Denote $G=\left( V, E \right)$ arboricity by $a\left( G \right)$. I'm trying to understand why $G$ is $2\cdot a \left( G \right)$-colorable. I came across this post. Both the OP and the answer say that $E$ can be partitioned to $F_1 , ..., F_a$. Then, color $V_1$ is $2$ colors because its a forest. then, color $V_2 \backslash V_1$ in $2$ new colors because this is still a forest, and so on.
I agree that the amount of colors to be used will be $2\cdot a \left( G \right)$. What bothers me, is that each coloring of a single forest is proper when you look at the forest itself, but might not be proper once you look at the entire graph.
Let me give a partial example. Say we have $\left( u_1 , u_2 \right), \left( u_2 , u_3 \right), \left( u_1 , u_3 \right)$, and say $\left( u_1 , u_2 \right), \left( u_2 , u_3 \right) \in F_1$ and that $\left( u_1 , u_3 \right) \in F_2$. So $u_1,u_2,u_3$ are spanned by $F_1$. Hence, they will be colored in $2$ colors: $1$ and $2$. WLOG, $f\left( u_1 \right) = 1$, $f\left( u_2 \right) = 2$, $f\left( u_3 \right) = 1$. Now as we approach $F_2$, the vertices $u_1,u_2,u_3 \notin V_2 \backslash V_1$. therefore their color will not be altered.
So, the coloring was proper when our world was $F_1$. However, upon looking at the entire graph, our coloring is not proper, and that's what bothers me.