(Theorem) The homology of the n-disk relative to its boundary is infinite cyclic for the nth group and 0 otherwise

Statement and proof

Homology of a disk relative to its boundary

$H_p(D^n,\partial D^n)\cong \{\begin{array}{ll}\mathbb{Z}& \text{if }p=n\\ 0& \text{otherwise}\end{array}$

Proof from Algebraic Topology (2025-03-25).   

Write ${\partial }_{+}{D}^n$ and ${\partial }_{-}{D}^n$ for the upper and lower hemispheres of the boundary $\partial D^n\cong S^{n-1}$, respectively. The computation of $H_p(D^n,\partial D^n)$ depends on the following isomorphisms:

• $D^n$ and ${\partial }_{-}{D}^n$ are both contractible, hence homotopy equivalent (to the one-point space), so the sequence of inclusions ${\partial }_{-}{D}^n\to \partial D^n\to D^n$ implies the connecting homomorphism $\delta :H_p(D^n,\partial D^n){\to }_{\cong }{H}_{p-1}(\partial D^n,{\partial }_{-}{D}^n)$ is an isomorphism for all $p$ and all $n\ge 1$.
• If $S$ is the “south pole”, excision implies the inclusion $(\partial D^n\backslash \{S\},{\partial }_{-}{D}^n\backslash \{S\})\to (\partial D^n,{\partial }_{-}{D}^n)$ induces an isomorphism $H_p(\partial D^n\backslash \{S\},{\partial }_{-}{D}^n\backslash \{S\}){\to }_{\cong }{H}_p(\partial D^n,{\partial }_{-}{D}^n).$
• There is a homeomorphism $\partial D^n\backslash \{S\}\to {\mathbb{R}}^{n-1}$ given by stereographic projection, which restricts to a homeomorphism ${\partial }_{-}{D}^n\backslash \{S\}\to {\mathbb{R}}^{n-1}\backslash \text{int}(D^{n-1})$. Thus this homeomorphism induces an isomorphism of groups $H_p(\partial D^n,{\partial }_{-}{D}^n){\to }_{\cong }{H}_p({\mathbb{R}}^{n-1},{\mathbb{R}}^{n-1}\backslash \text{int}(D^{n-1})).$
• Finally, homotopy invariance implies that the homotopy equivalence $(D^{n-1},\partial D^{n-1})\to ({\mathbb{R}}^{n-1},{\mathbb{R}}^{n-1}\backslash \text{int}(D^{n-1}))$ induces an isomorphism $H_{p-1}(D^{n-1},\partial D^{n-1}){\to }_{\cong }{H}_{p-1}({\mathbb{R}}^{n-1},{\mathbb{R}}^{n-1}\backslash \text{int}(D^{n-1})).$

Combining these gives an isomorphism

$$H_p(D^n,\partial D^n)\cong H_{p-1}(D^{n-1},\partial D^{n-1})$$

for all $p$ and all $n\ge 1$. By induction, we obtain

$$H_p(D^n,\partial D^n)\cong H_{p-n}(D^0,\partial D^0)=H_{p-n}(\{\ast \},\varnothing )=H_{p-n}(\{\ast \})$$ $$\cong \{\begin{array}{ll}\mathbb{Z}& \text{for }p=n\\ 0& \text{otherwise}\end{array}.$$

---

Applications

wip

Homology of the $n$-sphere

$S^n\cong S^m⟺n=m$.

${\mathbb{R}}^n\cong {\mathbb{R}}^m⟺n=m$.

Proof from Algebraic Topology (2025-03-25).  Suppose there exists a homeomorphism $h:{\mathbb{R}}^n\to {\mathbb{R}}^m$, and $h(0)=0$ without loss of generality. Then the restriction $h{|}_{{\mathbb{R}}^n\backslash \{0\}}:R^n\backslash \{0\}\to R^m\backslash \{0\}$ is also a homeomorphism, so it induces an isomorphism of relative homologies $h_{\ast }:H_p({\mathbb{R}}^n,R^n\backslash \{0\})\to H_p({\mathbb{R}}^m,R^m\backslash \{0\})$ which

• isomorphic for all $p$
• also can extract dimension
• Homeomorphism induces isomorphism on homology (why? singular homology of pairs is a functor?)