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A boundary component of a surface with boundary is a connected component of its manifold boundary. For a compact surface with boundary, these are finitely many circles. Their number $b$ enters the classification of compact surfaces: $\chi =2-2g-b$ in the orientable case and $\chi =2-k-b$ in the nonorientable surface case, where $k$ is the nonorientable genus.

The torus and Klein bottle are not homeomorphic because only the former is orientable. Adding a projective plane to either by connected sum of surfaces makes a nonorientable surface with Euler characteristic minus one. The classification of compact surfaces identifies both with three projective planes. Thus connected-sum cancellation can fail even for closed connected surfaces.

For a compact connected nonorientable surface, its nonorientable genus $k\ge 1$ is the number of real projective planes in its connected sum of surfaces representation after capping all boundary components of a surface by closed discs. The classification of compact surfaces gives $\chi =2-k-b$, with $b$ boundary components of a surface. The real projective plane has $k=1$ and the Klein bottle has $k=2$.

An orientable surface admits local coordinate orientations that are consistent on overlaps. Equivalently a triangulation can be oriented so that adjacent faces induce opposite directions on their common edge. Use a polygon word to record its successive boundary edges, identifying equal letters in their indicated directions; a negative exponent reverses that edge direction.

Square models are $P:abab$, $K:ab{a}^{-1}b$, and $T:ab{a}^{-1}{b}^{-1}$. The first identifies antipodal boundary points of a disk, giving the real projective plane. Only the torus among these three is orientable: its paired sides traverse the boundary oppositely, whereas each other model has a same-direction pair. The projective model has two vertex classes, and the other two one, giving Euler characteristics one and zero respectively.

The other possibility is the sphere, for example the word $a{a}^{-1}b{b}^{-1}$ with three vertex classes. There are no others: a paired square has one face, two edges and between one and three vertex classes, so its Euler characteristic is $V-1\in \{0,1,2\}$. The classification of compact surfaces leaves precisely the sphere, real projective plane, torus and Klein bottle at those values, with the indicated orientabilities.

For connected sums, remove a disk from each surface, join the resulting boundary circles, and cut a joining arc to produce a single polygon. Its word concatenates the two original words. The real projective plane also has the bigon word $cc$, soThese use respectively six and eight sides, proving the requested hexagon and octagon constructions. Their Euler characteristics are $-1$ and $-2$, also obtained from $\chi (X\#Y)=\chi (X)+\chi (Y)-2$.

The classification of compact surfaces says that every compact connected surface without boundary is homeomorphic either to the sphere with $g\ge 0$ handles, with Euler characteristic $2-2g$, or to a connected sum of $k\ge 1$ real projective planes, with Euler characteristic $2-k$. The first family is orientable and the second is nonorientable; the displayed invariants uniquely determine the type. With boundary one additionally removes a uniquely determined number of open disks.

Number the vertices of the fourteen-gon $v_0,\dots ,v_{13}$ in boundary order. The printed edge identifications implyThus there is one vertex, seven edges and one face in the quotient. ConsequentlyThe two occurrences of $a_1$ have the same sign, as do those of $a_5$, so the orientation on the polygon cannot extend consistently across those paired edges. Thus the surface is nonorientable, and classification identifies it as a connected sum of seven projective planes. The inverse on the first occurrence of $a_4$ and the positive second occurrence have been retained in the vertex calculation.

The classification of compact surfaces states that every compact connected topological surface without boundary is homeomorphic to exactly one of the sphere, a connected sum of surfaces consisting of $g\ge 1$ tori, or a connected sum of $h\ge 1$ real projective planes. The torus family is orientable and the projective-plane family is nonorientable; their respective genera are unique. Including $g=0$ makes the sphere the genus-zero orientable case. If boundaries are allowed, the number of removed open discs is an additional invariant.

Take $M=T^2$, $N={\mathbb{RP}}^2\#{\mathbb{RP}}^2$ and $S={\mathbb{RP}}^2$. The first is orientable, while $N$, the Klein bottle, is not, so $M$ and $N$ are not homeomorphic.

Both $M\#S$ and $N\#S$ are nonorientable and have Euler characteristic $-1$. By the classification of compact surfaces, both are homeomorphic to the connected sum of three projective planes. HenceThis is failure of cancellation for connected sums of surfaces.

The classification of compact surfaces says that every compact connected surface without boundary is homeomorphic to exactly one of the following: the sphere, an orientable surface obtained as the connected sum of $g\ge 1$ tori, or a nonorientable surface obtained as the connected sum of $k\ge 1$ real projective planes. The integers $g$ and $k$ are unique, and the orientable and nonorientable families do not overlap. Including $g=0$ makes the sphere the genus-zero member of the orientable family.

