Here a proof I am trying to make sense of.
Let $G$ be a graph in which each pair of odd cycles shares a common vertex. Show that $\chi(G)\leq 5$.
Let $C$ be any odd cycle of $G$ (if none exists, let $C=\emptyset$). Since each odd cycle of $G$ shared a vertex with $C$, we see that $G-C$ has no odd cycles. Therefore $G-C$ is bipartite and hence $2$-colorable. The vertices of $C$ can be colored with $3$ new colors, so we can color $G$ with $5$ colors. Therefore $\chi(G)\leq 5$.
If the number of vertices be even, can't $G-C$ become an odd cycle as well, therefore it cannot become a bipartite? I'm tripped by this fact that. What typical constraints does the $G-C$ must have in order for this property to be true? Planarity? There must exists a odd cycle of 3 or 5 in the graph then.
Could someone clarify some ambiguities with this simple proof?