(a) Let ? = (V, E, ?) be a finite graph. If all the vertices have the same degree d in ? then we say that ? is a regular graph of degree d. Use Euler's degree sum formula (=Handshaking Lemma) to prove that if ? is a regular graph of degree d and d is odd, then the number of edges in ? must be divisible by d.
(b) Let ? be as in (a). In Remark 62 in the Book we outlined the original Euler's proof of the degree-sum formula
$\sum_{v \in V} deg_?(v) = 2|E|$
in case when the graph ? is simple, i.e. it does not contain loops or multiple edges. When ? is simple, Euler's argument is simple too. When ? is maybe not simple but has no loops, Euler's argument is still visible. But when ? contains loops, then the argument is not that obvious at all. How would you strengthen Euler's argument to make it work in case when ? is a general finite graph, possibly containing loops inside. Please describe your approach in as much detail as possible.
Hint. Take any two element set, say {0, 1}, and consider the triple Cartesian product
V × E × {0, 1}.
The main idea is to duplicate loops by introducing two incident subsets
$I_0 = \{(v, e, i) \in V \times E \times \{0, 1\} \mid v \in \theta(e) \text{ and } e \text{ is not a loop at } v \text{ and } i = 0\}$
and
$I_1 = \{(v, e, i) \in V \times E \times \{0, 1\} \mid v \in \theta(e) \text{ and } e \text{ is a loop at } v \text{ and } i = 0 \text{ or } 1\}$.
Let
$I = I_0 \cup I_1$
be the total incident subset, and look at the following commutative diagram of natural maps
(diagram)
in which ? is the embedding of I in to the Cartesian product V × E × {0, 1}, ?v and ?E are the projections onto V and E respectively, ? = ?? ? ? and ? = ?v ? ?. You need to meditate on this diagram then.
