# Definition:Homology Group

## Definition

Let $X$ be a topological space.

Let the standard n-simplex be denoted:

$\ds \Delta^n := \set {\tuple {x_0, \ldots, x_n} \in {\R_{\ge 0} }^{n + 1}: \sum_{i \mathop = 1}^{n + 1} x_i = 1}$

Let $\map \CC {\Delta^n, X}$ be the set of continuous mappings from $\Delta^n$ to $X$.

For $n \ge 0$, let $\map {C_n} X$ be the free abelian group generated by $\map \CC {\Delta^n, X}$.

Then there is a boundary map $\partial_n: \map {C_n} X \to \map {C_{n - 1} } X$ defined as follows.

First, there are maps $s^i_n: \Delta^{n - 1} \to \Delta^n$, where $n > 0$ and $0 \le i \le n$, defined by:

$\map {s^i_n} {x_0, \ldots, x_{n - 1} } = \tuple {x_0, \ldots, x_{i - 1}, 0, x_i, \ldots, x_{n - 1} }$

These can be considered as the inclusion of $\Delta^{n - 1}$ as a 'face' of $\Delta^n$.

For a continuous function $\phi: \Delta^n \to X$ we define:

$\ds \map {\partial_n} \phi = \sum_{i \mathop = 0}^n \paren {-1}^i \phi \circ s^i_n$

This definition, along with the requirement that $\partial_n$ be a group homomorphism, uniquely specifies $\partial_n$.

In addition, one has $\partial_{n - 1} \partial_n = 0$, meaning that the sequence of groups and morphisms:

$0 \gets \map {C_0} X \stackrel {\partial_1} {\longleftarrow} \map {C_1} X \stackrel {\partial_2} {\longleftarrow} \map {C_2} X \stackrel {\partial_3} {\longleftarrow} \cdots$

are a chain complex.

This is demonstrated in Singular Chains form Chain Complex.

Let $\partial_0$ denote the map $\map {C_0} X \to 0$.

Thus the $n$th singular homology group of $X$ is defined as the $n$th homology group of this chain complex.

Explicitly:

Let $\map {B_n} X \subset \map {C_n} X$ denote the image of $\partial_{n+1}$.

Let $\map {Z_n} x$ denote the kernel of $\partial_n$.

Since $\partial_n \partial_{n + 1} = 0$:

$\map {B_n} X \subseteq \map {Z_n} X$

Then define:

$\map {H_n} X = \dfrac {\map {Z_n} X} {\map {B_n} X}$