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There are exactly four types: the two-sphere, real projective plane, torus, and Klein bottle. Four sides admit three pairings and four direction choices for each pairing. Up to cyclic relabelling, labels are adjacent, $a{a}^{s}b{b}^t$, or alternating, $ab{a}^s{b}^t$. For adjacent labels, the vertex counts for $(s,t)=(-1,-1)$, a mixed-sign pair, and $(1,1)$ are $3,2,1$. For alternating labels they are $1,1,2$. There are two edge classes and one face, so Euler characteristic is $V-1$. Orientability occurs exactly when $s=t=-1$. The classification theorem for surfaces identifies every case, leaving no additional closed surfaces from a square. Examples are $a{a}^{-1}b{b}^{-1}$ for the sphere, $aab{b}^{-1}$ for the projective plane, $ab{a}^{-1}{b}^{-1}$ for the torus and $ab{a}^{-1}b$ for the Klein bottle.

For the triangle quotient, the universal coefficient theorem for homology gives $H_j(X;{\mathbb{F}}_2)=0$ for every $j>0$, because its only positive integral homology is $\mathbb{Z}/3$. A connected positive-dimensional closed manifold has a nonzero top mod-two fundamental class, independently of orientability. Thus this quotient cannot be homotopy equivalent to such a manifold. A connected zero-dimensional closed manifold is a point, which its nonzero $H_1(X;\mathbb{Z})$ also excludes.

For the attached-band space, the same universal coefficient theorem for homology gives nonzero $H_2(X;{\mathbb{F}}_2)$ and zero homology above degree two. A connected closed manifold homotopy equivalent to it would consequently have dimension two. Its Euler characteristic would be one. The classification theorem for surfaces says that a closed connected surface of Euler characteristic one is the real projective plane. But its integral $H_1$ is $\mathbb{Z}/2$, differing from both possibilities computed in part 2. Thus this space is never homotopy equivalent to a closed manifold.

The antipodal mapping torus is already a closed manifold: gluing the two boundary components of $S^n\times [0,1]$ by a diffeomorphism gives a smooth $(n+1)$-manifold without boundary. For $n\ge 1$, it is orientable exactly when $(-1{)}^{n+1}=1$, namely when $n$ is odd. In that case $-I$ on ${\mathbb{R}}^{n+1}$ is isotopic to the identity through rotations, so the mapping torus is diffeomorphic to $S^n\times S^1$. The $n=0$ case is a circle as above. Only the third family can be homotopy equivalent to a closed manifold.

Put $A={\Sigma }_h^{\partial }$. This surface with boundary deformation retracts onto a wedge of $2h$ circles, so $H^1(A;\mathbb{R})$ has dimension $2h$ and $H^2(A;\mathbb{R})=0$. If $r$ were a retraction, ${\iota }^{\ast }{r}^{\ast }=\mathrm{id}$ would make $r^{\ast }$ injective. Its image $W\subset H^1({\Sigma }_g;\mathbb{R})$ would have dimension $2h$.

Every pair of classes $u,v$ in $H^1(A;\mathbb{R})$ has $u⌣v=0$, since $H^2(A;\mathbb{R})=0$. Naturality givesThus $W$ is an isotropic subspace of a symplectic vector space for the nondegenerate skew Poincare duality pairing on the $2g$-dimensional space $H^1({\Sigma }_g;\mathbb{R})$. The stated linear-algebra bound gives $2h\le g$. Hence

The bound is sharp. Double $A$ along its boundary: $D(A)=A{\cup }_{\partial A}A$ is the closed oriented surface of genus $2h$. Identify each copy with $A$ and fold them onto one copy. The two maps agree on the joining circle, so they give a continuous retraction fixing the first copy pointwise. This includes $h=0$, where the double of a disc is a sphere.

More generally, for $g\ge 2h$, add $g-2h$ handles in the interior of the second copy. Pinch those extra handles onto their connecting point, keeping the boundary fixed, and then fold onto $A$. This is still a retraction. In fact the construction works for any embedding in the question: the connected complement has one boundary component and genus $g-h$ by Euler-characteristic additivity, and the classification theorem for surfaces identifies it, relative to that boundary, with a copy of $A$ having $g-2h$ additional handles. Thus the exact existence criterion is $2h\le g$, as expressed by retraction onto a punctured oriented surface.

For a compact oriented regular surface $S\subset {\mathbb{R}}^3$ without boundary, the Gauss-Bonnet theorem statesHere $K=k_1{k}_2$ is the Gaussian curvature, the product of the two principal curvatures; $dA$ is the area element induced by the first fundamental form; and $\chi (S)$ is the Euler characteristic. If $S$ is connected and has genus $g$, the classification theorem for surfaces gives $\chi (S)=2-2g$, so

Since $K\ge 0$, Gauss-Bonnet gives $4\pi (1-g)\ge 0$, hence $g\le 1$. If $g=1$, the integral of the continuous nonnegative function $K$ is zero, so $K$ vanishes everywhere.