Their Euler characteristics are $2-2g$ and $2-k$, respectively. Thus orientability and the Euler characteristic determine the closed connected surface. If boundaries are allowed, one additionally specifies the number of boundary circles and removes that many open discs from a closed model.

Here the surface is the real projective plane, with Euler characteristic one. The bound gives $V\ge 6$. The following ten triangles on vertices $1,\dots ,6$ attain it:Take these as the maximal simplices of an abstract simplicial complex, identifying edges with the same vertex labels. Every one of the 15 pairs of vertices occurs as an edge in exactly two triangles. The links of the six vertices are single five-cycles; listed cyclically they areThus every point has a disk neighborhood, and the realization is a connected closed surface. Its Euler characteristic is $6-15+10=1$, so the classification of compact surfaces identifies it with the real projective plane. This is an explicit triangulation, proving $\boxed{V_{\min }=6}$.

The classification of compact surfaces says that every compact, connected topological surface without boundary is homeomorphic to exactly one of the following: the sphere $S^2$; a connected sum of surfaces consisting of $g\ge 1$ tori; or a connected sum of surfaces consisting of $k\ge 1$ real projective planes. The first two cases are orientable; the last case is a nonorientable surface. Their Euler characteristics areHere ${\Sigma }_g$ has genus $g$, and $N_k$ has $k$ crosscaps; $N_1$ is the real projective plane and $N_2$ the Klein bottle. Thus orientability together with the Euler characteristic determines the closed surface up to homeomorphism.

If the convention permits a surface with boundary, also remove the interiors of $b\ge 0$ disjoint closed discs from one of these closed surfaces. The Euler characteristic decreases by $b$, giving $2-2g-b$ in the orientable case and $2-k-b$ in the nonorientable surface case. The number of boundary components is then an additional invariant.

Number the polygon vertices $v_0,\dots ,v_9$ in the order of the given polygonal schema, with its $i$th polygon edge going from $v_i$ to $v_{i+1}$, and subscripts interpreted modulo ten. Equal labels identify equal arrow endpoints. The five pairings giveConsequently there are two vertex equivalence classes,The quotient space has one face, five edges and two vertices, so the polygonal-schema Euler count givesEach label occurs once with each sign. Hence every pairing reverses the direction induced by traversing the boundary of the polygon, so an orientation of a surface on its interior extends consistently across every glued edge. Thus $X$ is orientable. The classification of compact surfaces now gives $2-2g=-2$, or a closed orientable surface of genus two, equivalently the connected sum of surfaces of two tori.

We prove the Loop theorem by a finite tower of double covering spaces, keeping track of the boundary class during the descent. This works without assuming orientability or three-manifold irreducibility.

Let $F$ be the boundary component containing the basepoint. A nontrivial element in the kernel of ${\pi }_1(F)\to {\pi }_1(M)$ is represented by the boundary of a singular disk $u:D^2\to M$. Arrange that the interior of the disk maps into the interior of $M$, and make $u$ simplicial after simplicial subdivision. Let $V_0$ be a small regular neighbourhood of its image, retaining a subsurface $B_0\subset \partial V_0$ in $F$ containing the boundary loop. More generally, during the construction we require the boundary loop to stay outside the normal subgroup which is the kernel of the map from the current boundary subsurface to ${\pi }_1(F)$. This records exactly the information needed at the end.

If $V_0$ has a connected double cover, lift $u$ to that cover and take a regular neighbourhood $V_1$ of the lifted disk image. Repeat this whenever the current neighbourhood has a connected double cover. This procedure is finite. Indeed, let $C_i$ be the simplicial image of the disk at stage $i$. The full inverse image of $C_i$ in a connected double cover is connected, since the cover retracts to this inverse image. It is the union of the lifted disk image $C_{i+1}$ and its translate under the nontrivial deck transformation. These two images must meet. At least one image simplex downstairs consequently has two distinct preimages represented in $C_{i+1}$. No image simplex disappears, so the number of distinct image simplexes strictly increases. This number is at most the number of simplexes of the fixed domain disk. Thus the tower proof of the loop theorem terminates.

At the top, $V_r$ has no connected double cover. Equivalently, $H^1(V_r;{\mathbb{F}}_2)=0$. Mod-two Poincare-Lefschetz duality and the relative homology sequence giveThe classification of compact surfaces now makes every component of $\partial V_r$ a two-sphere. The lifted boundary subsurface $B_r$ is planar. Its fundamental group is normally generated by its boundary circles. Since the lifted loop is outside the prescribed normal subgroup, one boundary circle of $B_r$ is outside that subgroup. That circle bounds a disk on its boundary sphere; push the interior of this disk into $V_r$. We have obtained an embedded disk with the required boundary-class property at the top of the tower.

