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Choose a generator $[S^n]$ of ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree of a continuous map is the integer determined byFor $n\ge 1$ this is ordinary top-dimensional homology; using reduced homology also gives a convention for $n=0$. Domain and target use the same orientation, so reversing the chosen generator in both leaves the integer unchanged. Homotopy invariance of homology makes the mapping degree invariant under homotopy.

We interpret the manifold criterion for a suspension in the category of nonempty topological manifolds without boundary, and “sphere” as homeomorphic to the standard sphere. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit homeomorphism from its suspension to $S^{d+1}$ isIt is a continuous map that is a bijection: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The compact-to-Hausdorff continuous bijection theorem makes it a homeomorphism.

Conversely suppose $SX$ is a topological manifold. Its open part $X\times (0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is contractible, while $U\setminus \{v\}$ admits a deformation retraction onto $X$. The long exact sequence in relative homology and excision give the local homology from a link calculationA point of a $(d+1)$-dimensional topological manifold without boundary has local homology $\mathbb{Z}$ in degree $d+1$ and zero in all other degrees. ConsequentlyThus $X$ is an integral homology sphere. This conclusion alone does not identify its homeomorphism type.

If $d\ge 2$, it is also simply connected. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By compactness of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times \{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor asThe composite induces an isomorphism of fundamental groups, since both cone punctures are products of $X$ with intervals and their inclusion is a homotopy equivalence. The middle punctured ball has the homotopy type of $S^d$ and is simply connected for $d\ge 2$. Hence the composite is also zero on the fundamental group, forcing ${\pi }_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a sphere.

The Hurewicz theorem, applied successively to the vanishing lower homology groups, now gives ${\pi }_j(X)=0$ for $j<d$ and ${\pi }_d(X)\cong H_d(X)\cong \mathbb{Z}$. A representative of its generator is a continuous map $S^d\to X$ inducing an isomorphism in every integral homology degree. Topological manifolds have the homotopy type of CW complexes, so the homological Whitehead theorem makes it a homotopy equivalence. We have proved that $X$ is a homotopy sphere.

At this point the topological classification theorem is essential: the topological generalized Poincare theorem says that a closed homotopy sphere is homeomorphic to the standard sphere. Applying it completes the converse for $d\ge 2$. For $d=1$, the local calculation makes $X$ a connected compact one-dimensional manifold without boundary, hence a circle. For $d=0$, a compact zero-dimensional manifold is a finite set, and ${\tilde{H}}_0(X)\cong \mathbb{Z}$ forces exactly two points. Thus it is $S^0$. Altogether,The last classification step is not a consequence of homology alone, and we make no assertion of diffeomorphism with a prescribed smooth structure. If the word manifold allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto ((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\le 1$, although the interval is not a sphere.

To construct sphere maps of arbitrary integer degree, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta \mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\mathrm{deg}{h}_m=m$.

Two cone neighbourhoods cover a suspension; their intersection admits a deformation retraction onto the original space. The reduced Mayer–Vietoris theorem gives a natural suspension isomorphismNaturality means that the suspension of a continuous map induces the same integer multiplier. This proves degree under suspension. Using the standard identification $S(S^k)\cong S^{k+1}$, setThus every integer occurs in every positive sphere dimension, with negative degrees handled just as well as positive ones.

If “every $n$” includes zero, that final printed assertion is false. The generator of ${\tilde{H}}_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving preciselyThese degrees of maps of the zero-sphere are the explicit exception to the arbitrary-degree construction.

The weak-star topology $\sigma (X^{\ast },X)$ is generated by the seminorms $p_x(g)=|g(x)|$, $x\in X$. Thus every evaluation is continuous, so $f_n\stackrel{w^{\ast }}{\to }f$ implies $f_n(x)\to f(x)$ for each $x$. Conversely, a basic neighborhood of $f$ imposes finitely many conditions $|(g-f)(x_j)|<\epsilon$. Pointwise convergence eventually satisfies all of them simultaneously. HenceThe Uniform boundedness principle, applied on the Banach space $X$, gives $M={\sup }_n\Vert f_n\Vert <\infty$. If $x_n\to x$ in the norm topology, thenTherefore $\boxed{f_n(x_n)\to f(x)}$. Completeness of $X$ is used in this uniform boundedness step.

If $C\subseteq X^{\ast }$ is norm compact, the identity map from $(C,\Vert \cdot \Vert )$ to $(C,w^{\ast })$ is a continuous bijection. The target is Hausdorff, since distinct functionals differ at some $x\in X$. The compact-to-Hausdorff continuous bijection theorem gives

The Banach-Alaoglu theorem states that the closed dual unit ball $B_{X^{\ast }}$ is compact in the weak-star topology; in fact $X$ need only be a normed vector space. To prove it, work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and formEach $D_x$ is compact, so $Q$ is compact by the Tychonoff theorem. Embed $B_{X^{\ast }}$ by $g\mapsto (g(x){)}_{x\in X}$. Its image consists exactly of those families $(z_x)$ satisfyingThese are closed conditions in the product topology. Conversely, any family satisfying them defines a linear functional bounded by $|z_x|\le \Vert x\Vert$, hence an element of $B_{X^{\ast }}$. The image is therefore closed and compact. The induced product topology is precisely the weak-star topology, proving the theorem. This is a compactness proof for arbitrary nets, with no separability assumption.

