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The two singly marked sides of the first triangle are glued with their endpoints at $A$ matched. A cross-section parallel to the remaining side has its two endpoints identified, becoming a circle; the apex becomes its cone point. Thus this quotient triangle is a disc, whose boundary is the doubly marked edge.

In the second triangle, the two triply marked sides run consecutively along the boundary from $E$ through $D$ to $F$. Pairing these sides in the indicated directions introduces one reversal; its boundary word is $aa{b}^{-1}$, where $b$ is the remaining doubly marked edge. This is a Möbius band with boundary $b$. One can obtain the usual rectangle model by cutting from the identified triple-edge seam to the midpoint of $b$ and opening the triangle.

The double markings glue the disc boundary to the boundary of that Möbius band. Capping a Möbius band by a disc givesThis description also verifies that the quotient is a closed surface, including the vertices. As a check, the vertex classes are $\{A\}$ and $\{B,C,D,E,F\}$; there are three edge classes and two faces. The Euler characteristic is $2-3+2=1$, consistent with the projective plane. The Möbius band exhibits nonorientability directly.

The surface is a Klein bottle. Here is a cell-count justification that keeps all the arrow directions. Name the corners of the first square $a,b,c,d$, beginning at its upper left and proceeding clockwise; name those of the second $e,f,g,h$ in the same way. The single-arrow gluing identifies $a$ with $e$ and $b$ with $h$. The triple-arrow gluing identifies $d$ with $h$ and $c$ with $g$. The first square's quadruple-arrow gluing identifies $b$ with $d$ and $c$ with $a$. The second square's double-arrow gluing identifies $e,f,g$ with one another. There are exactly two resulting vertex classes:There are four edge classes and two faces, so the Euler characteristic is $2-4+2=0$. Every edge is paired and the corner links close into circles, giving a compact surface without boundary; the cross-square identifications make it connected. It is nonorientable: the quadruple-arrow identification glues two boundary edges of the same square in the same boundary-traversal direction, so no orientation of that face can extend across the gluing. The classification of compact connected surfaces now identifies the nonorientable surface of Euler characteristic zero as the Klein bottle.

Geometrically, the internal identification in each square produces a Möbius band. The remaining single- and triple-arrow edges comprise their two boundary circles, which are glued together. The double of a Möbius band is another description of the same Klein bottle.

Treat the two squares as a single edge-pairing construction, including the identifications between different squares. Number the corners of the left square counterclockwise from its lower left corner by $A,B,C,D$, and similarly use $E,F,G,H$ on the right. The single arrows identify $D$ with $H$ and $C$ with $E$. The double arrows identify $G$ with $F$ and $H$ with $G$. The triple arrows identify $A$ with $E$ and $B$ with $F$. The quadruple arrows identify $A$ with $C$ and $D$ with $B$.

Thus there are precisely two vertex classes, $\{A,C,E\}$ and $\{B,D,F,G,H\}$. There are four edge classes and two two-dimensional cells, so the resulting closed connected surface has Euler characteristicIt is a nonorientable surface: on the left square the quadruple-arrow sides have the same orientation relative to a traversal of its boundary, giving a twisted band after gluing. Equivalently, no choice of orientations for the two faces makes the induced directions opposite on every paired edge. The vertex links are circles, so these edge identifications produce a closed surface, rather than a singular cell space. The classification of compact surfaces now givesA torus also has Euler characteristic zero, but is excluded by the failed orientation test.

Choose an orientation for each simplex. The chain group $C_k$ is the free abelian group on the oriented $k$-simplices, with orientation reversal changing sign. On an ordered simplex setIts face maps respect the identifications in a Delta-complex. In ${\partial }_{k-1}{\partial }_k$, deleting any pair of vertices occurs in the two possible orders with signs $(-1{)}^{i+j-1}$ and $(-1{)}^{i+j}$, which cancel. Hence ${\partial }^2=0$. The groups of homological cycles and boundaries are $Z_k=\ker {\partial }_k$ and $B_k=\mathrm{im}{\partial }_{k+1}$; the identity just proved gives $B_k\subseteq Z_k$, so $H_k=Z_k/B_k$ is defined. Mod-two homology uses the same chain complex with coefficients reduced modulo two.

