\section{Simplicial and singular homology}

\begin{cislo}\label{HOexact}{\bf Exact sequences.}
A sequence of homomorphisms of Abelian groups or modules over a
ring
$$\dots\xrightarrow{\,f_{n+1}\,}
A_n\xrightarrow{\ f_n\ }A_{n-1}\xrightarrow{\,f_{n-1}\,} A_{n-2}
\xrightarrow{\,f_{n-2}\,}\dots$$
is called an \emph{exact sequence}\index{exact sequence} if
$$\im f_n=\ker f_{n-1}.$$

Exactness of the following sequences
$$O\xrightarrow{\ \ } A\xrightarrow{\ f\ }B,\quad B\xrightarrow{\ g\ } C\xrightarrow{\ \ } 0, \quad 0\xrightarrow{\ \ }
C\xrightarrow{\ h\ } D\xrightarrow{\ \ } 0$$
means that $f$ is a monomorphism, $g$ is an epimorphism and $h$ is
an isomorphism, respectively.

A \emph{short exact sequence}\index{short exact sequence} is an
exact sequence
$$0\xrightarrow{\ \ } A\xrightarrow{\ i\ } B\xrightarrow{\ j\ } C\xrightarrow{\ \ }0.$$
In this case $C\cong B/A$. We say that the short exact sequence
splits if one of the following three equivalent conditions
is satisfied:
\begin{enumerate}
\item There is a homomorphism $p:B\to A$ such that $p
i=\id_{A}$.
\item There is a homomorphism $q:C\to B$ such that $j
q=\id_{C}$.
\item There are  homomorphisms $p:B\to A$ and $q:C\to B$ such that
$i p+q j=\id_{B}$.
\end{enumerate}
The last condition means that $B\cong A\oplus C$ with isomorphism
$(p,q):B\to A\oplus C$.

\begin{ex*} Prove the equivalence of (1), (2) and (3).
\end{ex*}
\end{cislo}

\begin{cislo}\label{HOchain}{\bf Chain complexes.}
The \emph{chain complex}\index{chain complex}
$(C,\partial)$ is a  sequence of
Abelian groups (or modules over a ring) and their homomorphisms
indexed by integers
$$\dots\xrightarrow{\partial_{n+2}} C_{n+1}\xrightarrow{\partial_{n+1}}
C_n\xrightarrow{\ \partial_{n}\ }
C_{n-1}\xrightarrow{\partial_{n-1}}\dots$$
such that
$$\partial_{n-1}\partial_n=0.$$
This conditions means that
$\im\partial_n\subseteq\ker\partial_{n-1}$. The homomorphism
$\partial_n$ is called a
boundary operator. A \emph{chain homomorphism}\index{chain homomorphism}
of chain complexes $(C,\partial^C)$ and $(D,\partial^D)$  is a
sequence of homomorphisms of Abelian groups (or modules over a
ring) $f_n:C_n\to D_n$ which commute with the boundary operators
$$\partial^D_nf_{n}=f_{n-1}\partial^C_n.$$
\end{cislo}

\begin{cislo}\label{HOhomology}{\bf Homology of chain complexes.}
The $n$-th \emph{homology group}\index{homology group of a chain
complex} of the chain complex $(C,\partial)$ is the group
$$H_{n}(C)=\frac{\ker\partial_n}{\im\partial_{n+1}}.$$
The elements of $\ker\partial_n=Z_n$ are called
\emph{cycles}\index{cycle} of dimension $n$ and the elements
of $\im\partial_{n+1}=B_n$ are called
\emph{boundaries}\index{boundary} (of dimension $n$).
If a chain complex is exact, then its homology groups are trivial.

