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Every closed topological manifold with the homotopy type of $S^d$ is homeomorphic to $S^d$. This is a classification theorem about homeomorphisms, and does not assert a diffeomorphism between prescribed smooth structures. It is the precise final step needed after proving that a cone link is a simply connected integral homology sphere in the manifold criterion for a suspension.

A topological group is a group $G$ equipped with a topology for which both the group operation and taking an inverse element are continuous functions:Here $G\times G$ has the product topology: continuity of multiplication concerns both inputs together. Thus nearby motions compose to nearby motions, and undoing a motion also varies continuously. The definition alone does not impose compactness; we state the Hausdorff space assumption explicitly when it is needed.

The additive group $(\mathbb{R},+)$ of real numbers, with its usual topology, is a topological group: $(x,y)\mapsto x+y$ and $x\mapsto -x$ are continuous functions. Its identity element is $0$. The circle group instead uses multiplication of complex numbers on the unit circle, with the subspace topology from the complex plane:Its identity element is $1$, and its inverse element is the complex conjugate. Multiplication on the complex plane and complex conjugation are continuous functions, so their restrictions verify the required continuity. Angles add, but points multiply: a quarter-turn followed by another quarter-turn gives $i\cdot i=-1$, and the next two quarter-turns return to $1$. Ordinary addition would fail to stay on the unit circle, since $1+1=2$.

For any fixed $a\in G$, the translations $x\mapsto ax$ and $x\mapsto xa$ are homeomorphisms, with inverses given by translation by $a^{-1}$. Consequently the neighbourhoods of the identity element determine the entire topology. This is why local information near one motion can control a whole family of symmetries. See Tornier, Haar Measures, section 1.1 for these basic properties of topological groups.

A homeomorphism preserves every homology group, hence the Euler characteristic wherever the finite rational Betti number sum is defined. In fact, homotopy invariance of homology makes Euler characteristic invariant under homotopy equivalence. Thus different finite CW complex structures on the same space have equal alternating cell counts.^[ref]

A topological invariant is a property or mathematical object preserved by homeomorphisms of topological spaces. Connectedness and compactness are examples. The fundamental group is an invariant up to group isomorphism; it is also preserved up to isomorphism by homotopy equivalence, a stronger invariance property which does not hold for every topological invariant.

A topological $n$-manifold without boundary is a topological space $M$ satisfying three conditions: it is Hausdorff, it is second countable, and every point $p\in M$ lies in an open set $U\subseteq M$ with a homeomorphism $\phi :U\to V$ onto an open set $V\subseteq {\mathbb{R}}^n$. Here $n\ge 0$ is a fixed dimension of a manifold, and ${\mathbb{R}}^n$ has the usual topology of Euclidean space. The pair $(U,\phi )$ is a coordinate chart. Unless explicitly qualified, “topological manifold” here means without boundary.^[ref]

The Hausdorff condition requires disjoint open neighbourhoods for distinct points. Second countability requires a countable basis of a topology. Neither follows merely from the local homeomorphisms: non-Hausdorff manifolds fail the first condition, while an uncountable disjoint union of copies of the real line fails the second. Indeed, a basis of a topology must supply a different nonempty member inside each copy, so it cannot be countable.

The circle and the real line are both one-dimensional topological manifolds, with the same local model. Their global topologies differ: the circle is compact, while the real line is not. By contrast, the junction of a topological tripod already fails the local test. Thus local coordinate charts specify what kind of small pieces are allowed; global topological invariants distinguish how those pieces fit together.^[ref]

The usefulness of a topological manifold is that local questions about continuity can be transported into Euclidean space, while compactness, connectedness, and the fundamental group still record global information. For calculus, further structure is needed: a smooth atlas requires smooth transition maps, letting the chain rule reconcile derivatives computed in different coordinate charts. A topological atlas by itself only guarantees homeomorphisms between coordinates.^[ref]

A topological space is a set $X$ together with a topology $\tau \subseteq \mathcal{P}(X)$, a family of subsets satisfyingHere $\mathcal{P}(X)$ is the power set; the union may have arbitrarily many members, while the intersection must be finite. The members of $\tau$ are the open sets. These are axioms about subsets, with no assumed metric or surrounding Euclidean space. ^[ref]

For $X=\{a,b\}$, the discrete topology is ${\tau }_d=\{\varnothing ,\{a\},\{b\},X\}$, while the indiscrete topology is ${\tau }_i=\{\varnothing ,X\}$. Both satisfy the axioms. They describe different topological spaces on the same set: the first is Hausdorff and disconnected, while the second is connected and not Hausdorff. Each is compact because $X$ is a finite set.