We next descend one double cover. Projecting an embedded disk gives a singular disk with only double curves: triple points and branch points cannot arise from a two-sheeted cover. After general position, these double curves are circles and proper arcs. The following cut-and-paste procedure eliminates them while preserving a boundary class outside the prescribed normal subgroup.

First eliminate the circular double curves. Take a double circle innermost on one of its preimage sheets. If its two preimage circles bound disjoint subdisks, cut out the two subdisks and interchange small displaced copies of them; discard any resulting closed component and retain the disk carrying the original outer boundary. If the circles are nested, replace the disk bounded by the outer preimage circle by the innermost disk, pushed slightly onto the appropriate sheet. Along the common image circle the two transverse sheets are separated in a small product neighbourhood. In the nested case the intervening annulus may have its sides interchanged under projection; separating the two sheets on its double cover and then rounding the two opposite corners makes the same replacement downstairs. The outer boundary is unchanged, no new double curve is introduced, and the chosen double circle disappears. Repeating leaves only double arcs.

For a double arc, cut the domain along its two preimages and reglue the banks using the two possible resolutions. There are two disk choices. At least one has boundary outside the prescribed normal subgroup. Here is the boundary calculation that proves this, rather than assuming that an arbitrary resolution preserves essentiality. With connecting paths absorbed into four boundary words $\alpha ,\beta ,\gamma ,\delta$, the old word is $W=\alpha \beta \gamma \delta$. In one configuration the two possible new words areIn the other configuration they areChanging the connecting paths merely conjugates these words. In either configuration, if both new boundary words belonged to a normal subgroup, so would the old word. Choose a disk whose word stays outside the subgroup. The double arc used in the resolution disappears, and the remaining double curves are inherited from the old disk, so their number decreases. Remove any circular double curves again, and repeat. The resulting disk is embedded.

Apply this descent successively through the tower, and finally include $V_0$ in $M$. Its boundary lies in $F$ and represents a nontrivial element of ${\pi }_1(F)$; a simple closed curve in a surface is null-homotopic precisely when it bounds a disk in that surface. We have therefore provedThe argument also permits a boundary subsurface in place of the whole boundary component. In particular, if an embedded simple closed boundary curve $c$ bounds a singular disk, use a small annulus about $c$ as the boundary subsurface. The embedded disk produced has boundary an essential simple closed curve in that annulus, hence isotopic to $c$. Extend that isotopy through a boundary collar. This proves the fixed-boundary form of Dehn's lemma: the original curve $c$ itself bounds an embedded disk with interior in the manifold.

Now consider the knot map. Because the map is an embedding near the inverse image of $K$, its restriction to $\partial D^2$ is an embedding into $K$ and therefore maps onto $K$. No interior point can map to $K$, since it would duplicate a boundary image. Compactness away from a small domain collar permits a sufficiently small tubular neighbourhood of $K$ to be chosen whose inverse image lies in the embedded collar. Its intersection with this collar cuts off an annulus. The rest of the disk maps into the knot exterior, with boundary a simple closed curve $c$ on the boundary torus.

The annulus shows that $c$ runs once along $K$. The mapped disk in the exterior gives zero meridional coefficient, equivalently zero linking number with $K$. Thus $c$ is a longitude. Apply the fixed-boundary Dehn's lemma just proved to obtain an embedded disk in the exterior bounded by $c$. Attach the embedded collar annulus between $c$ and $K$. This gives an embedded spanning disk for $K$. A regular neighbourhood of that disk is a standard three-ball with $K$ its equatorial curve, so $K$ is the unknot.

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.

True. Label the pentagon polygon vertices clockwise from the upper left asUse $a$ for the upward single arrows and $b$ for the rightward double arrows in the original diagram. Their endpoint identifications areAll five polygon vertices are therefore identified. The remaining diagonal polygon edge, from $v_3$ to $v_4$, has its endpoints identified and is the only boundary component of a surface.

The paired polygon edges join half-disc neighbourhoods into disc neighbourhoods. At the common vertex, the five corner sectors join in one chain whose two free ends lie on the unpaired polygon edge: its vertex link in a polygonal schema is a closed interval. Hence that point has a half-disc neighbourhood. This verifies that the quotient space is a surface with boundary, rather than merely counting its CW cells. The paired polygon edges are oppositely traversed around the polygon, so this surface is orientable.

Its CW cells comprise one vertex, three edges (the $a$ pair, the $b$ pair, and the free polygon edge), and one face. ThereforeThe classification of compact surfaces identifies it as a torus with an open disc removed.

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.