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An augmented chain complex includes an augmentation $\epsilon :C_0\to \mathbb{Z}$ with $\epsilon d_1=0$. For a nonempty simplicial complex, the augmentation sends every vertex to $1$ and the augmented homology in degree zero is its reduced homology.

A compact smooth $n$-manifold with nonempty boundary embedded in $S^n$ is a proper compact locally contractible subset. Alexander duality therefore expresses the reduced homology of its complement by the reduced cohomology of the abstract manifold. Thus these groups do not depend on its particular smooth embedding.

Define mapping degree for $S^0$ using its rank-one reduced homology group ${\tilde{H}}_0(S^0;\mathbb{Z})$. Its generator is the difference of its two points. The four maps of that two-point set consist of the identity, the interchange, and two constant maps; their degree multipliers are $1,-1,0,0$. Thus sphere maps of arbitrary integer degree require positive sphere dimension.

Let $a,b\ge 2$ with $a+b=2n+1$. Collapse the lower cells of $S^a\times S^b$ to obtain $q:S^a\times S^b\to S^{2n+1}$ and compose with the Hopf fibration $p:S^{2n+1}\to {\mathbb{CP}}^n$. Both factor inclusions are killed by q, so every induced homotopy group map is zero; the induced reduced homology maps also vanish. If the composite were nullhomotopic, homotopy lifting property would deform q into the circle fibre. Every map from the simply connected product to that circle is nullhomotopic, contradicting the degree-one top homology map of q.

For nonempty $X$, the topological mapping cone has this long exact sequence in homology, with ordinary homology for $X,Y$ and reduced homology for $C_f$. In degree zero, the augmentation of $X$ is accounted for by the cone vertex. The sequence follows from the Mayer–Vietoris theorem applied to neighborhoods of the cone and $Y$.

A CW complex is built by attaching q-dimensional discs to the $(q-1)$-skeleton along maps of their boundary spheres, with the closure-finite and weak-topology conditions. Write $X^q$ for its q-skeleton and $X^{-1}=\varnothing$. The cellular chain group isThe cellular boundary is the composite of the connecting map and the relative quotient map:Consecutive maps in the relevant exact sequences compose to zero, so these maps make a cellular chain complex. Equivalently, a boundary coefficient is the degree of the attaching sphere followed by collapse onto the sphere of a selected lower cell. Its cellular homology is $\ker d_q/\mathrm{im}{d}_{q+1}$. The relative groups of successive skeleta vanish off their cell dimension; their exact sequences identify this homology with singular homology, the cellular homology theorem.

For $k\ge 1$, the sphere $S^k$ has a zero-cell and a k-cell with zero boundary. The Complex projective space ${\mathbb{CP}}^n$ has one cell in each dimension $0,2,\dots ,2n$ and no odd cells, so all cellular boundaries vanish. ConsequentlyAt $k=0$, the sphere consists of two points and $H_0(S^0)={\mathbb{Z}}^2$; at $n=0$, projective space is a point.

Put $a=2n-2$ and $Z=S^a\times S^3$. Its product CW complex has cells in dimensions $0,3,a,a+3=2n+1$. For $n>3$, these positive dimensions are distinct and never consecutive, so all cellular boundaries vanish. ThusIts $2n$-skeleton is $S^a\vee S^3$. Collapsing it gives the quotientIt induces an isomorphism on top homology, with degree $+1$ after choosing the quotient orientation. Let $p:S^{2n+1}\to {\mathbb{CP}}^n$ be the Hopf fibration and set $\varphi =pq$, the Hopf projection of a product smash quotient.

On reduced homology, the collapse kills the two lower-dimensional classes, and the remaining top class maps to zero because projective space has no odd-dimensional homology. Therefore ${\varphi }_{\ast }=0$ in every reduced degree. On homotopy groups, ${\pi }_j(Z)\cong {\pi }_j(S^a)\oplus {\pi }_j(S^3)$ for $j\ge 2$, with summands induced by the two factor inclusions. Both inclusions are collapsed by $q$, so $q_{\ast }$ and hence ${\varphi }_{\ast }$ vanish on both summands. The domain is simply connected and connected, which also handles the lower groups.