The component $f_n$ of the chain homomorphism $f:(C,\partial^C)\to
(D,\partial^D)$ maps cycles into cycles and
boundaries into boundaries. It enables us to define
$$H_n(f):H_{n}(C)\to H_n(D)$$
by the prescription $H_n(f)[c]=[f_n(c)]$ where $[c]\in H_n(C_*)$
and $[f_n(c)]\in H_n(D^*)$ are classes  represented by the
elements $c\in Z_n(C)$ and $f_n(c)\in Z_n(D)$, respectively.
\end{cislo}

\begin{cislo}\label{HOlesch}{\bf Long exact sequence in homology.}
A sequence of chain homomorphisms
$$\dots\xrightarrow{\ \ }
A\xrightarrow{\ f\ }B\xrightarrow{\ g\ }C\xrightarrow{\ \ }\dots$$
is exact if for every $n\in\mathbb Z$
$$\dots\xrightarrow{\ \ }
A_n\xrightarrow{\ f_n\ }B_n\xrightarrow{\ g_n\ }C_n\xrightarrow{\ \ }\dots$$
is an exact sequence of Abelian groups.

\begin{thm*} Let $0\to A\xrightarrow{i}B\xrightarrow{j}C\to 0$
be a short exact sequence of chain complexes. Then there is a
\emph{connecting homomorphism}\index{connecting
homomorphism} $\partial_*:H_n(C)\to H_{n-1}(A)$ such that the
sequence
$$\dots\xrightarrow{\ \partial_*\ } H_{n}(A)\xrightarrow{\,H_{n}(i)\,} H_n(B)\xrightarrow{\,H_n(j)\,}
H_{n}(C)\xrightarrow{\ \partial_*\ }H_{n-1}(A)\xrightarrow{\,H_{n-1}(i)\,}
\dots$$
is exact.
\end{thm*}

\begin{proof} Define the connecting homomorphism
$\partial_*$. Let $[c]\in H_n(C)$ where $c\in C_n$ is a cycle.
Since $j:B_n\to C_n$ is an epimorphism, there is $b\in B_n$ such
that $j(b)=c$. Further, $j(\partial b)=\partial j(b)=\partial
c=0$. From exactness there is $a\in A_{n-1}$ such that
$i(a)=\partial b$. Since $i(\partial a)=\partial
i(a)=\partial\partial b=0$ and $i$ is a monomorphism, $\partial
a=0$ and $a$ is a cycle in $A_{n-1}$. Put
$$\partial_*[c]=[a].$$
Now we have to show that the definition is correct, i.~e.
independent of the choice of $c$ and $b$, and to prove exactness.
For this purpose it is advantageous to use an appropriate
diagram. It is not difficult and we leave it as an exercise to the reader.
\end{proof}

\begin{cislo}\label{HOchhmtp}{\bf Chain homotopy.} Let $f,g:C\to
D$ be two chain homomorphisms. We say that they are \emph{chain
homotopic}\index{chain homotopic} if there are homomorphisms
$s_n:C_n\to D_{n+1}$ such that
$$\partial^D_{n+1}s_n+s_{n-1}\partial^C_{n}=f_n-g_n \quad
\text{for all }n.$$
The relation to be chain homotopic is an equivalence.
The sequence of maps $s_n$ is called a \emph{chain
homotopy}.\index{chain homotopy}
\end{cislo}

\begin{thm*} If two chain homomorphism $f,g:C\to D$ are chain
homotopic, then
$$H_n(f)=H_n(g).$$
\end{thm*}

\begin{ex*}  Prove the previous theorem from the definitions.
\end{ex*}
\end{cislo}

\begin{cislo}\label{HO5lemma}{\bf Five Lemma.} Consider the
diagram
$$
\xymatrix{
A \ar[r] \ar[d]_{f_1}^{\cong} & B \ar[r] \ar[d]_{f_2}^{\cong}&
C \ar[r] \ar[d]_{f_3}&D \ar[r] \ar[d]_{f_4}^{\cong}&E
\ar[d]_{f_5}^{\cong}   \\
\bar A \ar[r] & \bar B \ar[r] &
\bar C \ar[r] &\bar D \ar[r] &\bar E
}
$$
If the horizontal sequences are exact and $f_1$, $f_2$, $f_{4}$
and $f_{5}$ are isomorphisms, then $f_3$ is also an isomorphism.

\begin{ex*} Prove 5-lemma. 
\end{ex*}
\end{cislo}

\begin{cislo}\label{HOsimpl}{\bf Simplicial homology.} We
describe two basic ways how to define homology groups for topological spaces
-- simplicial homology which is closer to geometric intuition and singular
homology which is more general. For the definition of simplicial
homology we need the notion of $\Delta$-complex, which is a special
case of CW-complex.

