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Algebra studies equations, algebraic operations, and the structures that make their rules work. Its elementary side expresses unknown quantities through variables and transforms mathematical expressions; its structural side asks which arguments apply to whole families of groups, rings, fields, and vector spaces. The useful step is identifying exactly which properties a calculation needs, so that it can be reused in a new setting.

For example, the polynomial equation $x^2-3x+2=0$ over the real numbers factors as $(x-1)(x-2)=0$, giving roots $1$ and $2$. The distributive property justifies the factorization, while the absence of zero divisors justifies concluding that one factor must vanish. Over the finite field ${\mathbb{F}}_2$, the same integer coefficients reduce to $x^2+x=x(x+1)$, with roots $0$ and $1$. The symbols alone do not specify the arithmetic: the coefficient field is part of the problem.

In a ring with zero divisors, a familiar step can fail. Modulo $8$, all four residue classes $1,3,5,7$ satisfy $x^2=1$. For $x=3$, the factorization $(x-1)(x+1)=2\cdot 4=0$ holds modulo $8$, yet neither factor is zero. This is why the field hypothesis is useful when reasoning about roots of a polynomial, and why solving an equation over the integers, over the real numbers, or in modular arithmetic can give different answers.

The structure is also useful beyond scalar arithmetic. Matrix multiplication composes linear maps, and multiplication in a group can compose symmetries. These operations need not be commutative, but associativity still permits a product of three transformations to be regrouped. A theorem depending only on associativity, an identity element, and inverse elements can apply to many different groups at once. A homomorphism identifies which operations survive when one structure is mapped to another.

Historically, al-Khwārizmī's treatise organizes equations involving squares, unknown quantities, and numbers. Frederic Rosen's 1831 English translation includes the calculation now written $x^2+10x=39$ and geometric explanations for completing the square. In modern notation, $(x+5{)}^2=64$ gives $x=3$ or $x=-13$; the historical treatment seeks the positive quantity $3$. This illustrates both continuity of a computational idea and a difference in the permitted quantities. ^[ref]

An algebraic structure is a set equipped with specified algebraic operations satisfying selected axioms. An operation's domain is part of the structure: a binary operation on $S$ takes two elements of $S$ and produces another element of $S$. Groups specify one such operation, rings specify addition and multiplication, and vector spaces specify vector addition and scalar multiplication by a separate field.

Different choices of operations can give the same underlying set different structures. The integers form a group under addition, since every integer has an additive inverse. They do not form a group under multiplication: for example, $2$ has no multiplicative inverse among the integers. Even the nonzero integers fail that inverse requirement. In contrast, nonzero rational numbers form a group under multiplication.

Finite multiplication tables make such distinctions visible. Arithmetic modulo $5$ gives a field; arithmetic modulo $6$ gives a commutative ring with zero divisors. In the former, multiplying by any nonzero element permutes all residue classes. In the latter, multiplying by $2$ reaches only $0,2,4$, and $2\cdot 3=0$ even though both factors are nonzero.

The goal is to extract reusable properties rather than memorize isolated tables. A homomorphism preserves the specified operations, and an isomorphism shows that two structures have identical operation patterns after relabeling.

An embedding identifies one mathematical structure with a structure-preserving copy inside another. In topology it is a homeomorphism onto its image with the subspace topology; an algebraic embedding is an injective homomorphism. A smooth embedding additionally has injective differential everywhere, so its image is an embedded submanifold.

Realize a finite p-group $P$ as the Galois group of a totally ramified extension $E/{\mathbb{F}}_p((t))$, using Artin-Schreier embedding problems over a characteristic-p field. A uniformizer $u$ of $E$ gives $E\cong {\mathbb{F}}_p((u))$: every element has a unique convergent Laurent expansion with coefficients in the residue field. An automorphism has $u\mapsto a_{1}u+a_2{u}^2+\cdots$, and $a_1$ defines a homomorphism $P\to {\mathbb{F}}_p^{\times }$. Its image is trivial because $|P|$ is a $p$-power and $p\nmid |{\mathbb{F}}_p^{\times }|$. Thus the faithful Galois action lies in the Nottingham group, embedding $P$.

Write the suspension of an $(n-2)$-connected based CW complex as two cones meeting in the original space. Each cone pair is $(n-1)$-connected because its relative groups shift the original homotopy groups by one. Homotopy excision theorem compares the cone pair with the suspension relative to the other cone, giving the suspension homomorphism as an isomorphism for $i<2n-2$ and a surjection at $i=2n-2$ in the indexing ${\pi }_{i-1}X\to {\pi }_i\Sigma X$.