It remains to prove that $\varphi$ is not null-homotopic. The Hopf fibration has fibre $S^1$ and has the homotopy lifting property for the CW domain $Z$: a homotopy downstairs and a lift of its initial map have a lift with that specified initial value. If $\varphi$ were homotopic to a constant, lift this homotopy starting at $q$. The ending map would land in the single fibre $S^1$. Every map $Z\to S^1$ is null-homotopic, since the simply connected domain permits its lift to the contractible universal cover $\mathbb{R}\to S^1$. Thus $q$ would be null-homotopic in $S^{2n+1}$, contradicting its nonzero top homology map. Hence

Let $C_{\ast }(X)$ be the integral singular chain complex. Augment it by the homomorphism $C_0(X)\to \mathbb{Z}$ sending every point simplex to one. The homology of this augmented complex is reduced homology: for a nonempty space, it agrees with ordinary homology in positive degrees and ${\tilde{H}}_0(X)$ is the kernel of $H_0(X)\to \mathbb{Z}$. The relative homology $H_{\ast }(X,A)$ is the homology of $C_{\ast }(X)/C_{\ast }(A)$; a relative cycle has boundary in $A$, and chains differing by a boundary or a chain in $A$ represent the same class.

Excision theorem: if $U\subset A\subset X$ and $\overline{U}\subset {\mathrm{int}}_{X}A$, the inclusion $(X\setminus U,A\setminus U)\hookrightarrow (X,A)$ induces isomorphisms on all relative homology groups. In particular, for a nonempty CW subcomplex $A$, choose a neighborhood $N$ deforming onto $A$, with the deformation fixing $A$. Such neighborhoods exist for CW pairs. Then $N/A$ is contractible. The long exact sequence of a pair gives $H_{\ast }(X,A)\cong H_{\ast }(X,N)$ and $H_{\ast }(X/A,\{\ast \})\cong H_{\ast }(X/A,N/A)$. Excision removes $A$ and the collapsed point respectively; the resulting pairs $(X\setminus A,N\setminus A)$ and $((X/A)\setminus \{\ast \},(N/A)\setminus \{\ast \})$ are homeomorphic. ConsequentlyHere the nonempty-subcomplex condition is the usual quotient convention in this statement. If $A=\varnothing$, use the pointed cofiber $X_{+}$ with a disjoint basepoint; the literal unpointed quotient $X/\varnothing =X$ would not give the asserted degree-zero identity.

For one embedded circle, its long exact sequence of a pair reduces toA separating circle bounds a subsurface, so its homology class is zero. A nonseparating circle has a transverse closed curve meeting it once, showing that its class is nonzero and primitive. Thus $i_{\ast }$ is either zero or an injection with torsion-free cokernel. The resulting possibilities arewith all other relative groups zero. These ranks determine whether the circle is null-homologous.

For $g\ge 2$, a disk boundary and a separating circle bounding a once-punctured torus have the same relative groups. The first is null-homotopic; the second is not. For a direct verification, use the surface-group presentation $⟨a_1,b_1,\dots ,a_g,b_g\mid {\prod }_i[a_i,b_i]=1⟩$. Send $a_1,b_1$ to free generators $u,v$, send $a_2,b_2$ to $v,u$, and all remaining generators to one. The relation maps to $[u,v][v,u]=1$, so this defines a homomorphism to the free group. The separating circle represents $[a_1,b_1]$ and maps to the nonempty reduced word $uv{u}^{-1}{v}^{-1}$; it is therefore not null-homotopic. This proves the requested failure to detect nullhomotopy. There is a necessary low-genus qualification: on ${\Sigma }_0$ or ${\Sigma }_1$, an embedded circle is nullhomologous exactly when it is nullhomotopic. On the sphere every circle bounds a disk; on the torus every essential embedded circle is nonseparating. The printed nondetection assertion therefore needs $g\ge 2$.

For two disjoint embedded circles let $r$ be the rank of their image in $H_1({\Sigma }_g)$, and let $c$ be the number of connected components of the cut surface. Removing open annular neighborhoods and applying excision gives the compact cut surface $W$ relative to its boundary. Poincare-Lefschetz duality gives $H_2(W,\partial W)\cong H^0(W)={\mathbb{Z}}^c$ and $H_1(W,\partial W)\cong H^1(W)$, which is free. Thus there is no hidden torsion. The long exact sequence gives $c=3-r$, and the Euler characteristic of the pair is $2-2g$, givingThe possibilities are $r=0,1,2$: two separating circles give zero; a nonseparating circle with a disk boundary, or two parallel essential circles, give one; circles around two different handles give two. For genus zero only $r=0$ occurs; for genus one $r=0,1$ occur; for every genus at least two all three occur. This is relative homology of a surface modulo disjoint circles.