Let $v_0,v_1,\dots,v_n$ be points in $\mathbb R^m$ such that
$v_1-v_0,v_2-v_0, v_n-v_0$ are linearly independent. The
$n$-\emph{simplex}\index{$n$-simplex} $[v_0,v_1,\dots,v_n]$ with
the vertices $v_0,v_1,\dots, v_n$ is the subspace of
$\mathbb R^m$
$$\{\sum_{i=0}^n t_iv_i;\ \sum_{i=1}^nt_i=1,\ t_i\ge 0\}$$
with a given ordering of vertices. A \emph{face} of this simplex
is any simplex determined by a proper subset of vertices in
the given ordering.

Let $\Delta_{\alpha}$, $\alpha\in J$ be a collection of simplices.
Subdivide all their faces of dimension $i$ into sets $F^i_{\beta}$. A
$\Delta$-\emph{complex}\index{$\Delta$-complex} is a quotient
space of disjoint union $\coprod_{\alpha\in J} \Delta_{\alpha}$ obtained by
identifying simplices from every $F^i_{\beta}$ into one single simplex via
affine maps which preserve the ordering of vertices. Thus every
$\Delta$-complex is determined only by combinatorial data.

A special case of $\Delta$-complex is a \emph{finite simplicial
complex}.\index{finite simplicial complex}\index{simplicial
complex} It is a union of
simplices the vertices of which lie in a given finite set of
points $\{v_0,v_1,\dots,v_n\}$ in $\mathbb R^m$ such that
$v_1-v_0,v_2-v_0,\dots, v_n-v_0$ are linearly independent.

\begin{example*} Torus, real projective space of dimension 2
and Klein bottle are $\Delta$-complexes as one can see from the
following pictures.

%\begin{figure}[H]
%\centering
%\includegraphics[height=4cm,width=15cm]{fg3.1.eps}%
%\caption{}\label{fg3.1}
%\end{figure}

\begin{figure}[htb]
  \centering
  \def\svgwidth{15cm}
  \input{img/fig-3_1.pdf_tex}
  \caption{Torus, $\mathbb{R}P^2$ and Klein bottle as $\Delta$-complexes}
\end{figure}

In all the cases we have two sets $F^2$ whose elements are
triangles, three sets $F^1$ every with two segments and one set
$F^{0}$ containing all six vertices of both triangles.

These surfaces are also homeomorhic to finite simplicial
complexes, but their structure as simplicial complexes is more
complicated than their structure as $\Delta$-complexes.
\end{example*}

To every $\Delta$-complex $X$ we can assign the  chain complex
$(C,\partial)$ where $C_{n}(X)$ is a free Abelian group generated by
$n$-simplices of $X$ (i.~e. the rank of $C_n(X)$ is the number of
the sets $F^n$ and the boundary operator on generators is given
by
$$\partial[v_0,v_1,\dots,v_n]=\sum_{i=0}^n(-1)^i[v_0,\dots,\hat
v_i\dots,v_n].$$
Here the symbol $\hat v_i$ means that the vertex $v_i$ is
omitted. Prove that $\partial\partial=0$.

The \emph{simplicial homology groups}\index{simplicial homology}
of $\Delta$-complex $X$ are the homology groups of the chain
complex defined above. Later, we will show that these groups
are independent of $\Delta$-complex structure.


\begin{ex*} Compute simplicial homology of $S^2$ (find a
$\Delta$-complex structure),
$\mathbb{RP}^2$, torus and Klein bottle (with $\Delta$-complex
structures given in example above).
\end{ex*}

Let $X$ and $Y$ be two $\Delta$-complexes and $f:X\to Y$ a map
which maps every simplex of $X$ into a simplex of $Y$ and it is
affine on all simplexes. Using appropriate sign conventions
we can define the chain homomorphism $f_n:C_n(X)\to C_{n}(Y)$
induced by the map $f$. This chain map enables us to define
homomorphism of simplicial homology groups induced by $f$.

