My instructor asked me to prove that strongly connected component of a DAG is also a DAG. I don't get it. How can a graph be DAG and strongly connected at the same time?
Consider a graph on two vertices "u" and "v" and assume that it is strongly connected as well as a DAG. A path from u to v exists as well as a path from v to u which creates a cycle. No?
This is difficult to answer because in part we have to guess what your instructor actually meant, but that said, I can't see a way to reconcile the statement.
You could ask if the components that are strongly connected in the underlying undirected graph are DAGs, but this is trivial, any subgraph of a DAG is also a DAG - we're not adding any edges, so we can't introduce a cycle.
Alternatively, we could ask for strongly connected components of a DAG, but here your observation is correct, there are no non-trivial strongly connected components in a DAG. The trivial ones are just each vertex by itself, which is trivially a DAG too, so that kind of works, if your instructor is looking to test your understanding of the definitions.
The last thing I can think of is that if you take a directed graph (not necessarily acyclic), and shrink each strongly connected component to a single vertex, the resulting graph is a DAG (called the "condensation" of the original graph). The condensation of a DAG is the DAG itself (we're back to the trivial strongly connected components again).