The isomorphism theorems relate homomorphisms, their kernels and images, and quotient constructions in algebra. For groups, the first isomorphism theorem identifies a homomorphism's image with the quotient group by its group kernel. Analogous statements include the first isomorphism theorem for rings and the First isomorphism theorem for vector spaces: identifying precisely the inputs with the same output preserves the remaining algebraic structure.

A kernel records the information lost by a homomorphism. For a group homomorphism it is the preimage of the identity element, giving the kernel of a group homomorphism; for a ring homomorphism or linear map it is the preimage of zero, giving the kernel of a ring homomorphism or kernel of a linear map. More generally, the relation $a\sim b⟺f(a)=f(b)$ records which inputs the homomorphism identifies. In the group case, its equivalence classes are cosets of the group kernel, as explained by fibres of a group homomorphism are cosets of its kernel.

A Lie group homomorphism intertwines one-parameter subgroups and therefore obeys the displayed identity. A connected Lie group is generated by an identity neighbourhood in an exponential chart, so its homomorphism is uniquely determined by the differential at the identity. On a disconnected group, only the restriction to its identity component is determined this way; the images of the other components require extra data.

An isomorphism of Lie groups is a group isomorphism that is also a diffeomorphism. Both the homomorphism and its inverse preserve the smooth structures. Its identity differential is a Lie algebra isomorphism, but an isomorphism of Lie algebras determines a global Lie group isomorphism without extra period data only for connected simply connected groups.

An open normal subgroup $N$ of a topological group is simultaneously open in its topology and invariant under conjugation. For a profinite group, the quotient $G/N$ is a finite discrete group: its cosets form a disjoint open cover, so compactness gives finitely many. Conversely the group homomorphism kernel of a continuous homomorphism to a finite discrete group is open and normal. Such subgroups form an identity-neighbourhood basis of every profinite group.

Interpret the connectivity hypothesis in its usual based sense: $X$ is $(n-2)$-connected, with $n\ge 2$ for the nontrivial suspension range. Form the suspension as two cones $C_{-}X$ and $C_{+}X$ meeting in $X$. Both cones are contractible. Their relative homotopy sequences give ${\pi }_j(C_{\pm }X,X)\cong {\pi }_{j-1}(X)$ in the relevant degrees, so both cone pairs are $(n-1)$-connected.

The homotopy excision theorem says that if two CW pairs meeting in a connected subcomplex are $p$- and $q$-connected, $p,q\ge 1$, the map from the first relative pair to the union relative to the second part is an isomorphism in degrees $i<p+q$ and a surjection in degree $p+q$. Here $p=q=n-1$, givingas an isomorphism for $i<2n-2$ and a surjection for $i=2n-2$. Contractibility identifies the domain with ${\pi }_{i-1}(X)$ and the target with ${\pi }_i(\Sigma X)$. The comparison under these identifications is the suspension map: a sphere representative is coned in each half and the two cones give its suspended sphere representative. Hence the Freudenthal suspension from two cones provesFor connected CW complexes, reduced and unreduced suspension have the same based homotopy type, so either standard suspension convention gives this result. In the low-dimensional range, pointed sets are used where necessary; the displayed positive-degree homomorphism assertion concerns the usual groups.

For $S^r$ with $r\ge 2$, cellular approximation makes every sphere map of degree below $r$ null-homotopic: it can be homotoped into the point skeleton of the CW structure having cells only in degrees zero and $r$. The Hurewicz theorem therefore identifies its first possible nonzero homotopy group with homology. Its cellular chain complex gives $H_r(S^r;\mathbb{Z})=\mathbb{Z}$. For $r=1$, lifting a based loop to the universal cover $\mathbb{R}\to S^1$ classifies it by an integer endpoint displacement, so ${\pi }_1(S^1)=\mathbb{Z}$; the degree-one Hurewicz map is abelianization and is an isomorphism here. ThusAlternatively, after the $r=2$ Hurewicz base case, the suspension isomorphisms above propagate the generator to every higher sphere.

