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Your explanation is correct. And, you can not do better than $f(a) = 2a$. For example, take a complete graph on $4$ vertices: $a,b,c,d$. The Arboricity is $2$ since $(a,b), (b,c),$ and $(c,d)$ forms the first tree, and the remaining edges form the second tree. The graph is colorable with exactly four colors. In fact, the Arboricity of a complete graph on $n$ ...