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We construct both the required triangulation and its polygonal schema. Subdivide the square $Q=[0,1{]}^2$ into a $4\times 4$ grid. Insert the centre of each small square and join it by line segments to its four corners, making four triangles per small square. Identify opposite sides of $Q$ by translations: $(0,t)\sim (1,t)$ and $(t,0)\sim (t,1)$. This is the standard quotient space presentation of the torus.

The grid and centre vertices, with these identifications, define an abstract simplicial complex $K_1$. Every triangle contains its own small-square centre and two consecutive grid vertices, so no triangle has identified vertices and no distinct triangles acquire the same vertex set. Only intended common sides or vertices are shared. Thus this is a genuine finite triangulation, whose geometric realization is homeomorphic to $T^2$. There are $32$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic zero.

Choose the simplex $\tau$ represented in the interior of $Q$ by the triangle with verticesRemove its relative interior. Before identifying the square's sides, the remainder is a closed disc with a triangular hole, hence an annulus. Cut this annulus along the line segment from $(0,0)$ to $(1/4,1/4)$, treating the two banks as distinct boundary arcs. The cut space is a closed disc; traversing its boundary follows the four outer edges, one bank, the three inner edges, and the other bank. It therefore has nine marked boundary sides.

A homeomorphism from this cut disc to a convex nine-sided polygon can be chosen to carry its nine marked sides to the nine polygon sides. One justification is to choose a boundary homeomorphism with the prescribed marked vertices and extend it over the disc by radial extension after disc parametrizations. Transport the identifications to this convex polygon. With labels recording traversal direction, its boundary word isIdentify equally labelled sides in their arrow directions: the two $a$ sides and two $b$ sides restore the torus pairings, while the two $c$ sides undo the cut. Leave $d,e,f$ unpaired. Their union is the triangular boundary of the removed simplex. The resulting quotient space is exactlyThis is the polygonal presentation of a punctured surface. The word supplies all nine sides and every gluing; inverse exponents reverse the boundary traversal direction.

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.

For a space with finite-dimensional homology over a specified field and only finitely many nonzero Betti numbers, this polynomial records their graded dimensions. Its value at one is the total Betti number, and its value at minus one is the Euler characteristic. The Künneth theorem gives $P_{M\times N}(t)=P_M(t)P_N(t)$ over a field. For integer homology ranks, use the rational Betti numbers.

Consider assemblies of congruent polygonal tiles with same-labelled faces paired by reflection and pure Neumann boundary conditions on unpaired faces. An invertible intertwiner for all face permutation matrices preserves the Euler characteristic. The number of tiles is the dimension of the representation. For each face label the number of edge classes is $(n+\mathrm{tr}{M}_s)/2$. For each vertex label the number of vertex classes is the number of orbits of the subgroup generated by incident-face involutions, hence the dimension of its invariant vectors. Intertwining preserves all these numbers. Summing the vertex and edge counts proves equal $V-E+F$, including boundary corners. This argument uses genuine permutation matrices; signed matrices encoding mixed boundary conditions do not give the same topological count.

For a finite connected graph, collapse a spanning tree. Its $v-1$ edges disappear, and each of the remaining $e-v+1$ edges gives a circle in a bouquet of circles. The fundamental group of a bouquet of circles is free group, with one member of a free basis of a group for each circle. This gives the formula in terms of the Euler characteristic $\chi (X)=v-e$.

For a degree-$d$ holomorphic map, $2g_X-2=d(2g_Y-2)+\sum (e_p-1)$.

Let $f:X\to Y$ be a nonconstant holomorphic map between compact Riemann surfaces that are connected spaces, without boundary. Write $g_X,g_Y$ for their genera and $d\ge 1$ for the degree of a holomorphic map. With analytic ramification index $e_p$ at each $p\in X$, the formula isHere $\chi =2-2g$ is the Euler characteristic. Only finitely many ramification points contribute. The holomorphic map is automatically surjective; the valency theorem gives ${\sum }_{p\in f^{-1}(y)}{e}_p=d$ even at a branch value of a holomorphic map. Constant maps are excluded.^[ref]

The useful interpretation is that total ramification of a holomorphic map measures how much the source Euler characteristic falls short of $d$ copies of the target. The ramification of a power map of the Riemann sphere spends that entire correction at two points; the double cover of the sphere with four branch values spends it at four points and produces a genus-one surface.

For a closed point $p$ on a smooth projective curve, the quotient in $0\to L(-p)\to L\to L{|}_p\to 0$ has $k$-dimension $[k(p):k]$. Additivity of Euler characteristic therefore changes $\chi (L)$ by that residue-field degree. Every line bundle is associated with a divisor, so iteration proves the displayed formula. The dualizing sheaf on a smooth projective curve identifies $h^1(L)$ with $h^0({\Omega }_C^1\otimes L^{-1})$ and converts this Euler-characteristic formula into the usual divisor form of Riemann-Roch theorem.