Use finitely generated right projective modules throughout. The group K0 of a ring is the Grothendieck group of their direct-sum monoid, with $[P\oplus Q]=[P]+[Q]$. The group K1 of a ring is ${\mathrm{GL}}_{\infty }(A)/E_{\infty }(A)$, equivalently the abelianization of the stable general linear group. Stabilization is $g\mapsto \mathrm{diag}(g,1)$; $E_{\infty }$ is generated by elementary transvections. The stable elementary subgroup is the commutator subgroup: elementary transvections are commutators using a third index, and stable block interchange and elementary row identities make the quotient abelian.

Write $A_0$ for the lower-right ring, printed as $A^{\prime }$ in the source. A Milnor square has unital commuting maps, is a pullbackand has at least one of $j_1,j_2$ surjective. Interchange the two sides if necessary so that $j_1$ is surjective.

We first establish the patching fact used in the proof. For projective $P_i$ and an isomorphism $\varphi :P_1\otimes A_0\to P_2\otimes A_0$, setFor free modules denote the result by $M_g$. The block matrix identityexpresses $\mathrm{diag}(g,g^{-1})$ as elementary matrices. Their entries lift through $j_1$, and their elementary lifts remain invertible. A change of basis on the first side then trivializes this gluing. Thus $M_g\oplus M_{g^{-1}}\cong A^{2r}$, proving that $M_g$ is finitely generated projective. Notice that we lifted elementary matrices, not arbitrary invertible matrices.

For general $P_1,P_2$, choose $P_1\oplus Q_1\cong A_1^r$ and $P_2\oplus Q_2\cong A_2^s$. The reductions of $Q_1\oplus A_1^s$ and $Q_2\oplus A_2^r$ are isomorphic: insert the common reduced $P_i$ and reorder the two complements. Patch them by such an isomorphism. The sum with $M(P_1,P_2,\varphi )$ now patches two free modules of equal size, so is projective by the free case, making each summand projective. The same free change of basis followed by these direct-summand projections shows that extension to $A_i$ recovers $P_i$. Conversely, write an $A$-projective as the image of an idempotent over a finite free module. Since matrices over the pullback are compatible pairs of matrices, applying the compatible idempotents to the free pullback reconstructs this projective from its two reductions. Compatible module isomorphisms patch as well. This proves Milnor patching of projective modules in the form needed below.

The diagonal maps in the sequence are extension of scalars; the two maps to $A_0$ are differences $j_{1\ast }-j_{2\ast }$. Define the connecting map byIt is unchanged by stabilization or elementary multiplication: an elementary factor lifts through $j_1$ and changes a free basis in the gluing. The identity $\mathrm{diag}(g,h)=\mathrm{diag}(gh,I)\mathrm{diag}(h^{-1},h)$ and the preceding elementary factorization show $M_g\oplus M_h\cong M_{gh}\oplus A^r$ after matching sizes. Thus $\partial$ is an additive homomorphism on $K_1(A_0)$.

Exactness at $K_1(A_1)\oplus K_1(A_2)$. If the two reduced classes agree, choose stabilized representatives $g_1,g_2$ of the same size. Their reductions differ by an elementary matrix $e$ over $A_0$. Lift $e$ to an elementary matrix over $A_1$ and adjust $g_1$ without changing its $K_1$ class. Its reduction now equals that of $g_2$. The compatible pair and its compatible inverse give an invertible matrix over $A$, so the original pair of classes is in the diagonal image. The reverse containment follows from commutativity of the square.

Exactness at $K_1(A_0)$. A gluing obtained from a change of basis over either upper ring patches a free module, so the difference-map image is killed by $\partial$. Conversely, $\partial [g]=0$ means that $M_g$ becomes free after adding a common projective summand. Add a complement of that summand to obtain an actual isomorphism $M_g\oplus A^s\cong A^{r+s}$. Its two local bases express $g\oplus I_s=j_2(v_2)j_1(v_1{)}^{-1}$. Therefore $[g]=j_{2\ast }[v_2]-j_{1\ast }[v_1]$, which lies in the image of the difference map. This uses stable equality of projective classes, not cancellation of arbitrary projectives.

Exactness at $K_0(A)$. Connecting classes reduce to zero on both sides. Conversely, write a class in the kernel as $[P]-[A^r]$ by adding a complement to its negative projective term. Its reductions have the same classes as $A_i^r$. Equality in $K_0(A_i)$ becomes an isomorphism after a finite common free summand, by taking a complement to the auxiliary projective. Use one sufficiently large free summand on both sides. The stabilized $P$ is then locally free with gluing $g$, so the original class is $[M_g]-[A^{r+s}]=\partial [g]$.

