Prove a DAG can be obtained by an undirected graph's longest cycle

Question

Let G(V, E) be a finite undirected graph and let κ be its longest undirected cycle.

1. Prove that it is always possible to obtain an orientation ω(κ) in which κ is topologically sorted and hence its last edge (from node 1 to node N) is inverted.
2. Prove that given ω(κ) it is always possible to orient the remainder of the graph as to build a DAG.

For example, in a 5-node graph with its hamiltonian path as κ:

The proof of (1) seems intuitive: one can always direct the edges of an undirected cycle as to form a directed cycle and then invert one of the edges at random. This causes a topological sort starting at the node on the origin of the new edge. I have no clue on the proof of (2). I'm new to graph theory and still lack the knowledge to write proofs formally, so I'm hoping to learn a bit more on this with this question too.

Thanks in advance.

Answer 1

Assign different numbers as labels to the vertices, such that your chosen cycle visits vertices in increasing order.

It doesn't matter which labels you choose for vertices that are not in your cycle, as long as the labels are all distinct.

Orient each edge in the direction of increasing numbers.