Genus-two orientable polygon

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.