when cartesian product of graph is perfect 
What are the necessary and sufficient conditions for the following graph to be perfect?

*

*$ K_{1,n} \mathbin{\square} C_m $


*$ C_n \mathbin{\square} C_m $


*$ C_n \mathbin{\square} P_m $

I assume that necessary condition in every case would be that every factor must be perfect, but is that also the sufficient condition? How to prove it?
 A: Recognize that in all cases the resulting graph (and all its subgraphs) is triangle-free unless
one of the cycle-factors is a triangle.
Recognize that in all cases the resulting graph (and all its subgraphs) is bipartite unless one
of the cycle-factors is an odd cycle.
For the first example this means that for even $m$ the graph (and its induced subgraphs)
is bipartite and triangle-free, so both $\chi$ and $\omega$ are 2 (we can ignore the
edgeless subgraphs) so the graph must be perfect (regardless of $n$).
For odd $m>3$ we have an odd cycle (so we need at least 3 colors), but no triangle (so $\omega$ is 2),
so the graph cannot be perfect (regardless of $n$).
The only case left is $m=3$. It is easy to see that the largest clique is a triangle and
that we need three colors, but we still need to show that all induced subgraphs are perfect.
An induced subgraph can only be nonperfect if we succeed in breaking all triangles,
but keeping an odd cycle. When we break all triangles we need to remove at least one vertex
from each copy of $C_3$. Now first remove a vertex of the $C_3$-copy that is attached to
the center of the star and realize that each cycle must contain both of the other vertices of that
$C_3$. But then it can only be a 4-cycle or a 6-cycle. So if we break all triangles,
we also break all odd cycles, so this graph is also perfect (regardless of $n$).
You will probably be able to do the other two parts by yourself now.