Having a $\Delta$-subcomplex $A$ of a $\Delta$-complex $X$
(i.~e. subspace of $X$ formed by some of the simplices of $X$)
we can define simplicial homology groups $H_n(X,A)$. The
definition is the same as for singular homology in
paragraph \ref{HOsing2}. These groups fit into the long exact sequence
$$\dots\to H_{n}(A)\to H_{n}(X)\to H_n(X,A)\to H_{n-1}(A)\to\dots$$
See again \ref{HOsing2}.
\end{cislo}

\begin{cislo}\label{HOsing}{\bf Singular homology.} The \emph{standard}
$n$-\emph{simplex}\index{standard $n$-simplex} is the $n$-simplex
$$\Delta^n=\{(t_0,t_1,\dots,t_n)\in \mathbb R^{n+1};\
\sum_{i=0}^nt_i=1;\ t_i\ge 0\}.$$
The $j$-th face of this standard simplex is the $(n-1)$-dimensional
simplex $[e_0,\dots,\hat e_j,\dots,e_n]$ where $e_j$ is the
vertex with all coordinates $0$ with the exception of the $j$-th
one which is $1$. Define
$$\varepsilon_n^j:\Delta^{n-1}\to\Delta^n$$
as the affine map
$\varepsilon_n^j(t_0,t_1,\dots,t_{n-1})=(t_0,\dots,t_{j-1},0,t_j,
\dots, t_{n-1})$
which maps
$$e_0\to e_0,\ \dots,\ e_{j-1}\to
e_{j-1},\ e_{j}\to e_{j+1},\ \dots,\ e_{n-1}\to e_{n}.$$
It is not difficult to prove
\begin{lemma*}
$\varepsilon_{n+1}^k\varepsilon_n^{j}=\varepsilon_{n+1}^{j+1}
\varepsilon_n^{k}$ for $k<j$.
\end{lemma*}

A \emph{singular} $n$-\emph{simplex}\index{singular $n$-simplex}
in a space $X$ is a continuous map $\sigma:\Delta^n\to X$.
Denote the free Abelian group generated by all the singular
$n$-simplices by $C_n(X)$  and define the boundary operator
$\partial_n:C_{n}(X)\to C_{n-1}(X)$ by
$$\partial_{n}(\sigma)=\sum_{i=0}^n(-1)^i\sigma
\varepsilon_{n}^{i}$$
for $n\ge 0$. Put $C_n(X)=0$ for $n<0$.
Using the lemma above one can show that
$$\partial_{n+1}\partial_{n}=0.$$
The chain complex $(C_n,\partial_n)$ is called the
\emph{singular chain complex} of the space $X$. The \emph{singular homology
groups}\index{singular homology groups} $H_n(X)$ of the space $X$ are the
homology groups of the chain complex $(C_n(X),\partial_n)$, i.~e.
$$H_n(X)=\frac{\ker\partial_n}{\im\partial_{n+1}}.$$

Next consider a map $f:X\to Y$. Define the chain homomorhism
$C_n(f):C_n(X)\to C_{n}(Y)$  on singular $n$-simplices as the
composition
$$C_n(f)(\sigma)=f\sigma.$$
From definitions it is easy to show that these homomorphisms commute
with boundary operators. Hence this chain homomorphism induces
homomorphisms
$$f_*=H_n(f):H_n(X)\to H_{n}(Y).$$

Moreover, $H_n(\id_X)=\id_{H_n(X)}$ and $H_n(f g)=H_n(f)
H_n(g)$. It means that $H_n$ is a functor from the category $\top$
to the category $\ab$ of Abelian groups and their homomorphisms.
This functor is the composition of the functor $C$
from $\top$ to chain complexes and the $n$-th homology functor
from chain complexes to abelian groups.