For the dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$, each $x\in X$ satisfiesEvaluation at the fixed vector $Tx$ is weak-star continuous, so $\boxed{T^{\ast }\text{ is weak-star-to-weak-star continuous}}$.

Suppose $T$ is a compact operator. By the allowed Schauder theorem for compact operators, $T^{\ast }$ is compact. If $S\subseteq Y^{\ast }$ is bounded, the norm closure $C=\overline{T^{\ast }(S)}$ is norm compact. We already proved that its norm and weak-star topologies coincide. The weak-star continuous map $T^{\ast }{|}_S$ into $C$ is consequently norm continuous. Conversely, if $T^{\ast }$ is weak-star-to-norm continuous on bounded subsets, its restriction to $B_{Y^{\ast }}$ maps a weak-star compact set, by Banach-Alaoglu theorem, to a norm compact set. Thus $T^{\ast }$ is a compact operator, and the Schauder theorem for compact operators gives

For the first norm-one assertion, use the usual unital Banach algebra convention $\Vert 1\Vert =1$. A character of an algebra is a nonzero complex linear multiplicative functional, initially without any continuity assumption. It satisfies $\phi (1)=1$, since otherwise $\phi (a)=\phi (a1)=0$ for every $a$. Moreover, $a-\phi (a)1$ cannot be invertible: applying $\phi$ to an inverse identity would give $0=1$. ThusThis proves automatic continuity of characters and $\Vert \phi \Vert \le 1$. Evaluation at the normalized identity gives the reverse inequality, soIf a convention permits $\Vert 1\Vert >1$, only $\Vert \phi \Vert \le 1$ follows in general: on $\mathbb{C}$ with norm $2|z|$, the identity character has norm $1/2$. The later completion argument needs only the upper bound.

If $a$ is invertible, $\phi (a)\phi (a^{-1})=1$, so every character is nonzero on $a$. Conversely, if $a$ is not invertible, its principal ideal is proper and lies in a maximal ideal $M$ by Zorn lemma. The ideal $M$ is closed: if its closure were the whole algebra, it would contain $m$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series, which is impossible for a proper ideal. Maximality therefore gives $\overline{M}=M$. The quotient $A/M$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. The quotient map yields a character vanishing at $a$. Consequently

Now let $A=C(K)$. Its evaluation characters give the map $x\mapsto {\delta }_x\in {\Phi }_A$. Every character has this form. Indeed, its kernel is a proper ideal. If the functions in that kernel had no common zero, compactness would give finitely many $f_1,\dots ,f_m$ in the kernel with no common zero. Then $h={\sum }_j\overline{f_j}{f}_j$ belongs to the kernel and is everywhere positive, hence is invertible in $C(K)$, a contradiction. At a common zero $x$, the fact that $f-\phi (f)1$ belongs to the kernel gives $f(x)=\phi (f)$ for every $f$. Continuous functions distinguish points of the compact Hausdorff space $K$, so the map is bijective. It is continuous for the Gelfand topology, and the compact-to-Hausdorff continuous bijection theorem givesFor $K=\varnothing$ the norm comparison is trivial; assume henceforth that $K\ne \varnothing$.

Let $B$ be the completion for the other algebra norm $\Vert \cdot {\Vert }_1$. It remains commutative and unital, with $A$ embedded densely. The restriction of any $\psi \in {\Phi }_B$ is a nonzero character of $A$, since it takes $1$ to $1$, and is automatically continuous for the original supremum norm. Thus $R(\psi )=\psi {|}_A$ maps into ${\Phi }_A$. It is injective because two continuous characters agreeing on the dense subalgebra $A$ agree on $B$, and it is continuous because each coordinate $R(\psi )(f)=\psi (f)$ is weak-star continuous.

The character space ${\Phi }_B$ is weak-star compact even if its identity was not normalized: the spectral bound puts all its characters in $B_{B^{\ast }}$, and the conditions $\psi (1)=1$, $\psi (bc)=\psi (b)\psi (c)$ define a weak-star closed set. Apply Banach-Alaoglu theorem. Consequently $R$ is a homeomorphism onto a closed subset $L\subseteq K$.

