$χ(G) ≤ k_1k_2$ if and only if $G = G_1 ∪ G_2$, where $χ(Gi) ≤ k_i$, $i = 1, 2$.'' (S.A. Burr) I don't have an idea how to do this. Can anyone help?
''Show that $χ(G) ≤ k_1k_2$ if and only if $G = G_1 ∪ G_2$, where $χ(Gi) ≤ k_i$, $i = 1, 2$.'' (S.A. Burr)
 A: Ok here is my idea. You use the colors on $G_1$ and $G_2$ to make pairs of colors and use these pairs as colors on $G$.
Assume you got two graphs $G_i, i= 1,2$ on a vertex set $V$ with $\chi_i=\chi(G_i)$. Choose colorings $c_i(v)$. Define the coloring $c(v):=(c_1(v),c_2(v))$ on $G$. You got at most $\chi_1\chi_2$ different pairs.

The other way around was a bit more tricky. Lets consider $\chi(G)\leq k_1k_2$ and choose a coloring. Partition the set $C$ of colors on $G$ into $k_1$ sets $C_1,...,C_{k_1}$, each one of at most $k_2$ colors. Lets say the colors in set $C_j$ are $c_1^{(j)},...,c_{K}^{(j)},K\leq k_2$. Now you can interprete the color $c_i^{(j)}$ as the pair $(c_i^1,c_j^2)$ with new colors $c_m^1,m=1,...,K$ and $c_n^2,n=1,...,K',K'\leq k_2$. Spoken simply, you decompose a color into two colors dependent on the set in which you put it and which number you assigned it inside the set. Now you can go backwards and construct $G_1$ and $G_2$ using only the colors $c_m^1$ and $c_n^2$ as seen in the figure below.

The partition of the color set naturally induces a partition of the vertex set $V$: two vertices belong to the same class, if and only if their respective colors belong to the same set. Then choose $G_1$ to only contain the inner edges of these vertex partition classes, and $G_2$ to only the edges between the classes.
Of course a rigorous proof would have to contain some reasoning that all these constructions yield feasible colorings, but I think the main task was having this idea of pairing colors.
