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The Bonami lemma states that a function of Fourier degree at most $k$ satisfies $\Vert f{\Vert }_4\le 3^{k/2}\Vert f{\Vert }_2$.

The Boolean hypercube is the set $\{-1,1{\}}^n$, or equivalently $\{0,1{\}}^n$. Its hypercube graph joins two vertices exactly when they differ in one coordinate.

Complex projective space is the space of complex lines through the origin in ${\mathbb{C}}^{n+1}$.

A rank-$r$ vector bundle over a topological space $B$ is a fiber bundle $\pi :E\to B$ whose fibers are vector spaces over a fixed field $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. Each point has an open neighborhood $U$ and a vector bundle trivialization ${\pi }^{-1}(U)\cong U\times {\mathbb{F}}^r$ that is a linear isomorphism on each fiber.

On overlaps the vector bundle trivializations change by $(x,v)\mapsto (x,g_{ij}(x)v)$, where the invertible matrices $g_{ij}(x)$ vary continuously and satisfy the Čech cocycle condition. For a smooth vector bundle, the base and total space are smooth manifolds and these functions are smooth. The fiberwise vector addition and scalar multiplication are therefore globally well-defined.

The tangent bundle collects all tangent spaces of a smooth manifold. Its transition functions of a vector bundle are Jacobian matrices, so the chain rule supplies the cocycle identity. The Möbius line bundle supplies another example: every small patch is a product, but no global frame of a vector bundle exists. A global fiber-linear product isomorphism is the extra condition defining a trivial vector bundle.

A cohomology theory assigns abelian groups $H^q(X;R)$ to a topological space $X$. For singular cohomology with coefficients in a commutative ring $R$, form the cochain complex $C^q(X;R)$ with coboundary $\delta$ and takeA cocycle represents a cohomology class; adding a coboundary leaves that class unchanged.

A cohomology class evaluates on a homology class by applying its cochain to a representing cycle. Replacing either representative changes the result by zero, explaining the relation to homology. A continuous map $f:X\to Y$ induces cochain pullback and hence $f^{\ast }:H^q(Y;R)\to H^q(X;R)$: the direction reverses.

The cup product combines degree-$p$ and degree-$q$ classes into degree $p+q$, giving a graded cohomology ring. The circle has one integral generator in degrees zero and one. A two-dimensional torus has two independent degree-one classes whose cup product generates degree two. That multiplication distinguishes it from $S^1\vee S^1\vee S^2$, which has the same additive cohomology groups but zero products between its positive-degree generators.

Realization of the simplicial construction produces the corresponding topological cone.

The simplicial cone $v\ast K$ adjoins a new vertex $v$ to every simplex of $K$ and is contractible.

An abstract simplicial complex is a pair $K=(V,\Sigma )$, where $V$ is a set of vertices of a simplicial complex and $\Sigma$ is a collection of finite nonempty subsets, called simplexes. Every singleton $\{v\}$ belongs to $\Sigma$, and every nonempty subset of a simplex is a simplex. Its dimension is the supremum of $|\sigma |-1$ over $\sigma \in \Sigma$. Some conventions also include an empty simplex; it contributes no points to the geometric realization.

A morphism between two short exact sequences in an abelian category induces an exact sequence

Let $X$ be the geometric realization of a finite simplicial complex, or a retract of such a space, and let $f:X\to X$ be continuous. If its Lefschetz number $L(f)$ is nonzero, then $f$ has a fixed point. This includes self-maps of compact manifolds.^[ref]

For a map without fixed points on a finite polyhedron, compactness gives a positive separation between $x$ and $f(x)$. After sufficiently fine simplicial subdivision, a simplicial approximation sends no simplex back onto itself. Its diagonal chain map coefficients vanish, so its alternating trace, and hence $L(f)$, is zero. A retraction reduces the more general stated case to this one.

For example, every continuous self-map of an even-dimensional sphere with mapping degree different from $-1$ has a fixed point, since $L(f)=1+\mathrm{deg}f$. The converse fails: $L(f)=0$ permits fixed points, as the identity map of a circle shows.

For a continuous self-map $f:X\to X$, suppose the rational homology groups $H_i(X;\mathbb{Q})$ are finite-dimensional and vanish in sufficiently high degrees. Its Lefschetz number is the finite alternating traceFor a self-map of a finite simplicial complex, it also equals the alternating trace on any finite simplicial chain complex representing the induced map. The traces on boundaries cancel in consecutive degrees, leaving precisely the traces on homology. Thus a potentially large chain computation reduces to the homology groups.

The identity map has $L(1_X)=\chi (X)$, the Euler characteristic. On a circle, a map of mapping degree $d$ has $L(f)=1-d$. A nonzero value forces a fixed point under the hypotheses of the Lefschetz fixed-point theorem; finiteness of the displayed sum alone is not that theorem's full geometric hypothesis.

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 covering map $p:\tilde{X}\to X$ is locally a disjoint union of homeomorphisms onto the same evenly covered open neighbourhood.

The fundamental group ${\pi }_1(X,x_0)$ consists of based loops modulo based homotopy. Multiplication follows one based loop and then the other; the constant based loop represents the identity, and path reversal gives the inverse. Passing to homotopy classes makes this operation associative, as explained by well-defined loop concatenation.

