# Relative homology

In algebraic topology, a branch of mathematics, the (singular) homology of a topological space relative to a subspace is a construction in singular homology, for pairs of spaces. The relative homology is useful and important in several ways. Intuitively, it helps determine what part of an absolute homology group comes from which subspace.

## Definition

Given a subspace ${\displaystyle A\subset X}$, one may form the short exact sequence

${\displaystyle 0\to C_{\bullet }(A)\to C_{\bullet }(X)\to C_{\bullet }(X)/C_{\bullet }(A)\to 0}$

where ${\displaystyle C_{\bullet }(X)}$ denotes the singular chains on the space X. The boundary map on ${\displaystyle C_{\bullet }(X)}$ leaves ${\displaystyle C_{\bullet }(A)}$ invariant and therefore descends to a boundary map on the quotient. The corresponding homology is called relative homology:

${\displaystyle H_{n}(X,A)=H_{n}(C_{\bullet }(X)/C_{\bullet }(A)).}$

One says that relative homology is given by the relative cycles, chains whose boundaries are chains on A, modulo the relative boundaries (chains that are homologous to a chain on A, i.e. chains that would be boundaries, modulo A again).

## Properties

The above short exact sequences specifying the relative chain groups gives rise to a chain complex of short exact sequences. An application of the snake lemma then yields a long exact sequence

${\displaystyle \cdots \to H_{n}(A)\to H_{n}(X)\to H_{n}(X,A){\stackrel {\delta }{\to }}H_{n-1}(A)\to \cdots .}$

The connecting map δ takes a relative cycle, representing a homology class in Hn(X, A), to its boundary (which is a cycle in A).

It follows that Hn(X, x0), where x0 is a point in X, is the n-th reduced homology group of X. In other words, Hi(X, x0) = Hi(X) for all i > 0. When i = 0, H0(X, x0) is the free module of one rank less than H0(X). The connected component containing x0 becomes trivial in relative homology.

The excision theorem says that removing a sufficiently nice subset ZA leaves the relative homology groups Hn(X, A) unchanged. Using the long exact sequence of pairs and the excision theorem, one can show that Hn(X, A) is the same as the n-th reduced homology groups of the quotient space X/A.

The n-th local homology group of a space X at a point x0 is defined to be Hn(X, X - {x0}). Informally, this is the "local" homology of X close to x0.

Relative homology readily extends to the triple (X, Y, Z) for ZYX.

One can define the Euler characteristic for a pair YX by

${\displaystyle \chi (X,Y)=\sum _{j=0}^{n}(-1)^{j}\;{\mbox{rank}}\;H_{j}(X,Y).}$

The exactness of the sequence implies that the Euler characteristic is additive, i.e. if ZYX, one has

${\displaystyle \chi (X,Z)=\chi (X,Y)+\chi (Y,Z).\,}$

## Functoriality

The map ${\displaystyle C_{\bullet }}$ can be considered to be a functor

${\displaystyle C_{\bullet }:{\mathbf {Top}}^{2}\to {\mathcal {CC}}}$

where Top2 is the category of pairs of topological spaces and ${\displaystyle {\mathcal {CC}}}$ is the category chain complexes of abelian groups.

## Examples

One important use of relative homology is the computation of the homology groups of quotient spaces ${\displaystyle X/A}$. In the case that ${\displaystyle A}$ is a subspace of ${\displaystyle X}$ fulfilling the mild regularity condition that there exists a neighborhood of ${\displaystyle A}$ that has ${\displaystyle A}$ as a deformation retract, then the group ${\displaystyle {\tilde {H}}_{n}(X/A)}$ is isomorphic to ${\displaystyle H_{n}(X,A)}$. We can immediately use this fact to compute the homology of a sphere. We can realize ${\displaystyle S^{n}}$ as the quotient of an n-disk by its boundary, i.e. ${\displaystyle S^{n}=D^{n}/S^{n-1}}$. Applying the exact sequence of relative homology gives the following:
${\displaystyle \cdots \to H_{n}(D^{n})\rightarrow H_{n}(D^{n},S^{n-1})\rightarrow H_{n-1}(S^{n-1})\rightarrow H_{n-1}(D^{n})\to \cdots .}$

Because the disk is contractible, we know its homology groups vanish in all dimensions, so the above sequence collapses to the short exact sequence:

${\displaystyle 0\rightarrow H_{n}(D^{n},S^{n-1})\rightarrow H_{n-1}(S^{n-1})\rightarrow 0.}$

Therefore, we get isomorphisms ${\displaystyle H_{n}(D^{n},S^{n-1})\cong H_{n-1}(S^{n-1})}$. We can now proceed by induction to show that ${\displaystyle H_{n}(D^{n},S^{n-1})\cong \mathbb {Z} }$. Now because ${\displaystyle S^{n-1}}$ is the deformation retract of a suitable neighborhood of itself in ${\displaystyle D^{n}}$, we get that ${\displaystyle H_{n}(D^{n},S^{n-1})\cong H_{n}(S^{n})\cong \mathbb {Z} .}$