not CDH.