Write a point of the larger sphere as $(u,v)\in {\mathbb{R}}^{m+1}\times {\mathbb{R}}^{n-m}$. In the complement, $v\ne 0$. There is a homeomorphismwhose inverse sends $(u,w)$ to $(u,\sqrt{1-\Vert u{\Vert }^2}w)$. In particular the complement has a deformation retraction onto its transverse sphere. Explicitly,The denominator stays positive, the homotopy stays in the complement, and points with $u=0$ remain fixed. This establishes the homology of an equatorial sphere complement. By homotopy invariance of homology, its homology is therefore the homology of a sphere of dimension $d=n-m-1$:In the codimension-one case, the complement has two contractible components; this is why degree-zero homology has rank two. Equivalently, in every case its only nonzero reduced homology is ${\tilde{H}}_d(X;\mathbb{Z})=\mathbb{Z}$.

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 degree and its local signs. Orient $S^n$ and let $[S^n]$ generate its top reduced homology ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree is the integer characterized byFor $n\ge 1$ this is the usual top homology definition using the fundamental class; using reduced homology also covers $S^0$.

Suppose $f$ is smooth and $p$ is a regular value. Each $x\in f^{-1}(p)$ has invertible tangent map $D{f}_x$, so the inverse function theorem makes $f^{-1}(p)$ discrete. It is also closed in the compact sphere, hence finite. Choose disjoint small neighborhoods of these points on which $f$ is a local diffeomorphism. The Excision theorem identifies the source local homology with the direct sum of one copy of $\mathbb{Z}$ for each inverse image. The induced local map is multiplication by $+1$ or $-1$ according as $D{f}_x$ preserves or reverses orientation. The map from the global fundamental class to these local orientation classes therefore proves the degree as a sum of local degrees formula:Here the determinant is computed in positively oriented tangent bases. Thus the mapping degree counts inverse images with signs, rather than just their cardinality.

The quotient map. Put $m=n+1$ and write $f=F{|}_{S^{m-1}}$. Give $D^m$ its standard orientation and its boundary the induced orientation. The connecting homomorphismis an isomorphism: the disk has zero positive reduced homology. It sends the relative fundamental class $[D^m,S^{m-1}]$ to $[S^{m-1}]$. If the map on relative homology induced by $F$ multiplies this class by $d$, naturality givesThus $d=\mathrm{deg}f$. Collapsing the boundary gives a sphere $D^m/S^{m-1}\cong S^m$, and the quotient map identifies its top reduced homology with the top relative homology of the disk pair. Give this quotient sphere the orientation determined by that identification. The relation $qF=\tilde{F}q$ now showsThis is the quotient-sphere degree identity.

The graph intersection. Orient $D^m\times D^m$ by the product orientation, orient $N=D^m\times \{0\}$ by its first factor, and orient the graph of a function $\Gamma$ by $x\mapsto (x,F(x))$. An intersection is precisely a zero of $F$, and no such zero lies on the boundary because $F(S^{m-1})\subset S^{m-1}$. At an intersection, transverse intersection means that $D{F}_x$ is surjective, hence invertible. The zeros are consequently isolated and finite.

For the smooth intersection number use the ordered tangent spaces $T_{x}N$ first and $T_{(x,F(x))}\Gamma$ second. Relative to the product basis their concatenated basis has matrixIts determinant is $\det D{F}_x$, so the intersection sign is ${ϵ}_x=\mathrm{sgn}\det D{F}_x$. By the same local Excision theorem argument, now in relative homology at the interior point $0$, the sum of these signs equals the multiplier of $F_{\ast }$ on $H_m(D^m,S^{m-1};\mathbb{Z})$. Therefore the graph intersection formula for mapping degree isIf the tangent spaces are ordered $\Gamma$ first and $N$ second, every sign changes by $(-1{)}^m$; the order above specifies the appropriate convention.

The one-dimensional model $F(x)=2x^3-x$ on $[-1,1]$ illustrates the graph intersection formula for mapping degree: its three zeros have signs $+1,-1,+1$, and its endpoint map has mapping degree $1$ on reduced homology.

