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For a closed surface, $\chi =V-E+F$ for any finite cell decomposition. Gauss–Bonnet for a geodesic polygon saysand for a closed smooth surface it says ${\int }_{S}KdA=2\pi \chi (S)$.

For a large radius, the graph surface bounded over that circle is flat near its boundary. The boundary geodesic-curvature integral is $2\pi$, so Gauss–Bonnet on the disc gives $\int K=0$. Since $K\ge 0$, continuity forces $K\equiv 0$.

For the torus,Thus $T_{+}$ has $\cos u>0$ and $T_{-}$ has $\cos u<0$. Cap the two boundary circles of the outer half $T_{+}$ by flat discs. The result is a sphere, and the caps contribute no Gaussian curvature, so Gauss–Bonnet gives ${\int }_{T_{+}}K=4\pi$. The full torus has Euler characteristic zero, hence total curvature zero and

The universal coefficient theorem for homology and flatness of $\mathbb{Q}$ giveThus, ifwith $T_i$ torsion, then $H_i(X;\mathbb{Q})\cong {\mathbb{Q}}^{b_i}$.

For a finite triangulable space, its Euler characteristic isIf $K$ is a finite triangulation and $f_i$ is its number of $i$-simplices, then the Euler-Poincare formula isTo prove it over $\mathbb{Q}$, write $Z_i=\ker {\partial }_i$ and $B_i=\mathrm{im}{\partial }_{i+1}$. Rank-nullity and $H_i=Z_i/B_i$ giveAfter multiplying by $(-1{)}^i$ and summing, the two boundary sums cancel, leaving the claimed equality.

The standard simplex ${\Delta }^3$ is contractible, so

Write the invariant prime divisors in the order of the raysas $D_1,\dots ,D_4$. The toric divisor class sequence isand the two characters in the standard basis of $M$ giveThuswhere $S=D_2$, $F=D_1=D_3$, and $D_4=S+kF$. Since the fan is smooth, this is also the Picard group.

The orbit-cone correspondence decomposes ${\mathbb{F}}_k$ into one two-dimensional torus, four one-dimensional torus orbits, and four fixed points. Only the zero-dimensional orbits contribute to the compactly supported Euler characteristic, so

For a compact oriented smooth manifold $M$, the Euler class of its tangent bundle satisfiesEquivalently, the sum of the indices of the isolated zeros of a vector field equals the Euler characteristic.

Apply the oriented smoothing to every crossing of an oriented knot diagram, fill the resulting Seifert circles by disjoint disks at suitable heights, and reconnect them by twisted bands at the original crossings. The boundary is the original knot. If the connected surface has $s$ disks and $c$ bands, its Euler characteristic is $s-c$ and its genus is $(1-s+c)/2$.

For $f:M\to N$ between closed connected oriented manifolds of equal dimension, cyclic invariance of trace and $f^{!}{f}^{\ast }=(\mathrm{deg}f)\mathrm{id}$ give the formula. Thus if both the degree and the target Euler characteristic are nonzero, $f$ coincides somewhere with every map homotopic to it, by the Lefschetz coincidence theorem. For target ${\mathbb{CP}}^{2k}$ the Euler characteristic is $2k+1$.

For a finite group generated by $A,B$, glue sheets labeled by the group over the three-punctured Riemann sphere using left monodromy multiplication. Fill the punctures by the local power charts dictated by the orders $m,n,r$ of $A,B,AB$. Right multiplication by inverses defines the commuting deck action. Its nontrivial stabilizers are conjugates of the corresponding cyclic subgroups. The Riemann-Hurwitz formula gives the displayed Euler characteristic. A compatible metric can be averaged over the group; in genus at least two the invariant hyperbolic metric is canonical by the uniformization theorem.

A triangulation of a surface decomposes it into triangles meeting along full edges or vertices. On a closed surface, each edge is incident with two triangle faces, so $3F=2E$. Euler characteristic connects this local counting relation with the surface's topology.