That is impossible for a compact regular surface in ${\mathbb{R}}^3$. Choose $p\in S$ maximizing the Euclidean distance from the origin, after translating the origin off the surface if necessary. The sphere centered at the origin through $p$ is a supporting sphere. The supporting-sphere curvature bound shows that both principal curvatures at $p$ have the same sign and nonzero magnitude, so $K(p)>0$. Therefore $g\ne 1$, and $g=0$. The classification theorem for surfaces now shows that $S$ is diffeomorphic to the sphere.

False with the $\mathrm{limsup}$ printed in the paper. Let the even-indexed surfaces be one fixed unit sphere and the odd-indexed surfaces one fixed ring torus. Thenbut infinitely many surfaces are tori. The two fixed surfaces lie in one ball and have uniformly bounded area. Their injectivity radii have a positive common lower bound, and compactness gives a common bound on $|\nabla K|$, hence on every radial derivative $|{\partial }_{r}K|$. Thus this alternating sequence satisfies all four additional conditions as well.

The intended statement becomes true if $\mathrm{limsup}$ is replaced byHere is the proof. Put ${\delta }_n=\max (0,-{\inf }_{S_n}{K}_n)$, so ${\delta }_n\to 0$. Choose $p_n$ maximizing distance from the centre of the fixed containing ball. The supporting-sphere curvature bound givesWrite $K_0=K_n(p_n)$ and $j(r,\theta )=\sqrt{G(r,\theta )}$ in geodesic polar coordinates. Choose a fixed sufficiently small $c>0$ and set ${\rho }_n=c{K}_0^{-1/2}$. Conditions (3) and (4) allow $c$ to be chosen so that ${\rho }_n\le {ϵ}_0$ andThe Jacobi equation $j_{rr}=-K_{n}j$, with $j(0)=0$ and $j_r(0)=1$, and the Sturm comparison theorem then give $j(r,\theta )\ge c_{1}r$ on this ball for a constant $c_1>0$ independent of $n$. ConsequentlySince $K_n\ge -{\delta }_n$ elsewhere and $\mathrm{Area}(S_n)\le M$,for all sufficiently large $n$. Gauss-Bonnet forces $\chi (S_n)>0$, and the classification theorem for surfaces makes each such $S_n$ a sphere.

Apply the first extraction move to the two outer occurrences of $x$, then cyclically rotate the word to expose the two occurrences of $y^{-1}$:In the third line the repeated letter for the move is $y^{-1}$, and the intervening string is $z^{-1}{z}^{-1}$, whose inverse is $zz$. Rename $y^{-1}$ as a new edge label. The final word has the standard form $aabbcc$.

Each same-direction repeated pair represents a projective-plane summand, so the classification theorem for surfaces identifies this word with the connected sum of three real projective planes:Thus the surface has crosscap number three. As a check, the original identifications make all six polygon vertices equivalent, leaving one vertex, three edges and one face. Its Euler characteristic is $1-3+1=-1=2-3$, consistent with that crosscap count.

Here is an explicit simplicial complex on six vertices. Take every subset of one of the following ten triangular facets:For example, $135$ denotes $\{1,3,5\}$, with all its edges and vertices included. Its one-skeleton has all fifteen edges of the complete graph. Every edge belongs to exactly two triangles. The links of a simplex at the six vertices are the following cycles, with the first vertex understood to repeat at the end:Thus every vertex has a disk neighbourhood, as do edge and face interior points. The realization is a compact connected surface without boundary. Its Euler characteristic isThe classification theorem for surfaces identifies the unique closed connected surface with this Euler characteristic as the real projective plane. Consequently this complex triangulates ${\mathbb{RP}}^2$. The explicit facet and link checks make this a genuine simplicial complex, not a polygon whose boundary identifications have been mistaken for simplicial faces.

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.

The classification theorem for surfaces states that every connected closed combinatorial surface is homeomorphic to exactly one of:

The orientable family has Euler characteristic $2-2g$; the nonorientable family has Euler characteristic $2-h$. Orientability together with Euler characteristic distinguishes the entries.

Boundary words giving the required polygonal schemas areA two-sided polygon here means a disk with its boundary split into two arcs; subdivision produces an ordinary polygon with more sides without changing the quotient. Alternatively the square word $a{a}^{-1}b{b}^{-1}$ represents the sphere. Each four-letter block in the orientable word adds a torus handle; each repeated-letter block in the nonorientable word adds a projective-plane summand.

On vertices $1,\dots ,6$, take the simplicial complex generated by $123,124,135,146,156,236,245,256,345,346$. Every edge occurs in two facets. The vertex links, listed cyclically, are $23564$, $13654$, $12645$, $12536$, $13426$, and $14325$, respectively. Thus each vertex link is a cycle graph, and the realization is a compact connected surface without boundary. Its Euler characteristic is $6-15+10=1$, so the classification theorem for surfaces identifies it with the real projective plane. All fifteen possible edges occur, and the listed triangles are distinct three-element sets, so this is an honest abstract simplicial complex, not merely a polygon presentation.