The Euler number of a finite linear graph \(X\) equals the number of vertices of \(X\) minus the number of edges of \(X\). It is in fact a topological invariant of \(X\), as we shall see. What is the Euler number of an arc? a circle? a wedge of \(n\) circles? the complete graph on \(n\) vertices? If \(E\) is an \(n\)-fold covering space of \(X\), how are the Euler numbers of \(E\) and \(X\) related?
Solution
For a finite graph, \[ \chi(X)=V-E. \]
An arc has \(V=2,E=1\), hence \[ \chi=1. \] A circle has \(\chi=0\) (for example one may subdivide it into \(r\) vertices and \(r\) edges).
A wedge of \(n\) circles has one more vertex than the total excess contributed by its \(n\) cycles; using a model with one vertex and \(n\) loop-edges, or any subdivision, \[ \chi=1-n. \]
The complete graph \(K_n\) has \[ V=n,\qquad E=\binom n2, \] so \[ \chi(K_n)=n-\binom n2=\frac{n(3-n)}2. \]
Finally, if \(p:E\to X\) is an \(r\)-fold covering of a finite graph, every vertex of \(X\) has exactly \(r\) lifts and every edge has exactly \(r\) lifted edges. Therefore \[ V(E)=rV(X),\qquad E(E)=rE(X), \] and hence \[ \boxed{\chi(E)=r\chi(X).} \]