A singular $n$-cube in a topological space $X$ is a continuous map $T : I^n \to X (n \geq 0)$. A singular $n$-cube is degenerate if there exists an integer $i, 1 \leq i \leq n$, such that $T(x_1,x_2,...,x_n)$ does not depend on $x_i$.

$Q_n(x)$ denotes the free abelian group generated by the set of all singular $n$-cubes in $X$.
$D_n(x)$ denotes the subgroup of $Q_n(x)$ generated by the degenerate singular $n$-cubes.
$C_n(x)$ denotes the quotient group $Q_n(x)/D_n(x)$. This group is called the group of $n$-chains in $X$.

We can define face operators on $n$-cubes $A_i,B_i$; where given a singular $n$-cube represented by the continuous function $T(x_1,x_2,...,x_n)$, $A_i$ sets $x_i = 0$ and $B_i$ sets $x_i = 1$. We can now define the boundary operator $\partial_n : C_n(x) \to C_{n-1}(x)$, where $\partial_n$ defined on the generating elements of $C_n$: $$\partial_n(T) = \sum_{i = 1}^n (-1)^i [A_iT - B_iT]$$ We now define $$Z_n(X) = \text{kernel } \partial_n \qquad (n > 0)$$ $$B_n(X) = \text{image } \partial_{n+1} \qquad (n \geq 0)$$ $$H_n(X) = Z_n(X)/B_n(X) \qquad (n > 0)$$ Notice that $H_0(X)$ is not defined since $Z_0(X)$ is not defined. There are two canonical ways to complete our definition of $H_n(X)$:

Firstly, we could just define $Z_0(X) = C_0(X)$ and $H_0(X) = Z_0(X)/B_0(X)$.

Or, we could define a homomorphism $\epsilon : C_0(X) \to \mathbb{Z}$ where for a particular $0$-cube $T$, $\epsilon(T) = 1$ (i.e. $\epsilon$ is a function counting how many distinct points you have chosen in your formal sum). We now can define $\overset{\sim}{Z_0}(X) = \text{kernel } \epsilon$; and we are justified in defining $\overset{\sim}{H_0}(X) = \overset{\sim}{Z_0}(X)/B_0(X)$.

We now have the normal Homology group: $$H_0(X),H_1(X),H_2(X),...$$ and the reduced Homology group: $$\overset{\sim}{H_0}(X),\overset{\sim}{H_1}(X),\overset{\sim}{H_2}(X),...$$ where $\overset{\sim}{H_n}(X) = H_n(X)$ for $n > 0$.

Properties

The following sequence of groups and homomorphisms: $$0 \to \overset{\sim}{H_0}(X) \overset{{\xi}}{\to} H_0(X) \overset{\epsilon_*}{\to} \mathbb{Z} \to 0$$ is exact. Here, $\xi$ is an inclusion homomorphism, and $\epsilon_*$ is a homomorphism induced by $\epsilon$ (which was defined when we were defining $\overset{\sim}{Z_0}(X)$).
Then, we may identify $\overset{\sim}{H_0}(X)$ with the kernel of $\epsilon_*$.
Let $X$ be a nonempty arcwise-connected topological space. Then $\epsilon_* : H_0(X) \to \mathbb{Z}$ is an isomporphism, and $\overset{\sim}{H_0}(X) = \{0\}$.
Let $X_\gamma, \gamma \in \Gamma$, denote the set of arc components of the topological space $X$. Then the homology group $H_n(X)$ is naturally isomorphic to the direct sum of the groups $H_n(X_\gamma)$ for all $\gamma \in \Gamma$.
In particular, this implies that $H_0(X) \cong \mathbb{Z}^d$ where $d$ is the number of arc components of $X$.