Here the edge identifications put all six vertices into one vertex class. Let the three boundary edge classes be $a,b,c$ in cyclic order, and orient the internal diagonals as $d=[v_1,v_3]$, $e=[v_3,v_5]$, $f=[v_5,v_1]$. With the four face orientations listed in the source, the boundaries areThus $C_2={\mathbb{Z}}^4$, $C_1={\mathbb{Z}}^6$, $C_0=\mathbb{Z}$, and ${\partial }_1=0$. The Euler characteristic isThe face-boundary map is injective over the integers: coefficients of $a,b,c$ force the first three face coefficients to vanish, after which the diagonal coefficients force the fourth to vanish. Its cokernel eliminates $d,e,f$ and leaves generators $a,b,c$ with the single relation $2(a+b+c)=0$. Since $a+b+c$ is primitive in ${\mathbb{Z}}^3$, this cokernel is ${\mathbb{Z}}^2\oplus \mathbb{Z}/2$. ThereforeModulo two the first three face-boundary columns are $d,e,f$, and the last is their sum. The rank is three, the kernel is one-dimensional, and the edge quotient has dimension three. ConsequentlyThe top mod-two class is the sum of all four faces. Integral homology has no such top class, reflecting nonorientability.

No. Its top nonzero group would force dimension three. Poincare duality requires $H_2\cong H^1$, and the universal coefficient theorem gives $H^1\cong \mathrm{Hom}(H_1,\mathbb{Z})=\mathbb{Z}$, since $H_0$ is free. This contradicts the proposed vanishing of $H_2$. Equivalently its listed Betti numbers give Euler characteristic $-1$, impossible for a closed oriented odd-dimensional manifold.

With the usual second-countability convention, an $n$-manifold is a Hausdorff second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$. A local orientation of a manifold at $x$ is a choice of generator ofExcision and a coordinate ball identify this relative group with $H_n({\mathbb{R}}^n,{\mathbb{R}}^n\setminus \{0\})$. An orientation of a topological manifold is a coherent choice of such generators: near each point the choices are induced by one relative class on a neighbourhood, equivalently by an atlas with orientation-preserving transition maps.

For a connected closed oriented manifold, its fundamental class $[M]\in H_n(M;\mathbb{Z})$ is the unique class mapping to every chosen local generator. Here closed means compact without boundary. Poincare duality states that cap product with this class is an isomorphismOver rational coefficients it gives $b_k=b_{n-k}$. ThereforeWhen $n$ is odd this says $\chi =-\chi$, so $\chi (M)=0$. Torsion does not contribute to the Betti numbers or this Euler characteristic.

A finite tree is a nonempty connected graph with no graph cycles. We show by mathematical induction on its number $v$ of vertices that it has $v-1$ edges. For $v=1$ there are no edges, so the assertion holds.

If $v>1$, take a path in a graph of maximal length among those without repeated vertices. Each endpoint is a leaf of a tree: an additional incident edge would either extend the path in a graph or create a graph cycle. Delete one such leaf of a tree and its unique incident edge. The remaining graph stays connected, since a path in a graph between remaining vertices never needs that terminal edge, and it has no graph cycles. It is therefore a tree with $v-1$ vertices and, by mathematical induction, $v-2$ edges. Restoring the deleted edge gives $e=v-1$.

Thus its Euler characteristic is

Apply the first extraction move to the two outer occurrences of $x$, then cyclically rotate the word to expose the two occurrences of $y^{-1}$:In the third line the repeated letter for the move is $y^{-1}$, and the intervening string is $z^{-1}{z}^{-1}$, whose inverse is $zz$. Rename $y^{-1}$ as a new edge label. The final word has the standard form $aabbcc$.

Each same-direction repeated pair represents a projective-plane summand, so the classification theorem for surfaces identifies this word with the connected sum of three real projective planes:Thus the surface has crosscap number three. As a check, the original identifications make all six polygon vertices equivalent, leaving one vertex, three edges and one face. Its Euler characteristic is $1-3+1=-1=2-3$, consistent with that crosscap count.

Here is an explicit simplicial complex on six vertices. Take every subset of one of the following ten triangular facets:For example, $135$ denotes $\{1,3,5\}$, with all its edges and vertices included. Its one-skeleton has all fifteen edges of the complete graph. Every edge belongs to exactly two triangles. The links of a simplex at the six vertices are the following cycles, with the first vertex understood to repeat at the end:Thus every vertex has a disk neighbourhood, as do edge and face interior points. The realization is a compact connected surface without boundary. Its Euler characteristic isThe classification theorem for surfaces identifies the unique closed connected surface with this Euler characteristic as the real projective plane. Consequently this complex triangulates ${\mathbb{RP}}^2$. The explicit facet and link checks make this a genuine simplicial complex, not a polygon whose boundary identifications have been mistaken for simplicial faces.

