k colorability and complete graph intuition

Theorem: A graph $\mathbb{G}$ is $k$-colorable if and only if there exists a homeomorphism $\phi:\mathbb{G} \to \mathbb{C}_k $ such that if $(u,v)$ is an edge in $\mathbb{G}$, then $(\phi(u),\phi(v))$ is an edge in the complete graph $\mathbb{C}_k $.

I don't find this intuitive. Can someone make this theorem more intuitive for me?

Solution 1:

Assign a different colour to each vertex of $\mathbb C_k$, $k$ colours in all. Then the second half of the theorem is just a fancy way of expressing the condition that adjacent vertices get different colours, for $(\phi(u),\phi(v))$ is not an edge in $\mathbb C_k$ iff $\phi(u)=\phi(v)$ iff $u,v$ have the same colour. Thus the theorem may be reworded "a graph is $k$-colourable iff its vertices can be assigned one of $k$ colours such that no adjacent vertices have the same colour", which is essentially the standard definition.