Does every continuous map between compact metrizable spaces lift to the Cantor set? I'm interested in the universal properties of the Cantor set. It is well-know that the Cantor set $2^\mathbb{N}$ is "universal" in the category of metrizable compact spaces, in the sense that every compact metrizable space is a quotient of $2^\mathbb{N}$ (although the quotient map is not unique).
My main question is: Does every continuous map $f:K_1\to K_2$ between compact metrizable spaces lift to a continuous map of the Cantor space (together with some quotient maps)?
If so, what else can we say about such a lift? For example, if $f$ is a homeomorphism, can we lift it to a homeomorphism of the Cantor set? What about group actions: Does every action of a group on a compact metrizable space by homeomorphisms lift to an action on the Cantor set (with a quotient map)?
 A: The answer to your main question is yes.  Let $f\colon K_1\to K_2$ be a map between compact metrizable spaces.  Let $q_1\colon C_1\to K_1$ and $q_2\colon C_2\to K_2$  be quotient maps, where $C_1$ and $C_2$ are Cantor spaces. Then the following diagram commutes.
$$\require{AMScd}
\begin{CD}
    C_1 @>\mathrm{incl.}>> C_1 \uplus C_2\\
    @V q_1 V V @VV (f\;\circ\; q_1)\;\cup\; q_2 V\\
    K_1 @>>f> K_2
\end{CD}$$
Here $C_1\uplus C_2$ denotes the disjoint union (or coproduct) of $C_1$ and $C_2$.  Note that $C_1\uplus C_2$ is itself a Cantor space.  The map $(f\circ q_1)\cup q_2$ is automatically a quotient map since it's a surjective map between compact Hausdorff spaces.
The answer is also yes for actions of finite groups.  Given a finite group $G=\{g_1,\ldots,g_n\}$ of homeomorphisms of a compact metric space $K$, let $q\colon C\to K$ be a quotient map, and consider the map $q'\colon G\times C\to K$ defined by
$$
q'(g,x) \;=\; (g\circ q)(x).
$$
Note that $G\times C$ is a Cantor space, being the product of a Cantor space with a finite set, and the map $q'$ is clearly a quotient map.  If we let $G$ act on $G\times C$ by $g_1(g_2,x) = (g_1g_2,x)$, then the map $q'$ is equivariant.
Note that this argument actually works for any topological group $G$ such that $G\times C$ is itself a Cantor space.  In particular, it works whenever $G$ is a separable  profinite group.
Edit: I originally had an argument that it works for any group $G$ that embeds into a profinite group, but I no longer think that argument is correct.  Thus, I have no idea whether the statement holds even for the case $G=\mathbb{Z}$.
