Complement of Even Cycles are Perfect 
Show that complement $\overline{C_n}$ of an n-cycle ($n \ge 4$) is perfect if and only if $\overline{C_n}$ is even without
the weak perfect graph thm

So $\overline{C_n}$'s chromatic number is $\left\lceil
 {\large{\frac{n}{2}}}
 \right\rceil
 $ according to a well-written answer here. And $\overline{C_n}$'s clique number is $\left\lfloor
 {\large{\frac{n}{2}}}
 \right\rfloor
 $ so this shows that $\omega(\overline{C_n})=\chi(\overline{C_n})$ since floor only equals ceiling when n is even.
But this doesnt conclude the proof as I need to show the proper induced subgraphs H of $\overline{C_n}$ also has $\omega(H)=\chi(H)$. I have been stuck for this part for quite a long time so any help is appreciated.
 A: Lemma. If $G$ is the complement of a bipartite graph, then $\chi(G) = \omega(G)$.
Proof. Let $G$ have $n$ vertices, and let $M$ be a maximum matching in $\overline{G}$. By König's theorem, $\overline{G}$ has a vertex cover $U$ with $|M|$ vertices. The complement of $U$ is an independent set in $\overline{G}$ with $n - |M|$ vertices, so it is a clique in $G$ with $n - |M|$ vertices; therefore $n - |M| \le \omega(G)$.
We can color $G$ with $n-|M|$ colors as follows. Use $|M|$ colors to color all vertices covered by $M$, by using the same color on both endpoints of each edge in $M$ (which are adjacent in $\overline{G}$, so they are not adjacent in $G$). Then, use $n-2|M|$ more colors to give each remaining vertex its own color. This shows $\chi(G) \le n-|M| \le \omega(G)$; we already know $\chi(G) \ge \omega(G)$ for all graphs, so they are equal. $\qquad\square$
In the question, since $C_n$ is bipartite, any subgraph of $C_n$ is also bipartite, so the lemma applies to any induced subgraph of $\overline{C_n}$. If you do not want to apply König's theorem, we can skip it for this particular case, by finding $M$ and $U$ explicitly:

*

*Any induced subgraph of $C_n$ is a union of paths; let them have $n_1, n_2, \dots, n_k$ vertices.

*We can find a matching of size $\lfloor \frac{n_1}{2}\rfloor + \dots + \lfloor \frac{n_k}{2}\rfloor$ in this graph by choosing $\lfloor \frac{n_i}{2}\rfloor$ independent edges from the $i^{\text{th}}$ path. Use this for $M$ in the lemma.

*We can find a vertex cover of order $\lfloor \frac{n_1}{2}\rfloor + \dots + \lfloor \frac{n_k}{2}\rfloor$ by choosing the $2^{\text{nd}}$, $4^{\text{th}}$, and so on vertices from every path. Use this for $U$ in the lemma.