\begin{theex} Prove the lemma above and $\partial_{n+1}\partial_{n}=0$.
\end{theex}

\begin{theex} Show directly from the definition that the
singular homology groups of a point are $H_0(*)=\mathbb Z$
and $H_n(*)=0$ for $n\ne 0$.
\end{theex}
\end{cislo}

\begin{cislo}\label{HOsing2}{\bf Singular homology groups of a
pair.} Consider a pair of topological spaces $(X,A)$. Then
the $C_n(A)$ is a subgroup of $C_n(X)$. Hence we get this short
exact sequence
$$0\to C_n(A)\xrightarrow{i} C_n(X)\xrightarrow{j} \frac{C_n(X)}
{C_n(A)}\to 0.$$
Since the boundary operators in $C_n(A)$ are
restrictions of boundary operators in $C_n(X)$, we can define
boundary operators
$$\partial_n:\frac{C_n(X)}{C_n(A)}\to
\frac{C_{n-1}(X)}{C_{n-1}(A)}.$$
We will denote this chain complex as $(C(X,A),\partial)$ and
its homology groups as $H_n(X,A)$. Notice that the factor
$C_n(X)/C_{n}(A)$ is a free Abelian group generated by singular
simplices $\sigma:\Delta^n\to X$ such that
$\sigma(\Delta^n)\nsubseteq A$. We will need it later.

A map $f:(X,A)\to(Y,B)$ induces the chain homomorphism
$C_n(f):C_n(X)\to C_n(Y)$ which restricts to a chain
homomorphism $C_n(A)\to C_n(B)$ since $f(A)\subseteq B$.
Hence we can define the chain homomorphism
$$C_n(f):C_n(X,A)\to C_n(Y,B)$$
which in homology induces the homomorphism
$$f_*=H_n(f):H_n(X,A)\to H_n(Y,B).$$
We can again conclude that $H_n$ is a functor from the category
$\top^2$ into the category $\ab$ of Abelian groups. This functor
extends the functor defined on the category $\top$ since every
object $X$ and every morphism $f:X\to Y$ in $\top$ can be
considered as the object $(X,\emptyset)$ and the morphism
$\hat f=f:(X,\emptyset)\to(Y,\emptyset)$ in the category $\top^2$
and
$$H_n(X,\emptyset)=H_n(X),\quad H_n(\hat f)=H_n(f).$$
\end{cislo}

\begin{cislo}\label{HOles}{\bf Long exact sequence for singular
homology.} Consider inclusions of spaces $i:A\to X$, $i':B\to Y$
and maps $j:(X,\emptyset)\to (X,A)$,
$j':(Y,\emptyset)\to (Y,B)$ induced by
$\id_{X}$ and $\id_Y$, respectively.
Let $f:(X,A)\to (Y,B)$ be a map. Then there are connecting
homomorphisms $\partial^X_*$ and $\partial_*^Y$ such that the following
diagram
$$
\xymatrix{
{ } \ar[r]^-{\partial^X_*}  & H_n(A) \ar[r]^{i_*} \ar[d]^{(f/A)_*}&
H_n(X^1) \ar[r]^{j_*} \ar[d]^{f_*}&H_n(X,A)
\ar[r]^-{\partial^X_*}
\ar[d]^{f_*}& H_{n-1}(A) \ar[r]^-{i_*} \ar[d]^{(f/A)_*}&{ }   \\
{ } \ar[r]^-{\partial^Y_*}  & H_n(B) \ar[r]^{i'_*} &
H_n(Y) \ar[r]^{j'_*} &H_n(Y,B) \ar[r]^-{\partial^Y_*}
&H_{n-1}(B) \ar[r]^-{i'_*}&{ }
}
$$
commutes and its horizontal sequences are exact.

