Relative Homology Excision and Mayer Vietoris — Examples

The disk is contractible, and the sphere has reduced homology only in degree $n-1$.

For $n>1$, exactness makes $\delta :H_n(D^n,S^{n-1};G)\to H_{n-1}(S^{n-1};G)$ an isomorphism. For $n=1$, it identifies the relative group with the augmentation kernel ${\tilde{H}}_0(S^0;G)\subset H_0(S^0;G)$ rather than with all of $H_0(S^0;G)$. Exactness forces every other relative group to vanish, and the cycle formula gives $\delta [c_g]=[\mathrm{\partial }{c}_g]$ with the stated sign.

$I$ is contractible and $H_0(E;G)\cong G\oplus G$, while the map to $H_0(I;G)\cong G$ adds the two coordinates.

The kernel is $\{(-g,g):g\in G\}$, and $g\mapsto (-g,g)$ is an isomorphism from $G$ onto it. Exactness identifies this kernel with $H_1(I,E;G)$ and makes all remaining relative groups vanish. The boundary of $c_g$ is $g[1]-g[0]$.

Coning the three subdivided boundary edges to $b$ produces exactly six small triangles with their cone orientations.

Every interior radial or midpoint edge occurs twice with opposite orientations; the uncancelled edges are the once-subdivided boundary. Thus $\mathrm{\partial }S=S\mathrm{\partial }$ in this case.

Each proper open arc is an interval in the cyclic order. The interval inside $U$ joining the overlap points is one of the two closed subarcs between them, and the complementary subarc lies in $V$ because the omitted parts of $U$ must be covered by $V$. Both endpoints are in both open sets. Thus the entire images of $u$ and $v$, including their endpoints, lie in their respective cover members. Also $\mathrm{\partial }(u\otimes g)=([a_1]-[a_0])\otimes g$ and $\mathrm{\partial }(v\otimes g)=([a_0]-[a_1])\otimes g$, so $z_g$ is cover-small and is a cycle.

By Mayer–Vietoris connecting class, $\delta [z_g]=[([a_1]-[a_0])\otimes g]\in H_0(U\cap V;G)$. Identifying the two components in the order containing $a_0,a_1$ gives the coefficient pair $(-g,g)$. Indeed, paths within each arc component identify its point classes, and no path crosses the components. Replacing the oriented integral cycle by its negative negates this class. If $g=0$ (including $G=0$), both chain and connecting class are zero.

Both hemispherical neighborhoods are contractible and their intersection deformation retracts to $S^{n-1}$.

The reduced Mayer–Vietoris sequence has zero hemisphere terms, so its connector gives the displayed degree shift. Starting with $S^0$ gives $G$ in degree $n$ and zero elsewhere.

Choose two open arcs $A,B\subset S^1$ covering the first factor with $A\cap B=I_0\bigsqcup I_1$, two open arcs. Put $U=A\times S^1$ and $V=B\times S^1$. Each overlap component and each of $U,V$ retracts onto the second factor. After ordering $I_0,I_1$, the Mayer--Vietoris map $H_1(U\cap V;G)\to H_1(U;G)\oplus H_1(V;G)$ is $(a,b)\mapsto (a+b,-a-b)$, since inclusion preserves the second-factor circle and the sequence uses opposite signs for $U$ and $V$. The same formula holds in degree zero. Its kernel is $\{(g,-g):g\in G\}$ and its cokernel is $G$, with no freeness assumption on $G$.

Exactness gives $0\to G\to H_1(T^2;G)\stackrel{\delta }{\to }G\to 0$, where the right $G$ is the displayed degree-zero kernel. Fix $y_0\in S^1$ and orient $S^1\times \{y_0\}$ so that its arc in $U$ has boundary $g([x_0]-[x_1])$, with $x_j\in I_j$. Splitting its singular cycle into the arcs in $U$ and $V$, the chain-level Mayer--Vietoris connecting map sends its class with coefficient $g$ to $(g,-g)$. Hence this first-factor circle supplies a homomorphic section of $\delta$ for every abelian $G$. The second-factor circle at $x_0$ maps to $(g,0)$ in $H_1(U)\oplus H_1(V)$ and represents the cokernel generator of step 1.1. These two circle maps identify $H_1(T^2;G)$ with $G\oplus G$.

The excised pair is $((0,1],\mathrm{\varnothing })$, whose $H_0$ is $\mathbb{Z}$; the original pair has $H_0([0,1],\{0\})=0$ by its pair sequence.

Hence the inclusion cannot induce an isomorphism in degree zero. Here $\bar{Z}=\{0\}$ is not contained in the empty interior of $A$, exposing the missing hypothesis.

Choose $a\in U\setminus V$ and $b\in V\setminus U$. On each of the $2^r$ dyadic subintervals of the parameter interval, define $\sigma$ linearly from $a$ to $b$ or from $b$ to $a$, alternating orientations so the pieces join continuously.

Every singular $1$-simplex occurring in $S^r\sigma$ has image containing both $a$ and $b$, hence lies in neither $U$ nor $V$. In degree one all these subdivision summands have coefficient $+1$ over $\mathbb{Z}$; coincident summands add rather than cancel. Thus a non-small term survives and $S^r\sigma$ is not cover-small, disproving uniformity.

The relative disk computation gives $H_n(D^n,S^{n-1};\mathbb{Z})\cong \mathbb{Z}$.

The set difference is the open ball, which is contractible, so its $n$th homology is zero. These unequal groups disprove the proposed identification.