Write $v$ for the cone vertex. The quotient is the topological mapping cone $C_f=Y{\cup }_{f}CX$. First assume $X\ne \varnothing$, as is implicit in having this vertex; a map then also forces $Y\ne \varnothing$.

Take the open cover consisting of the image of $X\times [0,2/3)$ and the image of $Y\bigsqcup (X\times (1/3,1])$. Denote these sets by $U,V$. The first contracts to $v$, the second has a deformation retraction onto $Y$, and $U\cap V\cong X\times (1/3,2/3)$ retracts onto $X$. Under these identifications the two inclusion maps induce the zero map into positive-degree homology of $U$ and $f_{\ast }$ into the homology of $V$. The Mayer–Vietoris theorem consequently gives, in positive degrees,The same assertion holds in degree zero, but needs care: the ordinary Mayer-Vietoris map isSince ${\epsilon }_Y{f}_{\ast }a={\epsilon }_{X}a$, its kernel equals $\ker f_{\ast }$. On quotienting $H_0(Z;G)$ by the direct summand $G[v]$, its cokernel becomes $\mathrm{coker}{f}_{\ast }$. That quotient is canonically identified with reduced homology by subtracting the augmentation times $[v]$. Thus the last nonzero part isIn particular the map from $H_0(Y;G)$ sends a component class $[y]$ to $[y]-[v]$. This proves the mapping cone exact sequence; finite generation of $G$ is not needed for this construction.

There is a literal empty-space exception in the printed hypotheses. If $X=\varnothing$ and $Y$ is a point, the given quotient is just $Y$, and the asserted degree-zero exact sequence would be $0\to G\to 0$. Thus the displayed reduced sequence presupposes nonempty $X$. It holds also for empty $X$ if one defines its cone to include a new isolated vertex. The later homology-isomorphism claim itself remains true for empty finite complexes, with the cases involving one empty complex dealt with directly in degree zero.

Now take nonempty finite CW complexes $X,Y$. Put $A_q={\tilde{H}}_q(Z;\mathbb{Z})$. The exact sequence giveswhere $H_{-1}=0$. All $A_q$ are finitely generated abelian groups, since the groups for $X,Y$ are finitely generated. Moreover, by exactness,The reduced universal coefficient theorem for homology givesIf all $A_q$ vanish, so does the middle term for every prime, and the exact sequence with coefficients proves that $f_{\ast }$ is an isomorphism modulo every prime. Conversely, those isomorphisms imply that the middle terms vanish, hence $A_q\otimes {\mathbb{F}}_p=0$ for every prime $p$. By the Fundamental theorem of finitely generated abelian groups, a nonzero free summand would survive modulo every prime, and a cyclic torsion summand would survive modulo a prime dividing its order. Thus $A_q=0$ for every $q$, provingThis argument uses the exact sequence to obtain finite generation; it does not assume in advance that the cone of an arbitrary continuous map is a finite CW complex.

Finally suppose $f$ is a cellular map. Keep the cells of $Y$, add the vertex $v$, and add a $(q+1)$-cell for the cone on every $q$-cell of $X$. Its top face attaches to $Y$ through $f$, and its other faces attach to cones on lower-dimensional cells. The cellular condition puts these attachments in the appropriate skeleton. This constructs a finite CW complex structure on $Z$.

Let $A_{\ast }=C_{\ast }^{\mathrm{cell}}(X)$ and $B_{\ast }=C_{\ast }^{\mathrm{cell}}(Y)$ be the ordinary cellular chain complexes, with $A_{-1}=0$. Orient each coned cell so that its boundary has the form $f_{\#}(e)-c(d_{A}e)$. Using $[y]-[v]$ as the reduced generator for each vertex of $Y$ givesIn particular ${\tilde{C}}_0^{\mathrm{cell}}(Z)=B_0$, and the boundary of a cone edge is its image vertex minus $v$. The chain map identity $d_B{f}_{\#}=f_{\#}{d}_A$ givesThis is the cellular chain complex of a mapping cone, equivalently the algebraic mapping cone with the displayed summand and sign conventions.