An analogous theorem holds also for simplicial homology.

\begin{remark*} Consider the functor $I:\top^2\to\top^2$ which
assigns to every pair $(X,A)$ the pair $(A,\emptyset)$. The
commutativity of the last square in the diagram above means
that $\partial_*$ is a natural transformation of functors
$H_n$ and $H_{n-1}\circ I$ defined on $\top^2$.
\end{remark*}

\begin{proof} We have the following commutative diagram of chain
complexes
$$
\xymatrix{
0 \ar[r]  & C(A) \ar[r]^{C(i)} \ar[d]^{C(f/A)}&
C(X) \ar[r]^{C(j)} \ar[d]^{C(f)}&C(X,A)
\ar[r] \ar[d]^{C(f)} &0   \\
0 \ar[r]  & C(B) \ar[r]^{C(i')} &
C(Y) \ar[r]^{C(j')} &C(Y,B)
\ar[r] &0
}
$$
with exact horizontal rows. Then Theorem \ref{HOlesch} and the
construction of connecting homomorphism $\partial_*$ imply
the required statement.
\end{proof}

\begin{remark*} It is useful to realize how
$\partial_*:H_n(X,A)\to H_{n-1}(A)$ is defined. Every element of
$H_{n}(X,A)$ is represented by a chain $x\in C_n(X)$ with a
boundary $\partial x\in C_{n-1}(A)$. This is a cycle in $C_n(A)$
and from the definition in \ref{HOlesch} we have
$$\partial_*[x]=[\partial x].$$
\end{remark*}
\end{cislo}

\begin{cislo}\label{HOhmtp}{\bf Homotopy invariance.}
If two maps $f,g:(X,A)\to (Y,B)$ are homotopic, then they
induce the same homomorphisms
$$f_*=g_*:H_n(X,A)\to H_n(Y,B).$$

\begin{proof} We need to prove that the homotopy between $f$ and $g$
induces a chain homotopy between $C_*(f)$ and $C_*(g)$. For
the proof see [Hatcher],  %\cite{Ha},  
Theorem 2.10 and Proposition 2.19 or [Spanier], %\cite{Sp}, 
Chapter 4, Section 4.
\end{proof}

\begin{cor*} If $X$ and $Y$ are homotopy equivalent spaces, then
$$H_n(X)\cong H_n(Y).$$
\end{cor*}
\end{cislo}

\begin{cislo}\label{HOet}{\bf Excision Theorem.} There are two
equivalent versions of this theorem.

\begin{thm*}[Excision Theorem, 1st version] Consider spaces
$C\subseteq A\subseteq X$ and suppose  that
$\bar C\subseteq \inte A$. Then the inclusion
$$i:(X-C,A-C)\hookrightarrow (X,A)$$
induces the isomorphism
$$i_*:H_n(X-C,A-C)\xrightarrow{\cong} H_n(X,A).$$
\end{thm*}

\begin{thm*}[Excision Theorem, 2nd version] Consider two
subspaces $A$ and $B$ of a space $X$. Suppose that
$X=\inte A\cup\inte B$. Then the inclusion
$$i:(B,A\cap B)\hookrightarrow (X,A)$$
induces the isomorphism
$$i_*:H_n(B,A\cap B)\xrightarrow{\cong} H_n(X,A).$$
\end{thm*}

The second version of Excision Theorem holds also
for simplicial homology if we suppose that
$A$ and $B$ are $\Delta$-subcomplexes of a $\Delta$-complex
$X$ and $X=A\cup B$.
In this case the proof is easy since the inclusion
$$C_n(i):C_n(B,A\cap B)\to C_n(A\cup B,A)$$
is an isomorphism, namely the both chain complexes are generated by the same
$n$-simplices.

\begin{ex*} Show that the theorems above are equivalent.
\end{ex*}

The proof of Excision Theorem for singular homology can be found
in [Hatcher], %\cite{Ha}, 
pages 119 -- 124, or in [Spanier], %\cite{Sp}, 
Chapter 4, Sections 4 and 6.
The main step (a little bit technical for beginners)
is to prove the following lemma which we will need later.

