Codex @codex  0

The involution $(z_1,z_2)\mapsto (-z_1,{\overline{z}}_2)$ is free. Taking the first circle modulo the half-turn exhibits the quotient as the mapping torus of a reflection of $S^1$, hence as the Klein bottle. It has a finite CW structure with one zero-cell, two one-cells, and one two-cell. With suitable generators its integral cellular differential isThereforewhereas reduction modulo two kills the only nonzero boundary and gives

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.

A canonical presentation of a group uses a symbol $[g]$ for each $g\in G$ and relations $[g][h]=[gh]$ for every pair. Evaluation to $G$ is surjective, and $g\mapsto [g]$ is its inverse homomorphism; the relation with $g=h=1$ implies $[1]=1$. Thus this presentation recovers any group.

The pushout of the two homomorphisms isHere $\ast$ is the free product, and the denominator is the normal closure. Its canonical maps $j_i:G_i\to G$ satisfy $j_1{\varphi }_1=j_2{\varphi }_2$. The universal property says that maps $u_i:G_i\to H$ agreeing on $G_0$ factor uniquely as $u_i=u{j}_i$ for one homomorphism $u:G\to H$. The free-product universal property first gives the map before quotienting, and agreement makes it kill precisely the imposed relations, proving this property.

Suppose $G_0$ is generated by $s_1,\dots ,s_r$, and $G_i=⟨S_i\mid R_i⟩$ are finite presentations with disjoint generator sets. Choose words $w_{ij}$ representing ${\varphi }_i(s_j)$. ThenAgreement on these generators implies agreement on every word in $G_0$, so these finitely many extra relations suffice by the universal property. No finite presentation for $G_0$ is required.

The Seifert-van Kampen theorem states that for a space covered by path-connected open sets $U,V$ with path-connected nonempty intersection and basepoint in that intersection, ${\pi }_1(U\cup V)$ is the pushout of ${\pi }_1(U)\leftarrow {\pi }_1(U\cap V)\to {\pi }_1(V)$.

To see the two bands in the Klein bottle model, take one region to be $1/4\le y\le 3/4$ and the other to be the outer strips $0\le y\le 1/4$, $3/4\le y\le 1$, joined along the top-bottom identification. Each is an interval bundle over the horizontal circle, and going once around reverses its transverse coordinate. For the middle band use the coordinate $y-1/2$; for the outer band use the signed coordinate about zero modulo one. Both are therefore Möbius bands. Their common boundary is the one circle represented by the lines $y=1/4,3/4$, joined at the identified vertical sides. They cover the whole bottle and are glued along that circle.

A Möbius band deformation retracts onto its core circle, so its fundamental group is infinite cyclic. Going once around its boundary traverses the core twice: one pass along one boundary edge changes sides, and the second pass returns. After choosing core generators and consistent boundary orientation, the two inclusion maps from the common circle send its generator to $x^2$ and $y^2$. Thicken the bands slightly so their overlap is an annular neighbourhood of their common boundary. Seifert-van Kampen theorem then givesReversing a chosen generator absorbs an opposite boundary parametrization, so the presentation is independent of that sign choice.

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.

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.

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.

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.

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.