Zero Euler characteristic and $H_3=0$ give free rank $b_1-1$. The universal coefficient theorem for cohomology identifies $H^3(M;\mathbb{Z})=\mathbb{Z}/2$ with $\mathrm{Ext}(H_2(M;\mathbb{Z}),\mathbb{Z})$, so the only torsion in $H_2$ is one copy of $\mathbb{Z}/2$. In particular $b_1\ge 1$.

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 $2k+1$.

On vertices $1,\dots ,6$, take the simplicial complex generated by $123,124,135,146,156,236,245,256,345,346$. Every edge occurs in two facets. The vertex links, listed cyclically, are $23564$, $13654$, $12645$, $12536$, $13426$, and $14325$, respectively. Thus each vertex link is a cycle graph, and the realization is a compact connected surface without boundary. Its Euler characteristic is $6-15+10=1$, so the classification theorem for surfaces identifies it with the real projective plane. All fifteen possible edges occur, and the listed triangles are distinct three-element sets, so this is an honest abstract simplicial complex, not merely a polygon presentation.

Read a boundary word counterclockwise, giving every paired edge a label. A positive exponent means the edge arrow agrees with the boundary traversal, and an inverse means it disagrees. Identify equally labelled edges in their arrow directions. The sphere has a square word $ab{b}^{-1}{a}^{-1}$; the vertex count is three, giving Euler characteristic $3-2+1=2$. A closed orientable surface of genus $g\ge 1$ has wordA nonorientable surface which is the connected sum of surfaces of $k$ real projective planes has word $c_1{c}_1\cdots c_k{c}_k$. For $k=1$, this is a digon; the square word $abab$, identifying antipodal boundary points, is another model of the real projective plane. The Klein bottle has square word $ab{a}^{-1}b$, representing the nonorientable case $k=2$. The classification theorem for closed combinatorial surfaces says that these orientable and nonorientable families exhaust the connected closed cases.

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.

The sphere with three points removed is an orientable noncompact surface of genus zero with three ends. It can be realized by pairing the two sides at each of two opposite ideal vertices of an ideal hyperbolic quadrilateral in the hyperbolic disk. The four ideal vertices fall into three equivalence classes; after adding these three points the cell structure has one face, two edges and three vertices, hence Euler characteristic two. Removing these added vertices recovers the original quotient.

A homeomorphism preserves every homology group, hence the Euler characteristic wherever the finite rational Betti number sum is defined. In fact, homotopy invariance of homology makes Euler characteristic invariant under homotopy equivalence. Thus different finite CW complex structures on the same space have equal alternating cell counts.^[ref]

Consider a connected assembly of $n$ closed discs formed by identifying $e$ pairs of boundary arcs, each arc used once and all arcs in a given tile having disjoint endpoints. Its gluing graph has one vertex per tile and one edge per paired arc. A spanning tree joins the initial discs into a single closed disc; each of the remaining $e-n+1$ pairings attaches a band. This is a thickening of the gluing graph, so it is homotopy equivalent to that graph. Thus its Euler characteristic is $n-e$ and its fundamental group is the free group $F_{e-n+1}$. If the gluing graph is a tree, the assembled surface is itself a closed disc.

Suppose each pairing preserves the positive boundary parameter of the reference discs. A global orientation must then assign opposite orientation signs to adjacent tiles, because their induced boundary orientations on an interior seam must oppose one another. Such signs exist exactly when the gluing graph is bipartite. An odd gluing cycle therefore proves that the assembly is a nonorientable surface. A Dirichlet boundary condition or a Neumann boundary condition on an unpaired arc has no effect on this topological calculation.

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.

In a triangle whose two sides at an apex are glued with that apex fixed, cross-sections become circles and the apex becomes their cone point. This quotient is a disc. A second triangle with boundary word $aa{b}^{-1}$ becomes a Möbius band after the $a$ sides are identified. Gluing the remaining $b$ boundary to the first disc caps the band and gives the real projective plane. The resulting two-face cell structure has two vertices, three edges and Euler characteristic one.

In two worldsheet dimensions, ${\delta }_{\omega }(\sqrt{g}R)=-2\sqrt{g}{\Delta }_g\omega$. Thus the standard dilaton curvature coupling does not have classical Weyl invariance for a general spacetime-varying dilaton. A constant dilaton gives the topological Euler characteristic instead. In the quantum sigma model, this classical variation combines with the loop anomaly; vanishing improved sigma-model beta functions is the full background conformal condition.