Chromatic self-maps for almost disjoint families

Question

Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. We say that ${\cal A}\subseteq [\omega]^\omega$ is an almost disjoint family if $A \neq B \in {\cal A}$ implies $|A\cap B|< \aleph_0$.

Let $X\neq\varnothing$ be a set and let ${\cal E}\subseteq {\cal P}(X)\setminus\{\varnothing\}$ be a collection of non-empty subsets. We say that a map $f: {\cal E}\to X$ is a chromatic self-map if

1. $f(e) \in e$ for all $e\in {\cal E}$, and

2. if $e_1\neq e_2 \in {\cal E}$ and $e_1\cap e_2 \neq \varnothing$, then $f(e_1)\neq f(e_2)$.

Question. Does every almost disjoint family ${\cal A}\subseteq [\omega]^\omega$ have a chromatic self-map?

Remark. It suffices to answer the question for maximum almost disjoint families ("MAD families").

Answer 1

If such an $f$ exists then $\mathcal A_n=\{e\in\mathcal A:f(e)=n\}$ is a collection of pairwise disjoint subsets of $\omega$, so $\mathcal A_n$ is countable, so $\mathcal A=\bigcup_n\mathcal A_n$ is countable. So the answer is "no" if $\mathcal A$ is uncountable. Of course it is "yes" if $\mathcal A$ is countable.