\begin{lemma*} Let $\mathcal{U}=\{U_{\alpha};\ \alpha\in J\}$ be
a collection of subsets of $X$ such that $X=\bigcup_{\alpha\in J}
\inte U_{\alpha}$. Denote the free chain complex generated by
singular simplices $\sigma$ with $\sigma(\Delta^n)\in
U_{\alpha}$ for some $\alpha$ as $C_{n}^{\mathcal{U}}(X)$.
Then
        $$C_n^{\mathcal{U}}(X))\hookrightarrow C_{n}(X)$$
induces isomorphism in homology.
\end{lemma*}

\begin{proof} [Proof of Excision Theorem]
Consider $\mathcal{U}=\{A,B\}$. Then
the inclusion
$$C_n(i):C_n(B,A\cap B)\to \frac{C_n^{\mathcal{U}}(X)}{C_n(A)}$$
is an isomorphism and, moreover, according to the previous lemma,
the homology of the second chain complex is $H_n(X,A)$.
\end{proof}

\end{cislo}

\begin{cislo}\label{HOsum}{\bf Homology of disjoint union.}
Let $X=\coprod_{\alpha\in J} X_{\alpha}$ be a disjoint union.
Then
$$H_n(X)=\bigoplus_{\alpha\in J}H_n(X_{\alpha}).$$

The proof follows from the definition and connectivity of
$\sigma(\Delta^n)$ in $X$ for every singular $n$-simplex
$\sigma$.
\end{cislo}

\begin{cislo}\label{HOred}{\bf Reduced homology groups.} For
every space $X\ne\emptyset$ we define the \emph{augmented chain complex}
\index{augmented chain complex}
$(\tilde C(X),\tilde\partial)$ as follows
$$
\tilde C_n(X)=
\begin{cases}
C_n(X)\quad & \text{for }n\ne -1,\\
\mathbb Z\quad &
 \text{for }n=-1.
\end{cases}
$$
with $\tilde\partial_n=\partial_n$ for $n\ne 0$ and
$\partial_{0}
(\sum_{i=1}^{j}n_i\sigma_i)=\sum_{i=1}^j
n_i$.
The \emph{reduced homology groups}\index{reduced homology groups}
$\tilde H_n(X)$ are the homology groups of the augmented chain
complex. From the definition it is clear that
$$\tilde H_n(X)=H_n(X)\quad \text{for }n\ne 0$$
and
$$\tilde H_n(*)=0\quad\text{for all }n.$$

For pairs of spaces we define $\tilde H_n(X,A)=H_n(X,A)$ for all
$n$. Then theorems on long exact sequence, homotopy invariance
and excision hold for reduced homology groups as well.

Considering a space $X$ with distinguished point $*$ and applying
the long exact sequence for the pair $(X,*)$, we get that for all
$n$
$$\tilde H_n(X)=\tilde H_{n}(X,*)=H_{n}(X,*).$$
Using this equality and the long exact sequence for unreduced
homology we get that
$$H_0(X)\cong H_0(X,*)\oplus H_0(*)\cong \tilde H_0(X)\oplus
\mathbb Z.$$

\begin{lemma*} Let $(X,A)$ be a pair of CW-complexes,
$X\ne\emptyset$. Then
$$\tilde H_n(X/A)=H_n(X,A)$$
and we have the long exact sequence
$$\dots\to \tilde H_n(A)\to \tilde H_n(X)\to \tilde H_n(X/A)\to \tilde
H_{n-1}(A)\to\dots$$
\end{lemma*}

\begin{proof} According to example in Section 2 %\ref{CW2crit}
$$(X,A)\to (X\cup CA,CA)\to (X\cup CA/CA,*)=(X/A,*)$$
is the composition of an excision and a homotopy equivalence.
Hence $\tilde H_n(X/A)=H_{n}(X,A)$.
The rest folows from the long exact sequence of the pair $(X,A)$.
\end{proof}

\begin{ex*} Prove that $\tilde H_n(\bigvee X_{\alpha})\cong
\oplus \tilde H_n(X_{\alpha})$.
\end{ex*}

