Acyclic proper coloring of 2-degenerate graphs

Question

A proper vertex coloring of a graph $G$ is acyclic if there is no bicolored cycle. A graph is 2-degenerate if its every subgraph has a vertex of degree at most 2. I think every 2-degenerate graph has an acyclic proper vertex coloring using 3 colors, but I did not find any source stating this. Does anyone know this? Am I right? Thanks in advance.

Answer 1

Corollary 3 in https://dmtcs.episciences.org/344 says that for every graph $G$, if $G'$ is the 1-subdivision of $G$, then the acyclic chromatic number of $G'$ is at least $\sqrt{\frac12 \chi(G)}$. Apply this result with $G$ the complete graph $K_n$. Then the acyclic chromatic number of $K'_n$ is at least $\sqrt{\frac12 n}$. Since $K'_n$ is 2-degenerate, the acyclic chromatic number of 2-degenerate graphs is unbounded.

Answer 2

It is not true.

Let $G$ be a graph consisting of an independent set $I=\{v_1,v_2,v_3,v_4\}$ (blue below) and 18 additional vertices $u_{i,j}$, $u'_{i,j}$, and $u''_{i,j}$ where $i\ne j$ and each of the additional vertices is adjacent to the two vertices in $I$ with the same indexes (e.g. $u'_{1,3}$ is adjacent to $v_1$ and $v_3$; yellow below).

Then $G$ is 2-degenerate: each $u$ has degree two and after removing all of them the remaining set $I$ has degree zero.

For every proper 3-coloring of $G$ there exist $i\ne j$ such that $v_i$ and $v_j$ have the same color as each other, by the pigeonhole principle. For this $i$ and $j$, the three vertices $u_{i,j}$, $u'_{i,j}$, and $u''_{i,j}$ must have a different color than $v_i$ and $v_j$, but there are only two other colors to use for three vertices, so by the pigeonhole principle again two of them must have the same color. These two same-color vertices together with $v_i$ and $v_j$ form a bicolored 4-cycle.