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Every closed odd-dimensional manifold has Euler characteristic zero. For an orientable manifold this follows by pairing complementary Betti numbers using Poincare duality; the general case follows from the orientable double cover and Euler characteristic under a finite covering.

A covering map over a topological graph lifts its vertices to all their preimages and each open edge to disjoint open edges. Their endpoints are supplied by the path lifting theorem applied to each parametrizing path of a closed edge. The covering sheets over small vertex stars verify that the resulting graph topology agrees with the original topology upstairs. A constant finite degree of a covering map $d$ therefore multiplies both vertex and edge counts by $d$, proving the Euler characteristic under a finite covering formula for finite topological graphs.

For a free action of a group, every stabilizer subgroup is trivial. Hence every ramification index of a holomorphic map is one, and the projection is an unramified covering map of degree of a holomorphic map $|G|$. The Riemann-Hurwitz formula reduces toThis is also the multiplicativity of Euler characteristic under a finite covering.

For the closed orientable surface ${\Sigma }_g$, use the fundamental group presentationSince $g\ge 1$, define a homomorphism to $\mathbb{Z}/n$ by sending $a_1$ to $1$ and every other displayed generator to zero. All commutators map to zero, so the relation is respected, and the map is surjective. Its kernel has index $n$. The classification of connected covering spaces supplies a connected $n$-sheeted covering space, including the identity cover when $n=1$.

Every connected finite covering space of ${\Sigma }_g$ is a closed orientable surface: compactness follows from finite degree, and the base orientation lifts through local covering charts. Lift a finite cell decomposition of the base. Every cell has $n$ lifts, so the Euler characteristic under a finite covering gives $\chi (\tilde{\Sigma })=n\chi ({\Sigma }_g)$. If the covering surface has genus $h$, this reads $2-2h=n(2-2g)$. ThereforeThis genus of a finite cover of a closed orientable surface depends only on the sheet number, not on the particular subgroup used to construct the cover.

Call this simultaneous antipodal quotient of two spheres $M$. The simultaneous antipodal map acts freely on $S^2\times S^2$, so $M$ is a connected closed manifold of dimension four. Each antipodal factor has mapping degree $-1$; their product preserves orientation. Hence $M$ is orientable.

The product $S^2\times S^2$ is simply connected, so this double covering space is the universal cover. Consequentlyusing the abelianization of the fundamental group. The Euler characteristic under a finite covering gives $\chi (M)=4/2=2$. Its rational first Betti number is zero, and Poincare duality gives $b_3=b_1=0$ and $b_0=b_4=1$. Thus $b_2=0$ as well.

The universal coefficient theorem for cohomology now gives $H^1(M;\mathbb{Z})=0$ andHere $\mathrm{Hom}(H_2(M;\mathbb{Z}),\mathbb{Z})=0$ because $b_2=0$. Integral Poincare duality further gives $H^3(M;\mathbb{Z})\cong H_1(M;\mathbb{Z})=\mathbb{Z}/2$, and $H^4(M;\mathbb{Z})=\mathbb{Z}$.

Let $a,b,c$ generate these groups in degrees $2,3,4$ respectively. The only potentially nonzero positive-degree cup product is $a^2$. But $2a^2=(2a)a=0$, while $H^4(M;\mathbb{Z})$ has no nonzero torsion subgroup, so $a^2=0$. Every other positive product vanishes by dimension. Thuswith every product of positive-degree elements equal to zero. The degree-three torsion is essential: it would be lost by computing only rational cohomology.

Because the action is free, the quotient $S_g/G$ is a closed orientable surface $S_h$, and $S_g\to S_h$ is a covering of degree $|G|$. The Euler characteristic under a finite covering givesor equivalentlyThe quotient inherits a hyperbolic metric, so $h\ge 2$. Hence $h-1\ge 1$ and $|G|\le g-1$.

A closed manifold of odd dimension has Euler characteristic zero by the Euler characteristic of an odd-dimensional closed manifold. Decomposing $M$ into $P$ and the removed closed ball along $S^{2k}$ givesso $\chi (P)=1$.

If the antipodal boundary map extended to a fixed-point-free involution of $P$, the quotient map $P\to P/⟨\tau ⟩$ would be a double covering. The formula for Euler characteristic under a finite covering would implywhich is impossible. Hence no such extension exists.

Assume $p$ has a common finite degree of a covering map $d$. Declare the lifted vertices to be $p^{-1}(V(X))$, so there are exactly $dv(X)$ of them. Each open edge is homeomorphic to $(0,1)$, a simply connected interval. The restriction of the covering map above it is a disjoint union of $d$ copies of that open edge.

These copies acquire endpoints as follows. Lift the parametrizing path $[0,1]\to X$ of each oriented edge, starting at each of the $d$ points over its initial vertex. The path lifting theorem gives a unique lift all the way to $1$. Its endpoint lies over the terminal vertex, and the interiors of these lifted paths are exactly the $d$ open edges already described. This also handles a graph loop, whose two endpoints may coincide downstairs or upstairs. Local covering sheets above a small star at a vertex show that the lifted incidence relations have precisely the original topology. Thus the graph structure lifted through a covering map is a finite topological graph withSubtracting gives the Euler characteristic under a finite covering formulaThe formula holds even when $\tilde{X}$ is disconnected. If a disconnected base is allowed to have degrees $d_C$ that vary with its connected components $C$, the precise replacement isThe printed formula with a single $d$ uses the constant-degree convention explained in part (a)(ii).

For positive $p,q$, quotient $S^p\times S^q$ by $(x,y)\mapsto (-x,-y)$. This is a free double covering space action. Its closed manifold quotient is orientable exactly when $p+q$ is even, because the product mapping degree is $(-1{)}^{p+q+2}$. If $p,q\ge 2$, the product is simply connected and the quotient has fundamental group $\mathbb{Z}/2$. For $p=q=2$, the quotient has integral cohomology groups $\mathbb{Z}$ in degrees $0,4$, $\mathbb{Z}/2$ in degrees $2,3$, and zero elsewhere. Every positive-degree cup product is zero. These conclusions follow from Poincare duality, the Euler characteristic under a finite covering, and the universal coefficient theorem for cohomology.