Exactness at $K_0(A_1)\oplus K_0(A_2)$. Write a kernel pair as $([P_1]-[A_1^r],[P_2]-[A_2^r])$, using complements and padding to make the free sizes equal. Their reduced positive classes agree. Adding a common finite free summand makes those reductions isomorphic. Patch the resulting projectives, and subtract the matching free $A$-module. Its restrictions are the desired pair. The reverse containment is immediate because both reductions to $A_0$ agree.

These arguments prove the Mayer-Vietoris sequence in algebraic K-theoryNo zero is inserted at either end; this fragment does not assert injectivity at the first term or surjectivity at the last.

For the group ring of the cyclic group, take the cyclotomic Milnor squareThe maps send the cyclic generator to $1$ and to $\zeta$; the lower map sets $\zeta =1$, and the right map reduces modulo $p$. Since ${\Phi }_p(1)=p$, $\mathbb{Z}[\zeta ]/(\zeta -1)={\mathbb{F}}_p$. To verify the pullback, represent an element by a polynomial of degree at most $p-1$. If both evaluations vanish, it is a constant multiple of the minimal polynomial of an algebraic element ${\Phi }_p$, and evaluation at one forces that constant to be zero. Conversely, represent $b\in \mathbb{Z}[\zeta ]$ by $f$ of degree below $p-1$. For a compatible integer $a$, $a-f(1)$ is divisible by $p$; replacing $f$ by $f+[(a-f(1))/p]{\Phi }_p$ yields the pair $(a,b)$. The right map is surjective, so this is a Milnor square.

The hinted cyclotomic units are $u_k=1+\zeta +\cdots +{\zeta }^{k-1}$. If $kl\equiv 1(\mathrm{mod}p)$ with $1\le l<p$, the geometric sum ${\sum }_{j=0}^{l-1}{\zeta }^{kj}$ is their inverse, because ${\zeta }^{kl}=\zeta$. Their residues are $k\in {\mathbb{F}}_p^{\times }$, covering every unit of the common field. Gaussian elimination identifies $K_1({\mathbb{F}}_p)$ with ${\mathbb{F}}_p^{\times }$ by determinant. Thus $K_1(\mathbb{Z}[\zeta ])\to K_1({\mathbb{F}}_p)$ is surjective, and the connecting map vanishes.

The remaining exact sequence injects $K_0(\mathbb{Z}[C_p])$ into $\mathbb{Z}\oplus K_0(\mathbb{Z}[\zeta ])$, with image $\{(r,x):r=\mathrm{rank}x\}$. Here $K_0(\mathbb{Z})=K_0({\mathbb{F}}_p)=\mathbb{Z}$, and reduction of a projective over the integral domain $\mathbb{Z}[\zeta ]$ has dimension its rank: localize at $(\zeta -1)$, where the projective is free, and compare its residue-field dimension with its fraction-field dimension. Projection onto the second factor is therefore a bijection on this kernel. Hence extension along the cyclotomic evaluation givesThe essential step is reduction of cyclotomic units at the ramified prime; surjectivity of a ring map alone would not imply surjectivity on $K_1$.

Give $\mathcal{O}(U)$ the compact-open topology. Choose a finite polygonal cycle $\Gamma$ in $U\setminus {\sigma }_A(x)$ whose winding number is one on the Banach algebra spectrum and zero outside $U$. One can take the oriented boundary of a finite union of small squares surrounding the compact Banach algebra spectrum and contained in $U$. DefineThe integral is an algebra-norm integral of a continuous function on finitely many segments. Cauchy's theorem, applied after any bounded linear functional on $A$, makes it independent of the cycle. Moreoverso it is continuous for the compact-open topology.

Here is a direct verification of the algebra properties and uniqueness using the allowed Runge theorem. The version required is that every holomorphic function on an open plane set is a locally uniform limit of rational functions with poles outside that set, allowing a polynomial part. A rational function $r=p/q$ of this kind has the unambiguous value $r(x)=p(x)q(x{)}^{-1}$, since the roots of $q$ avoid the Banach algebra spectrum. These rational values form a unital algebra homomorphism.