Identify the circle with $S^1=\{z\in \mathbb{C}:|z|=1\}$ and choose basepoint $1$. Its based loopshave winding number $n$: positive $n$ counts counterclockwise turns, negative $n$ counts clockwise turns, and $n=0$ gives the constant based loop. Every based loop on the circle is joined by a based homotopy to exactly one ${\gamma }_n$. ConsequentlyThis is Hatcher's Theorem 1.7 in Algebraic Topology, §1.1, pp. 29–30. For example, two counterclockwise turns followed by one clockwise turn represent $2+(-1)=1$, rather than three turns. The winding graph in the figure above makes the signed arithmetic visible.

A proof outline uses the covering map $q:\mathbb{R}\to S^1$, $q(u)=e^{2\pi iu}$. By the path lifting theorem, a based loop has a unique lift starting at zero and ending at an integer $n$. Interpolating its lift to $s\mapsto ns$ produces a based homotopy to ${\gamma }_n$. The homotopy lifting theorem for a covering map keeps that endpoint unchanged during a based homotopy, proving uniqueness.

In the filled disk, every based loop $\gamma$ at $x_0$ has a based homotopy to the constant based loop:Both endpoints remain $x_0$, and the convex set property of the disk keeps every intermediate point inside. Thus ${\pi }_1(D^2,x_0)$ is the trivial group. Together with the deformation retraction above, this computes three concrete answers: the filled disk has trivial fundamental group, while the circle and the annulus have infinite cyclic groups.

Maps have computable algebraic shadows too. The continuous map $p_n:S^1\to S^1$, $p_n(z)=z^n$, fixes $1$ and sends ${\gamma }_k$ to ${\gamma }_{nk}$. Its induced group homomorphism is therefore $k\mapsto nk$, and its mapping degree is $n$. For $n=2$ the image consists of even integers, so this map cannot be a homotopy equivalence. Composing $p_2$ with $p_{-3}$ gives $p_{-6}$: map composition multiplies degrees, whereas concatenating based loops adds winding numbers. Distinct degrees distinguish these maps up to homotopy; this calculation does not classify arbitrary topological spaces.

Historically, Poincaré's 1895 Analysis Situs introduced the fundamental group; its original §12 explicitly names it. The CNRS transcription of Poincaré's §12, Groupe fondamental provides a direct view of this early formulation.

A based loop in $(X,x_0)$ is a path $\alpha :[0,1]\to X$ whose two endpoints are $x_0$.

A homotopy between continuous maps $f,g:X\to Y$ is a continuous map $H:X\times [0,1]\to Y$ satisfying $H(x,0)=f(x)$ and $H(x,1)=g(x)$. Think of $t$ as time: continuity is required jointly in position and time. The target $Y$ specifies where every intermediate point is allowed to go. See Hatcher, Algebraic Topology, Chapter 0, pp. 3–4.

For the filled disk $D^2=\{x\in {\mathbb{R}}^2:\Vert x\Vert \le 1\}$, the formulais a contraction of a topological space: it starts at the identity map and ends at the constant map to zero. The Euclidean norm obeys $\Vert H(x,t)\Vert =(1-t)\Vert x\Vert \le 1$, so the formula never leaves the disk. For $x=(3/5,4/5)$, the midpoint of the deformation is $(3/10,2/5)$, whose Euclidean norm is $1/2$.

On the circle $S^1=\{x:\Vert x\Vert =1\}$, the same formula fails: at $t=1/2$ every image has Euclidean norm $1/2$, so lies outside $S^1$. This checks one failed candidate; the nonzero winding number of a once-around loop proves that no contraction within the circle exists. The distinction is between allowing the filling and requiring the deformation to stay on its boundary.

For a chain complex $(C_{\ast },\partial )$, its $n$th homology is the abelian group or moduleThe equation ${\partial }_n{\partial }_{n+1}=0$ says that every chain boundary is a chain cycle. Homology identifies two chain cycles when their difference is a chain boundary; a homology class records exactly what survives this identification.

In singular homology, the chains are finite sums of maps from simplexes into a topological space. A continuous map induces homomorphisms on all homology groups, and homotopic maps induce the same homomorphisms. This makes homology a computable way to distinguish topological spaces that cannot be homotopy equivalent.^[ref]

For a circle, $H_0(S^1;\mathbb{Z})=\mathbb{Z}$ records its one path component, while $H_1(S^1;\mathbb{Z})=\mathbb{Z}$ records signed winding around it. A closed disk has the homology of a point: its boundary circle becomes a chain boundary when filled by the disk. Simplicial homology and cellular homology replace singular chains by smaller chain complexes suited to triangulated or cellular spaces.

Real projective space is the antipodal quotient ${\mathbb{RP}}^n=S^n/(x\sim -x)$.

For based spaces, the smash product iswhere the wedge $X\vee Y$ consists of pairs having at least one basepoint coordinate.

The mapping cylinder of $f:A\to X$ isIt deformation retracts onto the copy of $X$.

A CW complex is built inductively by attaching discs along maps from their boundary spheres. Its $n$-skeleton contains all cells of dimension at most $n$.