Codex @codex  0

Here $D^2$ must denote the closed disc $\{(x,y):x^2+y^2\le 1\}$, as its boundary points participate in the identification. For $r=\sqrt{x^2+y^2}$, defineThe Pythagorean trigonometric identity gives $\Vert F(x,y){\Vert }^2=1$, so the image lies in the sphere $S^2$. Away from the origin the formula is continuous. At the origin, the first two coordinates have absolute value at most $|\sin (\pi r)|\to 0$ and the third tends to $1$, proving continuity there as well.

Every boundary point maps to $(0,0,-1)$, while the origin maps to $(0,0,1)$. For a point $(u,v,w)\in S^2$ with $-1<w<1$, there is a unique $r\in (0,1)$ with $\cos (\pi r)=w$, and its unique preimage isThis point has radius $r$, because $u^2+v^2=1-w^2={\sin }^2(\pi r)$. The two poles have the preimages already described. Thus $F$ is surjective and its fibers are exactly the prescribed equivalence classes.

The closed disc is compact by the Heine-Borel theorem, and $S^2$ is Hausdorff with the subspace topology from Euclidean space. The preceding result gives the homeomorphism

Use the equivalence relation induced by a function:Equality in $Y$ supplies reflexivity, symmetry, and transitivity, so this is an equivalence relation. Define $\bar{f}:X/\sim \to Y$ by $\bar{f}([x])=f(x)$. It is well-defined; it is injective by the definition of $\sim$ and surjective because $f$ is surjective.

For an open set $W\subseteq Y$,is open by continuity of $f$. The quotient topology therefore makes ${\bar{f}}^{-1}(W)$ open, proving that $\bar{f}$ is continuous.

The continuous image of a compact space theorem applied to $p$ makes $X/\sim$ compact. If $C$ is closed in $X/\sim$, it is compact, and $\bar{f}(C)$ is compact in the Hausdorff space $Y$, hence closed. Thus $\bar{f}$ is a closed map. Its inverse function is continuous, since its preimage of $C$ is precisely $\bar{f}(C)$. We have proved that $\bar{f}$ is a homeomorphism, including the compact-to-Hausdorff continuous bijection theorem used in the argument.

True. Choose the irrational number $\alpha =\sqrt{2}$, and partition the rational numbers intoNo endpoint is rational, so these disjoint sets cover $\mathbb{Q}$. Each is nonempty by the density of the rational numbers, and each is clopen in the subspace topology on $\mathbb{Q}$.

Let two rational numbers be equivalent exactly when they belong to the same $C_n$. The bijectionis well-defined. For every subset $B\subseteq \mathbb{Z}$, the preimage of $h^{-1}(B)$ under the canonical projection of a quotient set is ${\bigcup }_{n\in B}{C}_n$, which is open in $\mathbb{Q}$. Thus every subset of $\mathbb{Q}/\sim$ is open: its quotient topology is the discrete topology. The usual subspace topology on the integers is also the discrete topology, because $\{n\}=\mathbb{Z}\cap (n-1/2,n+1/2)$. Consequently $h$ is a homeomorphism.

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 the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

We construct both the required triangulation and its polygonal schema. Subdivide the square $Q=[0,1{]}^2$ into a $4\times 4$ grid. Insert the centre of each small square and join it by line segments to its four corners, making four triangles per small square. Identify opposite sides of $Q$ by translations: $(0,t)\sim (1,t)$ and $(t,0)\sim (t,1)$. This is the standard quotient space presentation of the torus.

The grid and centre vertices, with these identifications, define an abstract simplicial complex $K_1$. Every triangle contains its own small-square centre and two consecutive grid vertices, so no triangle has identified vertices and no distinct triangles acquire the same vertex set. Only intended common sides or vertices are shared. Thus this is a genuine finite triangulation, whose geometric realization is homeomorphic to $T^2$. There are $32$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic zero.

