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.

With the two overlap circles ordered, the degree-zero and degree-one Mayer--Vietoris maps are both $(a,b)\mapsto (a+b,-a-b)$ from $G^2$ to $G^2$: each overlap cylinder retracts onto the second circle factor, and the fixed difference sign supplies the minus sign in the second coordinate. Thus each map has kernel and cokernel isomorphic to $G$.

Exactness gives $0\to G\to H_1(T^2;G)\to G\to 0$. This sequence splits: the first coordinate circle has connecting class $(g,-g)$ and hence gives a section of the right-hand $G$, while the second coordinate circle represents the left-hand cokernel generator. Therefore $H_1(T^2;G)\cong G\oplus G$.

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.