The identity function $i:(X,{\tau }_d)\to (X,{\tau }_i)$ is a continuous map: the only target open sets have inverse images $\varnothing$ and $X$. Its inverse is not a continuous map, since the target open set $\{a\}$ has inverse image $\{a\}$, which is not an open set in ${\tau }_i$. Thus even a continuous bijection need not be a homeomorphism.

For two coordinate charts in a topological atlas, the topological transition map is the homeomorphismBoth domains are open sets in Euclidean space. Function composition gives $g_{ij}=g_{ji}^{-1}$ and $g_{ki}=g_{kj}\circ g_{ji}$ wherever all three coordinate charts overlap. These identities express consistency of the coordinate descriptions; imposing smooth functions as the coordinate changes gives a smooth atlas.

A topological tripod is three copies of the closed interval $[0,1]$ with their zero endpoints identified to a single junction $o$. Equivalently, embed three line segments with distinct directions in the Euclidean plane and use the subspace topology. It is a compact, connected metric space, and is Hausdorff, second countable, and paracompact. Every interior arm point has a coordinate chart to an open interval, and each outer endpoint has a half-interval chart as in a topological manifold with boundary. The junction has neither kind of chart.

Here is a precise local obstruction. Let $S$ be a small open neighbourhood of $o$ consisting of initial segments of the three arms, excluding their outer endpoints. Every connected open neighbourhood $N$ of $o$ contained in $S$ contains an initial segment on every arm, because $N$ is open at $o$. Its intersection with each arm has no gaps: if a point between $o$ and a point of $N$ were missing, cutting there would separate the part beyond it from the rest of $N$, contradicting connectedness. Thus $N\setminus \{o\}$ has exactly three connected components, one interval per arm.

Suppose a coordinate chart $h$ to the real line existed at $o$. Choose a small open interval $I$ about $h(o)$ lying inside $h(S\cap \mathrm{domain}(h))$, and restrict the chart to $N=h^{-1}(I)$. This is a connected open neighbourhood of $o$ contained in $S$. Removing $o$ leaves three connected components in $N$, while removing $h(o)$ leaves two in $I$. A homeomorphism preserves connected components, so this is impossible. This restriction step is essential: the count is proved for any sufficiently small connected chart neighborhood, not merely one chosen drawing.

A higher dimension of a manifold cannot repair the junction. For $n\ge 2$, repeat the argument with an open ball about $h(o)$: a punctured open ball in ${\mathbb{R}}^n$ is path-connected, whereas $N\setminus \{o\}$ still has three connected components. A zero-dimensional chart would make $o$ an isolated point, which it is not. A chart at a boundary point of a topological manifold with boundary similarly gives a punctured relative half-ball: one connected component when $n=1$, and a path-connected space when $n\ge 2$. Consequently the topological tripod fails the local definition even when intrinsic boundary is allowed.

The circle provides a useful warning about scale: deleting a point from the whole circle leaves a connected space, but deleting it from a sufficiently small connected arc leaves two connected components. Local chart tests use the latter. The circle and interval are not homeomorphic article uses deletion from the whole spaces for a different, global obstruction.

A topological vector space is a vector space $E$ over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, equipped with a topology such that vector addition and scalar multiplication are jointly continuous:The domains carry the product topology, and the scalar field carries its usual topology. Joint continuity permits both inputs to vary at once. Here the definition does not impose the Hausdorff condition; we state it separately when needed. ^[ref].

Every normed vector space is a topological vector space: the triangle inequality gives $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$, whilecontrols scalar multiplication near any $({\lambda }_0,x_0)$. More generally, a seminorm topology supplies many measurements of smallness, rather than one norm. Translation by any vector is a homeomorphism, so a neighbourhood basis at zero determines all neighbourhoods by translation.

The compatibility matters: on a nonzero real vector space, the discrete topology makes vector addition continuous, but not scalar multiplication. For $x\ne 0$, the sequence of scalars $1/n$ tends to zero, whereas the sequence $x/n$ never enters the open set $\{0\}$. Thus choosing a topology arbitrarily need not respect the linear algebra.

This framework lets functional analysis choose the appropriate notion of limit of a sequence. Norm topology measures one global error, weak topology tests linear functionals, and the Fréchet space of smooth functions controls every order of derivative. In each case the operations forming linear combinations are continuous, making approximation compatible with the underlying equations.

Topology studies topological spaces through open sets and continuous maps. Its central equivalence is a homeomorphism: a bijection that preserves continuity in both directions. This lets us distinguish spaces by connectedness or compactness even when numerical metric values are irrelevant. The open interval $(0,1)$ and the real line have the same topology in this sense, although only the former is a bounded set with its usual metric.