There are exactly four homeomorphism types: the two-sphere, real projective plane, torus, and Klein bottle.

For completeness, enumerate all pairings. Four sides have three pairings; each pair can be matched in either direction, giving twelve signed pairings before symmetries. Up to cyclic relabelling, the repeated labels are either adjacent, of form $a{a}^{s}b{b}^t$, or alternating, of form $ab{a}^s{b}^t$, where $s,t\in \{1,-1\}$. The second adjacent pattern $abba$ rotates to the first. Endpoint identification gives the following table:Here the quotient always has two edge classes and one face, so its Euler characteristic is $V-2+1=V-1$. To reproduce the vertex count, label consecutive square vertices $0,1,2,3$. For paired sides $i,j$, a same-direction pairing identifies $i$ with $j$ and $i+1$ with $j+1$; a reversed-direction pairing identifies $i$ with $j+1$ and $i+1$ with $j$, with indices modulo four. Taking the generated equivalence classes gives exactly the displayed values of $V$.

The quotient is orientable precisely when each label is traversed once positively and once negatively around the boundary: then the face orientation extends across every paired edge. A same-direction pair obstructs such an orientation. Thus the two $\chi =0$ possibilities are distinguished as the orientable torus and nonorientable Klein bottle; the other rows are fixed by Euler characteristic and the classification theorem for surfaces.

Explicit square representatives areAll edge interior points have disk neighbourhoods by joining two half-disks, and each identified vertex has a cyclic sequence of corner sectors, again a disk neighbourhood. There are no leftover boundary edges. The enumeration therefore describes closed surfaces, and the twelve signed pairings leave no additional cases.

The classification theorem for surfaces states that every connected closed combinatorial surface is homeomorphic to exactly one of:

The orientable family has Euler characteristic $2-2g$; the nonorientable family has Euler characteristic $2-h$. Orientability together with Euler characteristic distinguishes the entries.

Boundary words giving the required polygonal schemas areA two-sided polygon here means a disk with its boundary split into two arcs; subdivision produces an ordinary polygon with more sides without changing the quotient. Alternatively the square word $a{a}^{-1}b{b}^{-1}$ represents the sphere. Each four-letter block in the orientable word adds a torus handle; each repeated-letter block in the nonorientable word adds a projective-plane summand.

A topological surface is a Hausdorff space with a countable basis of a topology in which each point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$. This definition has no boundary; allowing half-plane charts gives a surface with boundary. A closed surface is compact and has no boundary.

The classification theorem for closed combinatorial surfaces states that the realization of a finite connected closed combinatorial surface is homeomorphic to exactly one of:

The orientable family, including the sphere at $g=0$, has Euler characteristic $2-2g$; the nonorientable family has characteristic $2-k$. If disconnected complexes are permitted, apply the list to each connected component.

For polygon descriptions, read the boundary counterclockwise and pair equally labelled edges in the direction of their arrows. An inverse exponent means that the arrow runs against this boundary traversal. The following standard polygon words for closed surfaces give the required models:For $k=1$ the last model is a digon; alternatively the square word $abab$, identifying antipodal boundary points, gives the real projective plane. A square with word $ab{a}^{-1}b$ gives the Klein bottle, the case $k=2$. All polygon sides are paired, so these models have no boundary.

The quotient is a Klein bottle, equivalently a connected sum of two real projective planes.

Use $z\in S^1$ for the angular coordinate and $r\in [1/2,1]$ for the radius. Parametrize the first annulus by $\tau =2r-1\in [0,1]$, running from its inner to its outer boundary. Parametrize the second by $\tau =3-2r\in [1,2]$, running from its outer to its inner boundary. The outer-boundary identity gluing puts them together at $\tau =1$. The remaining inner-boundary identification reverses the angle, so the entire quotient isThis is the mapping torus of reflection of a circle, after rescaling the interval.