They agree with the contour integral. First $\frac{1}{2\pi i}{\int }_{\Gamma }(z1-x{)}^{-1}dz=1$, by replacing the cycle with a large circle and integrating its Neumann expansion. The identity $z^{j}1-x^j=(z1-x){\sum }_{k=0}^{j-1}{z}^{j-1-k}{x}^k$ then proves the assertion for polynomials. For $a\notin U$, the resolvent identity givesThe cycle has winding number zero at $a$, so integration gives $(x-a1{)}^{-1}$. This identity holds on a neighborhood of $a$ disjoint from the cycle and the Banach algebra spectrum. Differentiating with respect to $a$ gives the analogous result for every power $(z-a{)}^{-k}$, and partial fractions give it for all the rational functions under consideration.

If $r_n\to f$ and $s_n\to g$ locally uniformly, the contour estimate makes their evaluated sequences converge. Since $r_n{s}_n\to fg$ locally uniformly and multiplication in $A$ is continuous, their limits satisfy ${\Theta }_x(fg)={\Theta }_x(f){\Theta }_x(g)$. Also ${\Theta }_x(1)=1$ and ${\Theta }_x(Z)=x$. Conversely any continuous unital complex-algebra homomorphism sending $Z$ to $x$ must send $Z-a$ to $x-a1$ and its reciprocal to the inverse. It therefore agrees on rational functions and then, by Runge density and continuity, on all of $\mathcal{O}(U)$. This proves existence and uniqueness of the holomorphic functional calculus.

For its spectral mapping property, first note compatibility on a smaller open spectral neighborhood $W\subseteq U$: composing restriction to $W$ with the calculus there gives the calculus on $U$ by uniqueness. If $\mu \notin g({\sigma }_A(x))$, choose such a $W$ where $\mu -g$ never vanishes. Thenwith the right side evaluated on $W$. Conversely, for $\lambda \in {\sigma }_A(x)$, write $g(z)-g(\lambda )=(z-\lambda )q(z)$ with $q$ holomorphic on $U$. The evaluated factors commute. If their product were invertible, both factors would be invertible, contradicting $\lambda \in {\sigma }_A(x)$. ThusPullback $h\mapsto h\circ g$ is continuous from $\mathcal{O}(V)$ to $\mathcal{O}(U)$ because the image under $g$ of a compact set is compact. The composite $h\mapsto {\Theta }_x(h\circ g)$ is a continuous unital homomorphism taking the coordinate function to $y={\Theta }_x(g)$. Uniqueness of the calculus at $y$ provesFor the exponential conclusion, let $K={\sigma }_A(x)$. There is a simple polygonal arc from zero through its unbounded complementary component to outside a disk containing $K$, continued by a ray to infinity and avoiding $K$. Loops in an initial polygonal path can be removed to make the arc simple. The complement of this slit is a simply connected open neighborhood of $K$ avoiding zero, and hence admits a holomorphic logarithm $\ell$. Set $y={\Theta }_x(\ell )$. The composition rule givesThe calculus of the entire exponential agrees with the usual algebra exponential because its power-series partial sums converge locally uniformly. Finally an invertible matrix over the complex numbers has a finite Banach algebra spectrum avoiding zero. Removing finitely many plane points leaves a connected set, so the hypothesis always holds: every invertible matrix over the complex numbers is an exponential of a matrix over the complex numbers.

It suffices first to treat unital algebras and a unital homomorphism. A surjective homomorphism between unital algebras automatically preserves the identity. If $B=0$, continuity is immediate. In the nonunital case, extend to the forced unitizations by $T^{+}(a+\lambda 1)=T(a)+\lambda 1$. This is still surjective and the unitization of a semisimple algebra is semisimple: its radical projects to zero in its scalar quotient, and its remaining intersection with the original algebra is its original radical. Continuity of the extension implies continuity of the original map.

An algebra homomorphism decreases the spectral radius, even without any continuity assumption, because it takes an inverse of $\lambda 1-a$ to an inverse of $\lambda 1-T(a)$. Suppose $a_n\to 0$ in $A$ and $T(a_n)\to b$ in $B$. Fix any $c\in B$ and choose $u,v\in A$ with $T(u)=b$, $T(v)=c$. PutConsider the polynomial $q_n(z)=T((1-z)w_n+zw)=(1-z)d_n+zd$. Its value at one is exactly $d$. On the outer circle $|z|=R$,On the inner circle $|z|=1/R$, spectral-radius decrease givesApplying part (i) in $B$ and then letting $n\to \infty$, with $R$ fixed, givesThis holds for every $R>1$, so $r_B(cb)=0$ for every $c\in B$.