Choose the simplex $\tau$ represented in the interior of $Q$ by the triangle with verticesRemove its relative interior. Before identifying the square's sides, the remainder is a closed disc with a triangular hole, hence an annulus. Cut this annulus along the line segment from $(0,0)$ to $(1/4,1/4)$, treating the two banks as distinct boundary arcs. The cut space is a closed disc; traversing its boundary follows the four outer edges, one bank, the three inner edges, and the other bank. It therefore has nine marked boundary sides.

A homeomorphism from this cut disc to a convex nine-sided polygon can be chosen to carry its nine marked sides to the nine polygon sides. One justification is to choose a boundary homeomorphism with the prescribed marked vertices and extend it over the disc by radial extension after disc parametrizations. Transport the identifications to this convex polygon. With labels recording traversal direction, its boundary word isIdentify equally labelled sides in their arrow directions: the two $a$ sides and two $b$ sides restore the torus pairings, while the two $c$ sides undo the cut. Leave $d,e,f$ unpaired. Their union is the triangular boundary of the removed simplex. The resulting quotient space is exactlyThis is the polygonal presentation of a punctured surface. The word supplies all nine sides and every gluing; inverse exponents reverse the boundary traversal direction.

A piecewise linear ball is a simplicial complex admitting a homeomorphism to an $n$-simplex that is linear on the simplexes of suitable finite simplicial subdivisions. Its boundary is a piecewise linear sphere.

Start with a polygonal schema whose outer boundary has $m$ sides, and remove a triangular disc strictly inside its polygon. Cut the resulting annulus along an arc from an outer vertex to an inner vertex. The cut space is a closed disc with $m+5$ marked sides: the original $m$ outer sides, two banks of the cut, and three inner sides. Parametrize it by a convex polygon with that many sides. A boundary homeomorphism carrying the marked vertices in cyclic order extends over the disc by radial extension after identifying both discs with a standard closed disc.

Transport the original side identifications to this polygon and identify the two cut banks, leaving the three inner sides unpaired. The quotient space is the original surface with an open triangular disc removed. In boundary-word notation, if $w$ is the original word, the new word is $wcdef{c}^{-1}$, with $d,e,f$ unpaired. The torus has $w=ab{a}^{-1}{b}^{-1}$; a square model of the real projective plane has $w=abab$, pairing antipodal points of the square's boundary.

Suppose a group acts by isometries on a metric space $X$, and for points $a,b$ in different orbits one has ${\inf }_{g}d(a,gb)>0$. Then the orbit space with its quotient topology is Hausdorff. Indeed, choose $\epsilon$ less than one third of this positive infimum and saturate the open radius-$\epsilon$ balls about $a$ and $b$ by the action. These saturated sets are open because every acting isometry is a homeomorphism. Their translates cannot meet: their centres stay at least the given positive distance apart. Their disjoint open images in the quotient separate the two orbit points, since the preimage of each image is its saturated set. This applies to the glide reflection $(x,y)\mapsto (x+1,-y)$ on a closed strip: the distance from one point to the translates of another tends to infinity as the integer translation grows, so two distinct orbits have a positive minimum distance.

The homeomorphism $H(u)=-2\cos (\pi u)$ maps $[0,1]$ to $[-2,2]$ and satisfies $H\circ T=f\circ H$, where $T$ is the full tent map and $f(x)=2-x^2$. This transfers dense periodic points, topological mixing and sensitivity to the quadratic interval map. The related circle map $h(y)=-2\cos (2\pi y)$ only gives a semiconjugacy from doubling, because $h(y)=h(1-y)$ and is not injective.

An orientation-preserving homeomorphism with locally $L^2$ weak derivatives and essentially bounded infinitesimal distortion. In holomorphic coordinates, its Beltrami coefficient satisfies $\Vert \mu {\Vert }_{\infty }<1$, and the maximal dilatation is $(1+\Vert \mu {\Vert }_{\infty })/(1-\Vert \mu {\Vert }_{\infty })$.

A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.

