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Identify opposite sides of a square by translations to obtain a torus. All four corners become one vertex; the horizontal sides give one topological cell $a$ and the vertical sides one topological cell $b$. The open square supplies one two-dimensional topological cell. Thus $(c_0,c_1,c_2)=(1,2,1)$ and $\chi =0$. The cellular boundary of each one-dimensional topological cell is zero because its endpoints agree. Traversing the face boundary gives the group commutator $ab{a}^{-1}{b}^{-1}$, whose signed counts of both $a$ and $b$ vanish. Hence its cellular boundary is also zero, and $(b_0,b_1,b_2)=(1,2,1)$.^[ref]

The boundary of a tetrahedron is a two-sphere, with four vertices, six edges and four triangular topological cells: $\chi =4-6+4=2$. Its rational Betti numbers are $(1,0,1)$ by homology of a sphere. The torus's two independent one-dimensional homology classes cancel its degree-zero and degree-two contributions. The torus and annulus both have Euler characteristic zero, yet different homology groups; one integer cannot recover all of homology.

Let $M$ be the given closed three-dimensional manifold. It is connected by the meaning of simply connected. An orientation chosen at one point can be transported along paths, and its sign change around loops defines a homomorphism ${\pi }_1(M)\to \{\pm 1\}$. Since the fundamental group is trivial, this monodromy vanishes; thus $M$ is orientable.

The first homology group is the abelianization of the fundamental group, giving $H_1(M;\mathbb{Z})=0$. Also $H_0(M;\mathbb{Z})=\mathbb{Z}$. The universal coefficient theorem for cohomology in degree one givesBoth outer groups vanish: the first because $\mathbb{Z}$ is free and the second because $H_1=0$. Hence $H^1=0$. Integral Poincare duality now gives $H_2(M;\mathbb{Z})\cong H^1(M;\mathbb{Z})=0$ and $H_3(M;\mathbb{Z})\cong H^0(M;\mathbb{Z})=\mathbb{Z}$. Homology in degrees greater than three vanishes by the same duality with negative-degree cohomology. ConsequentlyThese are precisely the homology of a sphere for $S^3$, proving that simply connected closed three-manifolds are homology spheres. No homeomorphism classification of three-manifolds is needed.

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 ${\tilde{H}}_d(X;\mathbb{Z})=\mathbb{Z}$.

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 $H_k(K_n,L_{n-1})\cong {\tilde{H}}_{k-1}(S^{n-1})$.

Write $S^n$ as the union of slightly enlarged northern and southern hemispheres $U$ and $V$. Both are contractible, while $U\cap V$ deformation retracts onto $S^{n-1}$. The reduced Mayer–Vietoris theorem therefore givesStarting from ${\tilde{H}}_0(S^0;\mathbb{Z})\cong \mathbb{Z}$ proves the homology of a sphere:This uses no cellular homology. A reflection of $S^n$ reverses its orientation and has degree of a continuous mapping $-1$, so it induces the identity on $H_0$, multiplication by $-1$ on $H_n$, and the unique map between zero groups in every other degree.

For a CW complex with skeleta $X^k$, its cellular chain complex isThe differential is the connecting map to $H_{k-1}(X^{k-1})$ followed by passage to $H_{k-1}(X^{k-1},X^{k-2})$. Equivalently, the coefficient of a $(k-1)$-cell in the boundary of a $k$-cell is the degree obtained from its attaching map after collapsing the complement of that lower cell. This is the cellular boundary formula.

The quotient $D^k/(x\sim -x\text{ on }{S}^{k-1})$ builds ${\mathbb{RP}}^k$ from ${\mathbb{RP}}^{k-1}$ by one $k$-cell, so ${\mathbb{RP}}^n$ has one cell in each dimension $0,\dots ,n$. The two lifts of the attaching map contribute with relative sign $(-1{)}^k$, and the cellular homology of real projective space has differentialConsequentlyWith ${\mathbb{F}}_2=\mathbb{Z}/2$ coefficients every differential vanishes, and hencewith zero homology outside that range.

A local orientation of a manifold at $x\in M$ is a generator ofAn $R$-orientation is a locally coherent choice of such generators. An R-fundamental class is a class $[M{]}_R\in H_d(M;R)$ whose image in every one of these local homology groups is a generator. Those images vary coherently under the restriction maps between small coordinate balls, so an $R$-fundamental class determines an $R$-orientation.

For a closed $R$-oriented $d$-manifold, Poincare duality says that cap product with its fundamental class is an isomorphismfor every $k$.

Choose a generator $[S^d]\in H_d(S^d;\mathbb{Z})\cong \mathbb{Z}$ from the homology of a sphere. For every $x\in S^d$, the long exact sequence of $(S^d,S^d\setminus \{x\})$, together with the contractibility of $S^d\setminus \{x\}\cong {\mathbb{R}}^d$, shows thatThe chosen generator is therefore a local generator at every point and is a $\mathbb{Z}$-fundamental class.