This forces $b$ into the Jacobson radical. To see the criterion directly, if $b$ were outside a maximal left ideal $L$, then $L+Bb=B$, so $1=l+cb$ for some $l\in L$, $c\in B$. The equality $r_B(cb)=0$ makes $1-cb$ invertible, but this element equals $l\in L$, impossible in a proper left ideal. Hence $b$ lies in every maximal left ideal. Since $B$ is semisimple, $b=0$.

We have shown that the separating space of $T$ is zero. If $a_n\to a$ and $T(a_n)\to y$, apply the result to $a_n-a$ to obtain $y=T(a)$. Its graph is closed, and the closed graph theorem now provesThe crucial estimate uses part (i) with an exactly fixed central value; it does not assume the false general assertion that the spectral radius is continuous under norm limits.

Let $G$ be locally compact, Abelian and Hausdorff, and fix a Haar measure. Write $T_{x}f(y)=f(x^{-1}y)$ and use convolutionThe L1 group algebra is a commutative Banach algebra. Here a group character means a continuous homomorphism $\chi :G\to \mathbb{T}$, not the trace of an arbitrary group representation. A Gelfand character is a nonzero multiplicative complex-linear functional on the convolution algebra. We use the conventionWith the alternative standard Fourier convention involving $\overline{\chi }$, one simply conjugates every group character in this identification.

For such $\chi$, the functional is bounded by $\Vert f{\Vert }_1$, and Fubini's theorem, the substitution $y=xz$, and the homomorphism law giveIt is nonzero: if $u\ge 0$ is nonzero in $C_c(G)$, then $f=\overline{\chi }u$ has ${\Phi }_{\chi }(f)=\int u>0$.

Conversely let $\Phi$ be an algebra character. It is automatically bounded, with $|\Phi (f)|\le \Vert f{\Vert }_1$. Indeed, extend it to the unitization by $\tilde{\Phi }(f+\lambda 1)=\Phi (f)+\lambda$. If $|\Phi (f)|>\Vert f{\Vert }_1$, the Neumann series makes $f-\Phi (f)1$ invertible, whereas its character value is zero, contradicting multiplicativity of an invertible element and its inverse.

Choose $g$ with $\Phi (g)\ne 0$ and put $\chi (x)=\Phi (T_{x}g)/\Phi (g)$. Commutativity of the group gives $(T_{x}f)\ast g=f\ast (T_{x}g)$; applying $\Phi$ yieldsThus the ratio is independent of the chosen $g$. Applying the relation successively to $T_x{T}_{y}g=T_{xy}g$ gives $\chi (xy)=\chi (x)\chi (y)$, and $\chi (e)=1$. In particular it never vanishes.

The translation map $x\mapsto T_{x}g$ is continuous in $L^1$. For $g\in C_c(G)$, translated supports near a fixed $x$ lie in one compact set; continuity on that set and a finite cover give uniform convergence of the translated values, hence $L^1$ convergence because the compact set has finite Haar measure. For general $g$, approximate by $C_c(G)$ and use $\Vert T_{x}g{\Vert }_1=\Vert g{\Vert }_1$. Therefore $\chi$ is continuous. For all integers $n$,Positive powers exclude modulus greater than one; negative powers exclude modulus less than one. Hence $|\chi (x)|=1$.

Finally, the $L^1$-valued convolution formula $f\ast g=\int f(x)T_{x}gdx$ and boundedness of $\Phi$ giveCancel $\Phi (g)$ to obtain the displayed formula for every $f$. Different characters give different functionals: if their continuous difference is nonzero at a point, integration against a suitably phased compactly supported test function in a small neighbourhood detects that difference. Thus the correspondence is bijective.

It also identifies the character space of an Abelian L1 group algebra topologically with the Pontryagin dual group. For a net converging uniformly on compact subsets in $\hat{G}$, convergence of the functionals follows first for compactly supported $f$, then for all $L^1$ functions using the common norm bound one. Conversely suppose the functionals converge in the Gelfand topology. Choose $g$ whose limiting character value is nonzero; its values are eventually bounded away from zero. For compact $K\subset G$, the set $\{T_{x}g:x\in K\}$ is compact in $L^1$. Convergence of uniformly bounded linear functionals is uniform on this compact set: a finite norm-net reduces it to convergence at finitely many functions, with the common bound controlling the approximation error. The ratio formula then gives uniform convergence of the group characters on $K$. This proves equality with the compact-open topology, without assuming first countability of $G$.