The upper panels show an explicit homeomorphism and the inverse image test for a continuous map. The lower panels exhibit an obstruction to a homeomorphism: removing any point from a circle leaves a connected space, whereas removing an interior point from a closed interval gives two connected components.

Homotopy is a different notion: it varies a continuous map continuously with a parameter and may collapse points. For example, $H(x,t)=(1-t)x$ contracts $[0,1]$ to zero. Thus this closed interval and a singleton have the same homotopy type, but there is no homeomorphism between them because there is no bijection. Homotopy equivalence need not preserve compactness. ^[ref]

The abstract viewpoint already appears explicitly in Tietze's 1923 paper: he describes topology as the study of properties invariant under invertible continuous maps with continuous inverses, and discusses axioms for neighbourhoods, citing Hausdorff's 1914 book, page 213. ^[ref]

For a uniform pro-p group, the power map $x\mapsto x^{p^n}$ is a homeomorphism onto $G^{p^n}$. Transport multiplication back to $G$ byUnique roots make this an associative group operation. For odd $p$, group commutator collection gives $x{+}_{n}y\equiv y{+}_{n}x(\mathrm{mod}{G}^{p^{n+1}})$ and $x{+}_{n}y\equiv x{+}_{n-1}y(\mathrm{mod}{G}^{p^n})$. The operations converge to addition in the Lie algebra of a uniform pro-p group.

A triangulation of a topological space is a homeomorphism from the realization of a simplicial complex to that space. In three-manifold topology, thickening its one-skeleton and the dual one-skeleton constructs a Heegaard splitting.

Take the circle $S^1=\{(x,y)\in {\mathbb{R}}^2:x^2+y^2=1\}$ with its subspace topology. Write $W=(-1,0)$ for its west point and $E=(1,0)$ for its east point. Two coordinate charts obtained by stereographic projection areBoth take values in the entire real line. Geometrically, join $W$ to $(x,y)$ and intersect the resulting line with $x=0$: the height is $t$. Projecting from $E$ instead gives height $u$. Each projection omits its own starting point, but the other coordinate chart includes that point. Hence $U\cup V=S^1$.^[ref]

To verify the homeomorphisms, solve the displayed equations for $(x,y)$. The inverse functions areEach formula lies on the circle, avoids its excluded point, and is continuous for every real argument. Substitution into the projection formulas returns $t$ or $u$; substitution in the other direction returns $(x,y)$. Thus these are continuous bijections with continuous inverse functions, as required for coordinate charts. Since the Euclidean plane is Hausdorff and second countable, the subset $S^1$ with its subspace topology is too. This checks all three conditions for a topological manifold, rather than just drawing a smooth-looking curve.

On the overlap $U\cap V$, both $x=\pm 1$ are excluded, so $y\ne 0$. Using $y^2=1-x^2$, the topological transition map isFor example, $(x,y)=(3/5,4/5)$ has $t=1/2$ and $u=2$. The two connected components of the overlap, the upper and lower arcs, correspond to the positive and negative parts of the punctured real line. The values $t=0$ and $u=0$ represent $E$ and $W$, respectively, and belong to only one chart each. The reciprocal coordinate change and its inverse function are smooth functions away from zero, so this topological atlas also supplies a smooth atlas.

Why can neither chart be enlarged to a single coordinate chart on the whole circle? The circle is a closed set and a bounded set in the Euclidean plane, hence compact by the Heine-Borel theorem. A global coordinate chart would give a homeomorphism to a nonempty open set $O\subseteq \mathbb{R}$. Its image would be compact and therefore closed in the real line. By connectedness of the real line, this forces $O=\mathbb{R}$, which is not compact. This is the one-dimensional instance of no compact manifold has a single Euclidean chart. Locally the circle is indistinguishable from the real line; globally compactness obstructs a single chart.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra of $C(K)$, with the supremum norm, containing the constants and separating points. Evaluation embeds $K$ continuously and injectively in the character space of an algebra. The embedding is a homeomorphism onto a closed subset because its domain is compact and its target is Hausdorff. Extra characters need not be point evaluations on the specified $K$.

The last-column map has fibre the subgroup fixing the last standard unit vector. Unitary completion gives surjectivity, and compactness gives the quotient homeomorphism. Projecting the other standard vectors onto the orthogonal complement of a nearby unit vector and applying Gram-Schmidt gives a local section, so the projection is a fiber bundle. The bundle $U(n)\to U(n+1)\to S^{2n+1}$ then gives ${\pi }_{i}U(n)\cong {\pi }_{i}U(n+1)$ for $i\le 2n-1$ because both adjacent sphere homotopy groups vanish.