Definition:Homology Group

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Let $X$ be a topological space.

Let the standard n-simplex be denoted:

$\displaystyle \Delta^n := \left\{ {\left({x_0, \ldots, x_n}\right) \in {\R_{\ge 0} }^{n + 1}: \sum_{i \mathop = 1}^{n + 1} x_i = 1}\right\}$

Let $\mathcal C \left({\Delta^n, X}\right)$ be the set of continuous mappings from $\Delta^n$ to $X$.

For $n \ge 0$, let $C_n \left({X}\right)$ be the free abelian group generated by $\mathcal C \left({\Delta^n, X}\right)$.

Then there is a boundary map $\partial_n: C_n \left({X}\right) \to C_{n - 1} \left({X}\right)$ 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:

$s^i_n \left({x_0, \ldots, x_{n - 1} }\right) = \left({x_0, \ldots, x_{i - 1}, 0, x_i, \ldots, x_{n-1} }\right)$

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:

$\displaystyle \partial_n \left({\phi}\right) = \sum_{i \mathop = 0}^n \left({-1}\right)^i \phi \circ s^i_n$

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:

$0 \gets C_0 \left({X}\right) \stackrel {\partial_1} {\longleftarrow} C_1 \left({X}\right) \stackrel {\partial_2} {\longleftarrow} C_2 \left({X}\right) \stackrel {\partial_3} {\longleftarrow} \cdots$

are a chain complex.

This is demonstrated in Singular Chains form Chain Complex.

Let $\partial_0$ denote the map $C_0 \left({X}\right) \to 0$.

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


Let $B_n \left({X}\right) \subset C_n \left({X}\right)$ denote the image of $\partial_{n+1}$.

Let $Z_n \left({X}\right)$ denote the kernel of $\partial_n$.

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

$B_n \left({X}\right) \subseteq Z_n \left({X}\right)$

Then define:

$H_n \left({X}\right) = \dfrac {Z_n \left({X}\right)} {B_n \left({X}\right)}$

Also see