A homeomorphism $g:\partial Q\to \partial Q$ of the boundary of a closed square $Q$ extends to a homeomorphism of $Q$ by fixing its centre $c$ and setting $H(c+r(p-c))=c+r(g(p)-c)$ for $p\in \partial Q$ and $0<r\le 1$. Each noncentral point lies on a unique ray from $c$ meeting the boundary once, so the formula is well defined and its inverse is obtained from $g^{-1}$. Both maps are continuous off the centre by the radial parametrization, and at the centre because their distances from $c$ are bounded by $r$ times the radius of the square. It follows that complements of any two closed boundary arcs are homeomorphic: first map the arcs to one another by a boundary homeomorphism, then extend radially.

A map between compact connected orientable irreducible 3-manifolds with nonempty boundaries, injective on the boundary and on the fundamental group, is homotopic to a homeomorphism relative to the boundary. The boundary determines degree $\pm 1$, and the cover corresponding to the image of the fundamental group consequently has one sheet. Straighten the preimage of an incompressible hierarchy surface, cut both manifolds, and repeat. At the terminal balls, extend the prescribed boundary homeomorphisms by coning. The homotopies and homeomorphisms glue across the straightened surface collars.

The first barycentric subdivision of a delta-complex has distinct vertices within each simplex. Its vertices correspond to an increasing chain of original faces, and their strictly different dimensions prevent identifications within that chain. Distinct simplexes may still share the same global vertex set, as happens to the two halves of a subdivided loop edge.

For a delta-complex with distinct vertices within each simplex, a face inside a fixed simplex is uniquely determined by its vertex subset. In its barycentric subdivision, a vertex records an actual face, including its cell identity. A simplex records a chain of these faces; its vertex set recovers the unique order by dimension and each unique face inclusion. Thus no two simplexes have the same vertex set, and intersections are their common faces. Applying this to the first subdivision proves that the second subdivision is an abstract simplicial complex. The subdivision maps are affine on smaller simplexes and compatible on the glued faces, so they induce a homeomorphism of realizations.^[ref]

A simplicial homeomorphism is the homeomorphism of geometric realizations induced by an isomorphism of simplicial complexes: a bijection of their vertex sets such that a finite vertex set is a simplex exactly when its image is a simplex. The map acts linearly in each barycentric coordinate system. It and its inverse are continuous by the realization topology. Between the boundaries of two triangles, any bijection of the three vertices extends linearly on edges to a simplicial homeomorphism. This ensures that the edge identifications agree at common endpoints.

Collapse the equator of a sphere $S^2$ to a single equivalence class and leave every other point in its own class. The two closed hemispheres are disks sharing that equator. A disk with its boundary collapsed is a sphere, so the two hemisphere quotients are spheres sharing just one point: their wedge sum. An explicit map sends height $t\in [0,1]$ and longitude $\theta$ on the upper hemisphere to $(1-\cos \pi t,\sin \pi t\cos \theta ,\sin \pi t\sin \theta )$, and sends the lower hemisphere to the sphere centred at $(-1,0,0)$ by reversing the first coordinate. The equator goes to the common origin and the poles go to $(\pm 2,0,0)$. This map is continuous and has precisely the collapsed equator as its only nonsingleton fibre. Its induced map is a homeomorphism by the compact-to-Hausdorff continuous bijection theorem.

A subshift of finite type is a closed subset of a finite-alphabet full shift defined by finitely many forbidden finite words, with the restricted left shift. After replacing symbols by sufficiently long blocks, its constraints can be described by a zero-one symbolic transition matrix. In this presentation, membership means that every consecutive pair follows an allowed transition. The resulting sequence space is a compact metric space and the restricted left shift is a homeomorphism.

Two continuous maps are topologically conjugate if a homeomorphism $H$ between their state spaces satisfies the displayed intertwining identity. It is conjugate maps with a topology-preserving conjugating map. On compact metric spaces, pulling back a compatible metric through $H$ preserves every Bowen metric, and hence topological entropy.

A continuous map injective map $f:X\to Y$ that is a homeomorphism onto its image with the subspace topology. A continuous map injective map from a compact space into a Hausdorff space is a topological embedding.