All groups in the first calculation have integral coefficients. Let $U=S^n\setminus j(W)$ and $V=S^n\setminus \mathrm{int}j(W)$. A two-sided tubular neighborhood of $j(\partial W)$ identifies $V$ as a compact manifold with boundary, with interior $U$. Pushing its boundary inward in a collar neighborhood gives a homotopy equivalence $U\simeq V$. After thickening across the same collar, excision identifies the relative groupsThe orientation of $S^n$ gives $W$ an orientation. The Poincare-Lefschetz duality isomorphism therefore givesEach component of $W$ has boundary: a closed component would have open image by the local embedding condition and closed image by compactness, hence would occupy the whole connected sphere, leaving no room for the components with boundary. In particular $U$ is nonempty, and $H^n(W)=0$.

Apply the long exact sequence in relative homology of $(S^n,V)$, using reduced homology for the absolute terms. Away from the top-dimensional homology of the sphere, it identifiesThe exceptional map iswhere $r$ is the number of components of $W$. Local compatibility of the orientations sends $1$ to $(1,\dots ,1)$, the relative fundamental classes of all the components. Its kernel is zero and its cokernel is ${\tilde{H}}^0(W)$. Thus the exceptional term has exactly the reduced form required, and all higher groups vanish. We obtain the Alexander duality formulawith negative-index cohomology zero. Ordinary degree zero is recovered byBoth formulas depend on the abstract manifold $W$, so they establish the requested independence up to group isomorphism. They make no claim that the complements themselves are homeomorphic or have isomorphic fundamental groups. The argument also covers $n=1$, when the diagonal map handles the degree-zero exception. This is the complement homology of a compact codimension-zero submanifold.

For the second calculation write $F={\mathbb{F}}_2$, $C=S^n\setminus i(M)$ and $c=n-k$. First suppose $c>0$, with $n\ge 1$. The rank-$c$ normal bundle has a mod-two Thom class, regardless of orientability. A tubular neighborhood, excision and the homological Thom isomorphism theorem identifyThe long exact sequence in relative homology becomes the Gysin sequence of an embedding:The map $\gamma$ takes the mod-two fundamental class of $S^n$ to that of $M$: restricting to each normal fiber evaluates the Thom class as $1$. Since $M$ is closed and connected, the degree-$n$ mapis an isomorphism. This cancels the exceptional top term and gives $H_i(C;F)=0$ for $i\ge n-1$ when $i>0$. In all the intervening degrees the sphere groups vanish. In degree zero, the remaining $F$ is removed by the augmentation. Consequently the mod-two homology of a submanifold complement iswhere negative-index homology of $M$ is zero. To recover ordinary homology, add one copy of $F$ in degree zero and change nothing in positive degrees. For $n=1$ the first range is empty, so the complement of a connected zero-manifold has the homology of a point. For example, codimension at least two gives $H_0(C;F)=F$, while codimension one gives $H_0(C;F)=F^2$ when $n\ge 2$.

If $k=n\ge 1$, an embedding of the closed connected manifold has image both open and closed in $S^n$, hence is onto; its complement is empty and all its ordinary homology groups vanish. The positive-codimension formula is not asserted in that case. The standard sphere calculations here assume $n\ge 1$; in ambient dimension zero the complement of the embedded connected point in $S^0$ is the other point.

Let $i:L\hookrightarrow M$, $j:L\hookrightarrow N$, $r:M\hookrightarrow K$, and $s:N\hookrightarrow K$ be the inclusions. The Mayer–Vietoris sequence is the long exact sequence in homologyThe same sequence continues through degree zero, or uniformly in all degrees when written with reduced homology.

For the pair from part (c), every simplex of dimension below $n$ lies in $L_{n-1}$. Consequently the relative simplicial chain complex haswhere the degree-$n$ generator is the class of $[v_0\cdots v_n]$. Its relative boundary is zero because every codimension-one face lies in $L_{n-1}$. All differentials therefore vanish, andThis is the relative homology of a simplex and its boundary. Equivalently, the long exact sequence in relative homology, together with the fact that $|K_n|$ is a contractible space and the homology of a sphere $|L_{n-1}|\cong S^{n-1}$, gives the reduced homology identification $H_k(K_n,L_{n-1})\cong {\tilde{H}}_{k-1}(S^{n-1})$.

Collapsing the boundary identifies the top relative homology of a disk pair with the top reduced homology of its quotient sphere. Together with the boundary connecting homomorphism, this shows that the induced quotient map and the boundary map have the same mapping degree, using the corresponding orientations.