Proving that $ \chi(G) = \omega(G) $ if $ \bar{G} $ is bipartite. I know that $ \chi \! \left( \bar{G} \right) = 2 $ and that $ \chi(G) \geq \alpha \! \left( \bar{G} \right) $, but how can I conclude that $ \chi(G) = \omega(G) $?
First, it’s always true that $\chi(G)\ge\omega(G)$, so your real problem is to show that $\chi(G)\le\omega(G)$ if $\overline{G}$ is bipartite. Note that it need not be true that $\chi(\overline{G})=2$: the chromatic number of a bipartite graph is $1$ if the graph has no edges. In that case $G$ is a complete graph, and the result is trivial, so I’ll assume that $G$ is not complete.

Suppose that $\overline{G}$ is bipartite with parts $V$ and $W$. Without loss of generality we may assume that all isolated vertices of $\overline{G}$ belong to $V$, and that $|V|\ge|W|$. Clearly $V$ is a clique in $G$. Color each vertex of $V$ with a different color. Now observe that for each $w\in W$ there is at least one $v\in V$ such that $vw$ is an edge of $\overline{G}$ and hence such that $wv$ is not an edge of $G$, so that $w$ can be given the same color as $v$. Thus, $\chi(G)\le|V|\le\omega(G)$.
suppose if $w \in W$ is adjacent to every vertex of V in G then  we cannot color that w with that and  cardinality of V and W are less tahn or equal to clique of G complement not exactly equal to . @gcolucci\in