Given a continuous map $f : X \to Y$, we can define a sequence of homomorphisms $f_* : H_n(X) \to H_n(Y)$ between their Homology groups in the following manner. First of all, we define homomorphisms $f_\#: Q_n(X) \to Q_n(Y)$ by the simple rule: $$f_\#(T) = f(T)$$ Firstly, we note that $f_\#(D_n(X)) \subset D_n(Y)$, i.e. it maps degenerate cubes into degenerate cubes, hence it induces a homomorphism: $$f_\# : C_n(X) \to C_n(Y)$$ Notice also that the following diagram is commutative for $n > 0$: $$ \require{AMScd} \begin{CD} Q_n(X) @>{f_\#}>> Q_n(Y)\\ @VV{\partial_n}V @VV{\partial_{n}}V \\ Q_{n-1}(X) @>{f_\#}>> Q_{n-1}(Y) \end{CD}$$ It follows that the following diagram is also commutative for $n > 0$: $$ \require{AMScd} \begin{CD} C_n(X) @>{f_\#}>> C_n(Y)\\ @VV{\partial_n}V @VV{\partial_{n}}V \\ C_{n-1}(X) @>{f_\#}>> C_{n-1}(Y) \end{CD}$$ By considering elements in both $Z_n(X)$ and $B_n(X)$, you can see that the commutativity of the above diagram implies that $f_\#$ maps $Z_n(X)$ to $Z_n(Y)$ and $B_n(X)$ to $B_n(Y)$ for all $n \geq 0$ and hence induces a homomorphism of quotient groups, denoted by: $$f_* : H_n(X) \to H_n(Y) ;\qquad n = 0, 1, 2, ..$$ It can also be seen that $f_*$ induces a homomorphism of $\overset{\sim}{H_0}(X)$ into $\overset{\sim}{H_0}(Y)$. We can see quite quickly that if $f$ is the identity map, then $f_*$ is the identity map on the Homology groups, and also that if we have maps $g : X \to Y$ and $f : Y \to Z $, with composition $fg : X \to Z$ then we have the identity: $$(fg)_* = f_*g_*$$

Properties

Let $f,g$ be continuous maps of $X$ into $Y$. If $f$ and $g$ are homotopic, then the induced homomorphisms, $f_*$ and $g_*$, of $H_n(X)$ into $H_n(Y)$ are the same. Also, $f_* = g_* : \overset{\sim}{H_0}(X) \to \overset{\sim}{H_0}(Y)$.
If $f : X \to Y$ is a homotopy equivalence, then $f_* : H_n(X) \to H_n(Y)$ for $n \geq 0$ and $f_* : \overset{\sim}{H_0}(X) \to \overset{\sim}{H_0}(Y)$ are isomorphisms.

Let $A$ be a subspace of $X$. Let $C_n(X,A) := C_n(X)/C_n(A)$ (i.e. we are now treating anything in $A$ as negligible). Notice that $\partial_n : C_n(X) \to C_{n-1}(X)$ has the property that it takes $C_n(A)$ to $C_{n-1}(A)$, hence induces a map $\partial_n : C_n(X,A) \to C_{n-1}(X,A)$, and hence we can define in the usual manner $Z_n(X,A)$ and $B_n(X,A)$ as kernels and images of this boundary operator respectively, and also we can define the relative Homology group as $H_n(X,A) = Z_n(X,A)/B_n(X,A)$.

Properties
The infinite sequence: $$\cdot \cdot \cdot \overset{j_*}{\to} H_{n+1}(X,A) \overset{\partial_*}{\to} H_n(A) \overset{i_*}{\to} H_n(X) \overset{j_*}{\to} H_n(X,A) \overset{\partial_*}{\to} \cdot \cdot \cdot$$ called the homology sequence of the pair $(X,A)$ is exact. Here, $j_*$ is the map induced by the quotient map, $i_*$ is the map induced by the inclusion map. To define $\partial_*$, for any $u \in H_n(X,A)$, pick a representative $u' \in C_n(X,A)$, and since $j_*$ is an epimorphism, pick $u'' \in C_n(X)$ such that $j_* [u''] = [u']$. By the commutativity of the following diagram: $$ \require{AMScd} \begin{CD} C_n(X) @>{j_\#}>> C_n(X,A)\\ @VV{\partial_n}V @VV{\partial_{n}}V \\ C_{n-1}(X) @>{i_\#}>> C_{n-1}(X,A) \end{CD}$$ and by noticing that $u'$ was picked to be a boundary-less cycle (since it represents something in $H_n(X,A)$), then $\partial_*(u'') = 0$ in $C_{n-1}(X,A)$, so it actually belongs to the subgroup $C_{n-1}(A)$ of $C_{n-1}(X)$. Hence we have started with $u \in H_n(X,A)$ and ended up with $[\partial_*(u'') \in H_n(A)$.

