Suppose I orient the edges of the power of cycle graph $G=C_n^k$ where $n=16$ and $k=4$ in such a way that all the generated cycles by the elements $1,2,3,4$ are given the standard lexical orientation. For example, the cycle $0-1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-0$ is given the orientation in the usual sense $0\leftarrow1\leftarrow2\leftarrow3\leftarrow4\leftarrow5\leftarrow6\leftarrow7$ $\leftarrow8\leftarrow9\leftarrow10\leftarrow11\leftarrow12\leftarrow13\leftarrow14\leftarrow15\leftarrow0$; similarly for all other generated cycles from the elements $2,3,4$. Then, as I see, all cliques are having kernels, and, overall, i think the graph $G$ is kernel-perfect. But, in this orientation, I observe that maximum outdegree is equal to $k=4$. Then, does it imply that choice number, or, list chromatic number of $G$ is equal to $k+1=5$, by the kernel lemma. Clearly, this should be impossible, as the chromatic number of graph is $6$.
What am I missing here? Thanks beforehand.
