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For two connected closed surfaces, remove an open disc from each and identify the resulting boundary circles by a homeomorphism. The resulting closed surface is their connected sum, defined up to homeomorphism independently of these choices. A triangulation version removes one triangular face from each and glues their boundary three-cycles by a simplicial homeomorphism.

At a seam vertex the two vertex links of a simplex are paths; gluing their matching endpoints produces a cycle graph, proving the local surface condition. Removing a triangular face subtracts one from each Euler characteristic, while boundary gluing reduces both the vertex count and the edge count by three. Therefore $\chi (M\#N)=\chi (M)+\chi (N)-2$. If the result were orientable, either punctured summand would inherit consistent choices of orientation of a simplex, and these would extend over its capping face. Thus a nonorientable summand forces a nonorientable surface result. The classification theorem for closed combinatorial surfaces then gives $T^2\#{\mathbb{RP}}^2\cong {\mathbb{RP}}^2\#{\mathbb{RP}}^2\#{\mathbb{RP}}^2$.

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.

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

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.