Cograph is a well-know graph that does not have induced $P_4$. My questions are about determining the chromatic number of graphs in the class cograph+v.
Notations:
Denote by $\chi(G)$ the chromatic number of a graph $G$.
Let $V(G)$ denote the vertices of $G$. For any vertex $v\in V(G)$, denote by $N(v)$ the neighbors of $v$ and $N[v]=N(v)+v$ the closed neighborhood of $v$. Let $A\subseteq V(G)$, denote by $\langle A \rangle_{G}$ be the induced subgraph of $A$ of $G$.
Let $G$ be an undirected simple graph, and $H=G+v$ be a graph such that $G=H-v$, i.e., $G$ is the graph obtained by deleting $v$ from $H$. This type of notation is introduced in the paper Parameterized complexity of vertex colouring. For a graph class $\mathcal{F}$, let $\mathcal{F}+kv$ denote the graphs which can be built by adding at most $k$ vertices to a graph in $\mathcal{F}$.
Question 1: Is the following statement true?
$\chi(G+v)=\chi(G)+1$ if and only if $\chi(\langle N(v) \rangle)=\chi(G)$, where $G$ is a cograph.
The "if" direction is easy, because $\chi(G+v) \le \chi(G)+1 $ and $\chi(\langle N[v] \rangle) = \chi(\langle N(v) \rangle) + 1$. For general simple graph $G$, the "only if" direction is not true. For example, let $G$ be a $P_4$ (path of 4 vertices) and $G+v$ be a $C_5$ (cycle of 5 vertices).
Question 2: If the proposition in question 1 is false. Is it easy to determine $\chi(G+v)$, where $G$ is a cograph.
Update: Jim showed than cograph+v is a subset of the class of perfect graph. Also he actually proved another more general theorem.
For any perfect graph $G$, $\chi(G)=\chi(G-v)+1$ if and only if $\chi(\langle N(v)\rangle) = \chi(G-v)$.