To verify the homeomorphism, send $rz$ in the first annulus to $[z,2r-1]$ in $M$, and $rz$ in the second to $[z,3-2r]$. Both formulas are continuous, agree under the outer gluing, and agree under the reflected inner gluing. They therefore descend by the universal property of the quotient topology. Conversely, on the lower half of the cylinder send $(z,\tau )$ to the class of $((1+\tau )/2)z$ in the first annulus; on the upper half send it to the class of $((3-\tau )/2)z$ in the second. The formulas agree at $\tau =1$, and at the identified endpoints. The pasting lemma and the quotient property give a continuous inverse. Thus the two quotient descriptions are homeomorphic.

The quotient $M$ is compact. Its equivalence relation on the compact Hausdorff cylinder is closed: it is the union of the diagonal and the two graphs of the boundary reflection. Therefore $M$ is Hausdorff by compact Hausdorff quotient by a closed equivalence relation. Away from the seam its local charts are circle arcs times open intervals. At the seam, an arc near $z$ at $\tau =0$ glues to its conjugate arc at $\tau =2$; using the reflection as the transition gives another open rectangle. These are disc neighbourhoods, and a finite choice of such charts gives a countable basis. Hence $M$ is a closed surface.

Cutting the circle coordinate gives a rectangle with its vertical sides matched and its horizontal sides reflected. The reflection changes the sign of the angular coordinate while leaving the interval direction unchanged, so an orientation transported around the interval loop comes back reversed. Thus $M$ is a nonorientable surface. The rectangle quotient has one vertex, two edges and one face, giving Euler characteristic zero. By the classification theorem for closed combinatorial surfaces, the nonorientable genus is $k=2-\chi (M)=2$, so $M\cong {\mathbb{RP}}^2\#{\mathbb{RP}}^2$. This is the gluing annuli with one reflected boundary construction.

False: the result is again a torus with one open disc removed. Label the hexagon polygon vertices clockwise from the upper left asWith the same arrow labels as in part (ii),There are two vertex equivalence classes, $P=\{v_0,v_2,v_4\}$ and $Q=\{v_1,v_3,v_5\}$. The free diagonal polygon edges are $v_1{v}_2$ and $v_4{v}_5$. Each joins $P$ to $Q$, so together they form one boundary component of a surface, not two. At each vertex equivalence class the three corner sectors join in a single chain between these two free polygon edges. Thus each vertex link in a polygonal schema is a closed interval, and the quotient space is an orientable surface with boundary.

Its CW complex has CW cells comprising two vertices, four edges and one face, givingIn contrast, a torus with two disjoint open discs removed has two boundary components of a surface and Euler characteristic $2-2-2=-2$. Either topological invariant rules out the proposed homeomorphism.

Label the hexagon polygon vertices $v_0,\dots ,v_5$ in the direction in which its boundary word is $abc{a}^{-1}{b}^{-1}{c}^{-1}$, with the first side running from $v_0$ to $v_1$. Matching the endpoints according to the arrows givesThus the two vertex equivalence classes are $\{v_0,v_2,v_4\}$ and $\{v_1,v_3,v_5\}$.

The polygonal schema gives a compact connected surface with empty manifold boundary. Interior points have disc neighbourhoods; a paired polygon edge glues two half-disc neighbourhoods into a disc. At each of the two vertex equivalence classes, its three corner sectors join cyclically, so the vertex link in a polygonal schema is a circle, again giving a disc neighbourhood. Pairing every label once positively and once negatively makes these gluings consistent with an orientation, so the resulting surface is orientable.

The resulting CW complex has CW cells comprising two vertices, three edges and one face. Its Euler characteristic isBy the classification of compact surfaces, a connected closed orientable surface with $\chi =0$ has genus $g$ satisfying $2-2g=0$, hence $g=1$. The quotient space is a torus.

Every complete polygon edge pairing gives a compact connected surface with empty manifold boundary. Here is the local reason at the vertices. Replace each polygon corner by a small link interval. The polygon edge pairings identify all endpoints of these link intervals in pairs. In a given vertex equivalence class, the resulting graph is connected: the endpoint identifications generating that vertex equivalence class also connect its corner intervals. Every link graph vertex has degree of a vertex two, so the graph is a circle. Its topological cone is a disc, giving the required local surface model. Edge and face CW cell interiors have the usual disc neighbourhoods as well. A finite triangulation compatible with the polygon edge pairings realizes the result as a finite simplicial complex, giving a Hausdorff, second-countable space.

