Classification of Compact Connected Surfaces: Examples

Both pairs of the word $aabb$ carry equal exponents: the bottom side $a$ is glued to the right side $a$ with matching parameters, $(t,0)\sim (1,t)$, and the top side $b$ is glued to the left side $b$ with matching parameters, $(1-t,1)\sim (0,1-t)$. The a-pair identifies the corners $v_0\sim v_1$ and $v_1\sim v_2$, and the b-pair identifies $v_2\sim v_3$ and $v_3\sim v_0$; hence all four corners form one vertex class, so the quotient $Y$ has $V=1$ vertex class, $E=2$ paired side classes and $F=1$ face, and these are the cells of its finite CW structure.

Read the word of the Klein-bottle presentation of [L3] cyclically from its middle letter, $b{a}^{-1}ba$; in the crosscap identity of [L2] with the letter $a$ renamed $b$, the separating arc $X=a^{-1}$ and the complementary arc $Y=a$ give a quotient homeomorphic to that of $bb{Y}^{-1}X=bb{a}^{-1}{a}^{-1}$. Cyclically rotating this word and renaming the letter $a^{-1}$ to $a$ turns it into $aabb$; since the presentation of [L3] realizes the Klein bottle $K$, the quotient $Y$ of the word $aabb$ is homeomorphic to $K$.

Both equal-exponent pairings preserve boundary direction: their formulas in step 1.1 use matching boundary parameters. Choose collars around paired interior points of the $a$-sides, with $t$ increasing along the boundary and $r\ge 0$ pointing inward. These coordinates give the same face orientation on the two collars. A chart across the paired edge uses $(t,r)$ on the first collar and $(t,-r)$ on the second, so its generator has opposite signs relative to the face generators on the two sides. The sign is the local-homology sign of a reflection, which reverses the cyclic orientation of the boundary of a small disk, exactly as in the seam calculation for [L4]. A path through the open face joining the collar interiors, closed by one crossing of the seam, therefore transports its generator to its negative. A continuous orientation is unchanged by loop transport [L5], which is impossible for a generator of an infinite cyclic group. Hence $Y\cong K$ is nonorientable. The two equal-exponent pairs are crosscap blocks, and [L4] identifies the single block with ${\mathbb{R}\mathbb{P}}^2$.

The cell counts of step 1.1 give $\chi (Y)=V-E+F=1-2+1=0$ by [L6], and this value is that of the Klein bottle by step 1.2. Together with steps 1.2 and 2.1 the word $aabb$ realizes the Klein bottle, is nonorientable, and has Euler characteristic $0$. Every construction used is a finite explicit pairing, rotation or crosscap move of polygons, so no choice axiom is used.

Each of the four side pairs identifies two corners: pairing $a$ with $a^{-1}$ gives $v_0\sim v_3$ and $v_1\sim v_2$, pairing $b$ with $b^{-1}$ gives $v_1\sim v_4$ and $v_2\sim v_3$, pairing $c$ with $c^{-1}$ gives $v_4\sim v_7$ and $v_5\sim v_6$, and pairing $d$ with $d^{-1}$ gives $v_5\sim v_0$ and $v_6\sim v_7$; the chain $v_0\sim v_5\sim v_6\sim v_7\sim v_4\sim v_1\sim v_2\sim v_3\sim v_0$ shows that all eight corners form a single vertex class. Hence $Y$ has $V=1$ vertex class, $E=4$ paired side classes and $F=1$ face, and these are the cells of its finite CW structure.

Let $\delta$ be the straight diagonal from $v_0$ to $v_4$; its relative interior lies in the interior of the convex octagon, and cutting $P$ along $\delta$ produces the pentagon $P_1$ with vertices $v_0,v_1,v_2,v_3,v_4$ and sides $a,b,a^{-1},b^{-1},\delta$, together with the pentagon $P_2$ with vertices $v_4,v_5,v_6,v_7,v_0$ and sides $c,d,c^{-1},d^{-1},{\delta }^{\mathrm{\prime }}$. By the split identity of [L3] the quotient of $P_1\bigsqcup P_2$ by the pentagon pairings $R_1,R_2$ together with the identification $S$ of the two copies of $\delta$ by the natural reversal is homeomorphic to $Y$; write $Y_1=P_1\mathrm{/}{R}_1$ and $Y_2=P_2\mathrm{/}{R}_2$ for the two punctured pieces.