Suppose $x\in U=K\setminus L$. The compact Hausdorff space $K$ is normal, so choose an open $V$ with $x\in V\subseteq \overline{V}\subseteq U$. By the Urysohn lemma, choose $g\in C(K)$ with $g(x)=1$ and $g=0$ on $K\setminus V$, and choose $f\in C(K)$ with $f=0$ on $\overline{V}$ and $f=1$ on $L$. Then $fg=0$. Every character $\psi$ of $B$ restricts to evaluation at a point of $L$, so $\psi (f)=1$. The invertibility criterion proved above makes $f$ invertible in $B$. Multiplying $fg=0$ by $f^{-1}$ in $B$ gives $g=0$, contradicting $g(x)=1$. Therefore $U=\varnothing$ and $R$ is onto $K$.

For each $x\in K$, choose a character ${\psi }_x\in {\Phi }_B$ extending ${\delta }_x$. Its spectral bound givesTaking the supremum over $x$ proves the minimality of the supremum norm on C(K):

A closed set $C\subseteq X$ is compact because it is closed in the compact space $X$. Its image $f(C)$ is compact by the fact that a continuous image of a compact space is compact, and every compact subset of the Hausdorff space $Y$ is closed. Thus the continuous bijection $f$ is a closed map. Its inverse $f^{-1}:Y\to X$ is consequently continuous, so $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

Define a continuous map $F:S\to T$ by the standard embedded torus of revolution parametrizationIts image lies in $T$ becauseThe periodicity of the trigonometric functions makes $F(s,0)=F(s,1)$ and $F(0,t)=F(1,t)$, so $F$ is constant on every $\sim$-equivalence class. The universal property of the quotient topology therefore gives a unique continuous map

Every point of $T$ has cylindrical radius between $1$ and $3$, so its polar angle determines $s$ modulo one, while the pair $(\sqrt{x^2+y^2}-2,z)$ on the unit circle determines $t$ modulo one. Hence $\overline{F}$ is surjective, and two parameters have the same image exactly when they differ by the endpoint identifications defining $\sim$; thus it is injective.

The quotient $S/\sim$ is compact as the continuous image of the given compact space $S$ under the quotient map. The surface $T$ is Hausdorff because it has the subspace topology inherited from ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now shows that $\overline{F}$ is a homeomorphism. Therefore

The identity mapis a continuous bijection by $d_1\le d_{\infty }$. Its domain is compact by part (i), and its codomain is a Hausdorff space because every metric space is Hausdorff. The compact-to-Hausdorff continuous bijection theorem therefore makes the identity a homeomorphism. Consequently

DefineThis is continuous and takes values on the unit sphere. It agrees at $(0,y)$ and $(1,y)$, is constant on the bottom edge, and is constant on the top edge. Thus it is constant on every $R$-class, so the universal property of the quotient topology gives a continuous map

For $0<y<1$, the third coordinate $\cos (\pi y)$ determines $y$, and the first two coordinates determine $x$ modulo one. Consequently the only equal values of $f$ in the open strip arise from $x=0$ and $x=1$. At $y=0$ the whole edge maps to the north pole, and at $y=1$ the whole edge maps to the south pole. These are exactly the identifications defining $R$, so $\overline{f}$ is injective. The spherical-coordinate formula also shows that it is surjective.

The square is compact, hence its quotient $X/R$ is compact by part (ii), while $S^2$ is Hausdorff as a subspace of ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now makes $\overline{f}$ a homeomorphism. Geometrically, identifying the vertical sides produces a cylinder and collapsing each boundary circle to a point produces the suspension of a topological space $S^1$, which is $S^2$. This proves the square quotient model of the two-sphere.

Let $X$ be compact and $q:X\to X/R$ the quotient map. If $(U_i{)}_{i\in I}$ is an open cover of $X/R$, then $(q^{-1}(U_i){)}_{i\in I}$ is an open cover of $X$. A finite subfamily covers $X$; because $q$ is surjective, the corresponding $U_i$ cover $X/R$. Thus every quotient of a compact space is compact.

The Hausdorff property need not survive. On the real line, defineThe quotient $\mathbb{R}/\mathbb{Q}$ has more than one point. If two nonempty open subsets of the quotient were disjoint, their inverse images would be disjoint nonempty open subsets of $\mathbb{R}$ invariant under rational translation. But any two nonempty open intervals acquire an intersection after one is translated by a suitably chosen rational number, so two such saturated open sets cannot be disjoint. Distinct quotient points cannot be separated, and the quotient is not Hausdorff. This is the non-Hausdorff quotient of the real line by rational translation.

Finally let $f:X\to Y$ be a continuous bijection, with $X$ compact and $Y$ Hausdorff. Every closed subset $C\subseteq X$ is compact. Its continuous image $f(C)$ is compact, and every compact subset of a Hausdorff space is closed. Hence $f$ is a closed map. For every closed $C\subseteq X$,is closed in $Y$, so $f^{-1}$ is continuous. Therefore $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.