For later use, these transforms form a self-adjoint subalgebra dense in $C_0(\hat{G})$. The set of algebra characters together with the zero functional is weak-star compact by Banach-Alaoglu theorem and the closed multiplicativity equations. Removing zero gives a locally compact Hausdorff character space; each evaluation $\Phi \mapsto \Phi (f)$ vanishes at infinity, since its modulus-at-least-$ϵ$ set is compact and avoids zero. Transforms separate characters, and none vanishes identically at a given character. They are closed under conjugation because $f^{\ast }(x)=\overline{f(x^{-1})}$ satisfies ${\Phi }_{\chi }(f^{\ast })=\overline{{\Phi }_{\chi }(f)}$. The locally compact Stone-Weierstrass theorem therefore gives the claimed density.

A chain complex consists of abelian groups $C_q$ and homomorphisms ${\partial }_q:C_q\to C_{q-1}$ with ${\partial }_{q-1}{\partial }_q=0$. Its groups of chain cycles and chain boundaries are $Z_q=\ker {\partial }_q$ and $B_q=\mathrm{im}{\partial }_{q+1}$; the identity ensures $B_q\subseteq Z_q$. Its homology groups areFor a nonnegative complex, take $C_q=0$ for $q<0$.

The singular chain group $C_q(X)$ is the free abelian group on continuous maps $\sigma :{\Delta }^q\to X$, called singular simplices. With ${\delta }_i$ the inclusion of the face opposite vertex $i$, defineDeleting any two vertices in the two possible orders gives the same face with opposite signs, proving ${\partial }^2=0$. This defines the singular chain complex and its singular homology. A continuous map $f$ induces the chain map $f_{\#}(\sigma )=f\circ \sigma$; it commutes with the boundary because composition commutes with restriction to faces.

To prove homotopy invariance of homology, let $F:X\times I\to Y$ have endpoints $f,g$. Triangulate ${\Delta }^q\times I$ by the $q+1$ simplices with ordered verticesLet $p_i:{\Delta }^{q+1}\to {\Delta }^q\times I$ be their affine parameterizations. The singular prism operator isOn taking boundaries, shared internal faces of consecutive prism simplices cancel. The top and bottom faces give $g_{\#}\sigma -f_{\#}\sigma$, and the remaining faces are $-P_{q-1}(\partial \sigma )$ with their induced signs. ThusFor a chain cycle $z$, $g_{\#}z-f_{\#}z=\partial Pz$ is a chain boundary, so $f_{\ast }=g_{\ast }$ on homology. If $f$ is a homotopy equivalence, its homotopy inverse therefore induces an inverse to $f_{\ast }$, proving that singular homology depends only on homotopy type.

Use the standard CW complex for an orientable genus-three surface: one vertex, six edges $a_1,b_1,a_2,b_2,a_3,b_3$, and a two-cell whose attaching word is ${\prod }_{j=1}^3[a_j,b_j]$. The cellular homology theorem identifies this cellular calculation with singular homology. The boundary of every edge is zero, and each edge's total exponent in the attaching word is zero, so the two-cell boundary is zero. HenceChoose $C_1$ to be the nonseparating curve represented by $a_1$, and choose $C_2$ as a small circle bounding a disc in a surface patch. Then $[C_1]$ is a nonzero primitive homology class, whereas the inclusion of $C_2$ induces zero on $H_1$. If a homotopy equivalence carried $C_1$ onto $C_2$, its restriction would factor through $C_2$ and kill $[C_1]$, contradicting injectivity on first homology. This is the homology obstruction to carrying surface curves onto one another.

There is nevertheless such a continuous map. Map the edge $a_1$ homeomorphically onto $S^1$ and all other edges to its base point. The two-cell attaching word then has winding number zero in $S^1$, hence is null-homotopic and the map extends to $r:{\Sigma }_3\to S^1$. Compose $r$ with a homeomorphism $S^1\to C_2$ and the inclusion of $C_2$ in the surface. This gives $g(C_1)=C_2$. Thus an arbitrary map can carry the first curve onto the second, but a homotopy equivalence cannot.

For an unambiguous drawing, realize ${\Sigma }_3$ as the double of a manifold formed from a disc with three holes. Its two copies are glued along all four boundary circles. The red curve below follows an arc from the outer boundary to the first hole on the front copy and returns along the matching arc on the back copy; it is a nonseparating curve. The blue circle is entirely in a disc patch on the front copy and is contractible.