In the notation of step 1.2, each pentagon quotient $Y_i$ is homeomorphic to $T^2$ minus an open disk. For $i=1$, subdivide the free boundary side $\delta$ of $P_1$ at an interior point into consecutive sides $e,f$, and glue a bigon $M$ with boundary word $e^{-1}{f}^{-1}$ to $P_1$ by the affine reversals pairing $e$ with $e^{-1}$ and $f$ with $f^{-1}$. Merging the two faces along $e$ by [L3] gives a one-face schema with cyclic boundary word $fab{a}^{-1}{b}^{-1}{f}^{-1}$; it contains the adjacent inverse pair $f^{-1}f$ after cyclic rotation, and since the pairs $a,a^{-1}$ and $b,b^{-1}$ remain, cancellation by [L3] gives the square schema $ab{a}^{-1}{b}^{-1}$, whose quotient is $T^2$ by [L4]. Let $q:P_1\bigsqcup M\to T^2$ be the quotient map followed by this homeomorphism. The interior $M^{\circ }$ is a saturated open subset of the source, and $q$ is injective on it, so $D_1:=q(M^{\circ })$ is an open disk in $T^2$ and $q(P_1)=T^2\setminus D_1$. The summand $P_1$ is closed in $P_1\bigsqcup M$, though it is not saturated because each point of the subdivided boundary side is also represented on $\mathrm{\partial }M$. The finite-schema quotient map $q$ is closed by [L1], so its restriction $q{\mathrm{\mid }}_{P_1}:P_1\to q(P_1)=T^2\setminus D_1$ is a closed continuous surjection and hence a quotient map. Its fibers are exactly the $R_1$-classes: the bigon pairings identify each subarc of $\delta$ with its corresponding bigon side, and the two endpoints of $\delta$ are already in one $R_1$-class by the $a,b$ pairings. Consequently $Y_1=P_1\mathrm{/}{R}_1\cong T^2\setminus D_1$, with boundary circle the image of $\delta$. The same construction with $c,d,{\delta }^{\mathrm{\prime }}$ in place of $a,b,\delta$ gives $Y_2\cong T^2\setminus D_2$ for an open disk $D_2$, with boundary circle the image of ${\delta }^{\mathrm{\prime }}$.

The identification $S$ glues the boundary circle of $Y_1$ to the boundary circle of $Y_2$ by a homeomorphism, because both are images of the same diagonal arcs; iterating the quotient by the pairings on each summand gives $Y\cong (Y_1\bigsqcup Y_2)\mathrm{/}S$, and by step 2.1 this is exactly the connected-sum model $((T^2\setminus D_1)\bigsqcup (T^2\setminus D_2))\mathrm{/}\text{\hspace{0.17em}⁣}{\sim }_{\varphi }$ of [L5], with the disks and the boundary homeomorphism just constructed. Hence the octagon word realizes the connected sum of two tori.

Each of the four letters occurs once with exponent $+1$ and once with exponent $-1$, so by [L1] all four pairings are orientation compatible; the generator carried by the oriented face continues unchanged across each edge class, and its locally constant generator classes supply a continuous generating section as in [L6]. Hence $Y$ is orientable.

The cell counts of step 1.1 give $\chi (Y)=V-E+F=1-4+1=-2$ by [L7], and this is the Euler characteristic of the connected sum of two tori by step 3.1. Every construction used is a finite explicit quotient, cut or gluing of polygons, so no choice axiom is used.

The torus: by [L1] the quotient of the square by the pairings of the word $ab{a}^{-1}{b}^{-1}$ is homeomorphic to $T^2=(\mathbb{R}\mathrm{/}\mathbb{Z}{)}^2$ and is a connected surface schema, so $T^2$ is a nonempty compact connected boundaryless surface, it is orientable, and its Euler characteristic is $\chi (T^2)=0$.

The Klein bottle: by [L2] the quotient of the square by the pairings of the word $aabb$ is the Klein bottle $K$ and is a connected surface schema, so $K$ is a nonempty compact connected boundaryless surface, it is nonorientable, and $\chi (K)=0$.

Orientability is a homeomorphism invariant. Let $h:M\to N$ be a homeomorphism of boundaryless $n$-manifolds and let $\nu$ be an integral orientation of $N$. For $x\in M$ the map $h$ is a homeomorphism of pairs $(M,M\setminus \{x\})\to (N,N\setminus \{h(x)\})$, so by [L4] it induces an isomorphism $h_{\ast x}:H_n(M,M\setminus \{x\};\mathbb{Z})\to H_n(N,N\setminus \{h(x)\};\mathbb{Z})$, and ${\mu }_x:=h_{\ast x}^{-1}({\nu }_{h(x)})$ generates the infinite cyclic fiber at $x$ because ${\nu }_{h(x)}$ generates its fiber and an isomorphism carries generators to generators. For continuity let $x\in M$: by [L3] there are a closed coordinate ball $K$ around $h(x)$ in $N$ and a generator $c$ of $G_K=H_n(N,N\setminus K;\mathbb{Z})$ with $\nu =\pm s_{K,c}$ on $\mathrm{int}K$; applying [L4] to the homeomorphism of pairs $(M,M\setminus h^{-1}(K))\to (N,N\setminus K)$ and then to $(M,M\setminus \{y\})\to (N,N\setminus \{h(y)\})$ for $y\in \mathrm{int}{h}^{-1}(K)$ gives $r_{K,h(y)}(c)=h_{\ast y}(r_{h^{-1}(K),y}(h_{\ast K}^{-1}(c)))$, so on the ball interior $h^{-1}(K)$ the section $\mu$ equals $\pm s_{h^{-1}(K),h_{\ast K}^{-1}(c)}$, which is a basic generator section and hence continuous. Thus $\mu$ is an integral orientation of $M$, and a homeomorphism transports orientability from its target to its source.

Conclusion. If there were a homeomorphism $h:T^2\to K$, then applying step 1.3 to $h^{-1}:K\to T^2$ and an integral orientation of $T^2$, which exists by [L1], would produce an integral orientation of $K$, contradicting that $K$ is nonorientable by [L2]. Hence $T^2$ and $K$ are not homeomorphic, while both are nonempty compact connected boundaryless surfaces with $\chi =0$ by [L1] and [L2]. Therefore equal Euler characteristic does not determine the homeomorphism class of a compact connected surface. Only the two explicit schema computations, the given orientation data and the functoriality of relative homology are used, so no choice principle is used.