For every such surface, $\chi =V-2\ge -1$. If it is orientable, the classification of compact surfaces gives $\chi =2-2g$, so $g=0$ or $1$. If it is a nonorientable surface, write $k\ge 1$ for its nonorientable genus; then $\chi =2-k\ge -1$ gives $k\le 3$. The exhaustive list is therefore

To prove that every listed possibility actually occurs, the following six-letter boundary words specify complete pairings of a hexagon:The vertex counts follow by matching starts and ends of equally directed arrows, just as in part (b). The first and third words pair every letter with its inverse, giving orientable surfaces. Each other word contains a letter twice with the same sign, giving a nonorientable surface: an orientation on the face would induce equal, rather than opposite, boundary directions along that glued pair. With the displayed Euler characteristics, the classification of compact surfaces identifies the examples exactly as listed. All five types occur, and no others occur.

A complete pairing of the six polygon edges leaves three edges and one face as CW cells, together with the vertex equivalence classes. If $V$ denotes their number, then $V\ge 1$ and the polygonal-schema Euler count givesA closed orientable surface of genus two has Euler characteristic $2-2\cdot 2=-2$. Since the Euler characteristic is a topological invariant, no complete pairing of a hexagon can give that surface. Equivalently, the polygonal-schema Euler count bounds the genus of an orientable result by $2g\le 3$, so $g\le 1$.

The classification theorem for closed combinatorial surfaces states that the geometric realization of every closed combinatorial surface is homeomorphic to exactly one of the following:

With ${\Sigma }_0=S^2$, the complete distinguishing data are orientability and the Euler characteristic:Here $\chi (K)=V-E+F$, where $V,E,F$ count the vertices of a simplicial complex, edges, and triangles of a finite triangulation. Thus two such connected closed surfaces are homeomorphic if and only if they have the same orientability and Euler characteristic. In particular $N_1$ is the real projective plane and $N_2$ is the Klein bottle. Every listed type admits a finite triangulation. For disconnected finite closed combinatorial surfaces, apply the theorem to each of the finitely many connected components; the resulting unordered collection of types determines the homeomorphism type.

Use compatible edge identifications: choose a bijection between the three boundary vertices of $\tau$ and those of $\sigma$, and extend it linearly on each edge. This defines a simplicial homeomorphism of the two boundary circles. This compatibility is the usual meaning of gluing the two boundary triangles; independently chosen edge reversals need not agree at their endpoints. The classification needed here is 3(a)(iv); the printed reference to 3(a)(iii) is a source error.

Delete the two triangular simplexes but retain all their proper faces of a simplex, and identify the two resulting boundary simplicial subcomplexes by this map. The result is a finite abstract simplicial complex $L$: vertices from different sides are identified only in the boundary, and the only shared edges are boundary edges. There is no triangle consisting of all three boundary vertices after the two faces have been removed. Consequently no other simplexes become duplicates. Its geometric realization has the specified quotient topology and is homeomorphic to $Z$.

We check the closed combinatorial surface conditions. Away from the seam, vertex links of a simplex remain cycle graphs. At a boundary vertex of $X$ or $Y$, removing the face changes its link from a cycle graph to a graph path, with endpoints corresponding to its two boundary edges. Gluing the boundaries identifies the two paths' respective endpoints, making one cycle graph. Each seam edge now meets one triangle from $X$ and one from $Y$, and all other edges retain their original two incident triangles. Thus every vertex link is a cycle graph and every simplex lies in a triangle. The simplicial complex is connected, since the two connected punctured surfaces meet along their boundary circle. Hence $L$ is a closed combinatorial surface.

This construction is the connected sum of surfaces $T^2\#{\mathbb{RP}}^2$. For completeness, compute its Euler characteristic directly. Removing a triangular face while retaining its edges and vertices subtracts one, soGluing identifies three vertex pairs and three edge pairs, givingFor the explicit triangulations above, this is also $62-189+126=-1$.

Finally $Z$ is a nonorientable surface. If $Z$ were orientable, its restriction would make $Y$ orientable. The consistent choices of orientation of a simplex on its triangles extend across a capping disc: choose the cap's orientation of a simplex so that the induced directions on the common boundary oppose. Capping $Y$ by $\sigma$ would therefore make the real projective plane orientable, a contradiction. The obstruction to orientability in the real projective plane can also be seen from a Möbius band neighbourhood of a projective line.

By the classification theorem for closed combinatorial surfaces, a nonorientable surface with Euler characteristic $-1$ has $2-k=-1$, hence $k=3$. Therefore

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.