Intuitively, elements of $H_n(X,A)$ are like "closed n-dimensional manifolds that have no boundary if we ignore $A$, and are non-trivial in the sense that they bound some hole". If we take the boundary in the $(X,A)$ setting, then we end up with $0$ by definition. However, these objects could have some boundary in $A$ (that we just can't see, because we are ignoring $A$ in $H_n(X,A)$). Hence we have a map associating objects in $H_n(X,A)$ with their potential boundaries in $A$.

Also, if $A$ is nonempty, then we can replace $H_0(A)$ with $\overset{\sim}{H_0}(A)$ and replace $H_0(X)$ with $\overset{\sim}{H_0}(X)$, and the sequence remains exact.

Properties
Suppose we have two pairs $(X,A)$ and $(Y,B)$, and a map $f : X \to Y$ that 'respects' the subspaces, i.e. that satisfies $f(A) \subset B$. $f$ naturally induces a homomorphism $f_\# : C_n(X) \to C_n(Y)$, and since $f$ respects subspaces, it also induces a homomorphism of quotient groups $f_\# : C_n(X,A) \to C_n(Y,B)$. These induced homomorphisms commute with the boundary operators: $$ \require{AMScd} \begin{CD} C_n(X,A) @>{f_\#}>> C_n(Y,B)\\ @VV{\partial}V @VV{\partial}V \\ C_{n-1}(X,A) @>{f_\#}>> C_{n-1}(Y,B) \end{CD}$$ and hence it follows just as before that $f_\#$ induces a homomorphism $f_* : H_n(X,A) \to H_n(Y,B)$.

If we extend our notion of homotopic maps, by saying that two maps $f,g : (X,A) \to (Y,B)$ are homotopic if there exists a continuous map $F : (I \times X, I \times A$ \to (Y,B)\) such that $F(0,x) = f(x)$ and $F(1,x) = g(x)$ for all $x \in X$. (the point is that we require that $F(I \times A) \subset B$). This additional condition allows us to prove that the induced homomorphisms of these homotopic maps $f_*$ and $g_*$ of $H_n(X,A)$ into $H_n(Y,B)$ are the same.

Now, consider the effect of a map $f : (X,A) \to (Y,B)$ on the exact homology sequence of the pairs: $$ \require{AMScd} \begin{CD} \cdot\cdot\cdot @>>> H_n(A) @>>> H_n(X) @>>> H_n(X,A) @>>> H_{n-1}(A) @>>> \cdot\cdot\cdot\\ @. @VVV @VVV @VVV @VVV @.\\ \cdot\cdot\cdot @>>> H_n(B) @>>> H_n(Y) @>>> H_n(Y,B) @>>> H_{n-1}(B) @>>> \cdot\cdot\cdot \end{CD}$$ Each square in this diagram is commutative.

The Excision Property
Let $(X,A)$ be a pair, and let $W$ be a subset of $A$ such that $\bar{W}$ is contained in the interior of $A$. Then the inclusion map $(X-W,A-W) \to (X,A)$ induces an isomorphism of relative homology groups: $$H_n(X-W,A-W) \approx H_n(X,A), \qquad n = 0,1,2,...$$ i.e. under the given hypothesis, we can excise the set $W$ without affecting the relative homology groups.

Smallness
In proving the excision property above, it is crucial that we are able to restrict our consideration to singular cubes which are 'arbitrarily small'. We formalise this now. Let $\mathscr{U} = \{U_\lambda | \lambda \in \Lambda\}$ be a family of subsets of the topological space $X$ such that the interiors of the sets $U_\lambda$ cover $X$. A singular $n$-cube $T : I^n \to X$ is said to be small of order $\mathscr{U}$ if there exists an index $\lambda \in \Lambda$ such that $T(I^n) \in U_\lambda$. Then: $$H_n(X,A,\mathscr{U}) \approx H_n(X,A)$$ are isomorphic, where $H_n(X,A,\mathscr{U})$ is produced by only considering cubes which are small of order $\mathscr{U}$.