$\tilde H_n$ can be considered as a functor from $\top_*$ to
Abelian groups.
\end{cislo}

\begin{cislo}\label{HOtriple}{\bf The long exact sequence of a
triple.} Three spaces $(X,B,A)$ with the property
$A\subseteq B\subseteq X$ are called a \emph{triple}\index{triple}.
Denote $i:(B,A)\to (X,A)$ and $j:(X,A)\to (X,B)$ maps induced by
the inclusion $B\hookrightarrow X$ and $\id_X$, respectively.
Analogously as for pairs one can derive the following long exact
sequence:
$$\dots\xrightarrow{\partial_*}
H_n(B,A)\xrightarrow{i_*} H_n(X,A)\xrightarrow{j_*}
H_{n}(X,B)\xrightarrow{\partial_*}H_{n-1}(B,A)\xrightarrow{i_*}\dots$$
\end{cislo}

\begin{cislo}\label{HOsphere}{\bf Singular homology groups of
spheres.} Consider the long exact sequence of the triple
$(\Delta^n,\partial\Delta^n,\Lambda^{n-1}=\partial\Delta^n-\Delta^{n-1})$:
$$\dots\to
H_{i}(\Delta^n,\Lambda^{n-1})\to H_{i}(\Delta^n,\partial\Delta^n)
\xrightarrow{\partial*}
H_{i-1}(\partial\Delta^n,\Lambda^{n-1})\to H_{i-1}(\Delta^n,\Lambda^{n-1})\to\dots$$
The pair $(\Delta^n,\Lambda^{n-1})$ is homotopy equivalent to $(*,*)$
and hence its homology groups are zeroes. Next using Excision
Theorem and homotopy invariance we get that
$H_i(\Delta^{n},\Lambda^{n-1})\cong H_i(\Delta^{n-1},\partial
\Delta^{n-1})$. Consequently, we get an isomorphism
$$H_i(\Delta^n,\partial\Delta^n)\cong
H_{i-1}(\Delta^{n-1},\partial\Delta^{n-1}).$$
Using induction and computing $H_i(\Delta^1,\partial\Delta^1)=
H_i([0,1],\{0,1\})\cong H_{i-1}(\{0,1\},\{0\})$
we get that
$$H_{i}(\Delta^n,\partial\Delta^n)=
\begin{cases}
\mathbb Z\quad& \text{for }i=n,\\
0\quad& \text{for }i\ne n.
\end{cases}$$
Doing the induction carefully we can find that the generator of
the group $H_{n}(\Delta^n,\partial\Delta^n)=\mathbb Z$ is
determined by the singular $n$-simplex $\id_{\Delta^n}$.

The pair $(D^n,S^{n-1})$ is homeomorphic to
$(\Delta^n,\partial\Delta^n)$. Hence it has the same homology
groups. Using the long exact sequence for this pair we obtain
$$\tilde H_{i-1}(S_{n-1})=H_{i}(D^n,S^{n-1})=
\begin{cases} 0\quad & \text{for }i\ne n,\\
\mathbb Z\quad& \text{for }i=n.
\end{cases}$$
\end{cislo}

\begin{cislo}\label{HOmves}{\bf Mayer-Vietoris exact sequence.}
\index{Mayer-Vietoris exact sequence}
Denote inclusions $A\cap B\hookrightarrow A$,
$A\cap B\hookrightarrow B$, $A\hookrightarrow X$, $B\hookrightarrow X$
by $i_A$, $i_B$, $j_A$, $j_B$, respectively. Let $C\hookrightarrow
A\cap B$ and suppose that $X=\inte A\cup\inte B$. Then the
following sequence
\begin{multline*}
\dots\xrightarrow{\ \partial_*\ }
H_n(A\cap B,C)\xrightarrow{(i_{A*},i_{B*})} H_n(A,C)\oplus
H_n(B,C)\\
\xrightarrow{j_{A*}-j_{B*}}
H_{n}(X,C)\xrightarrow{\ \partial_*\ }H_{n-1}(A\cap
B,C)\xrightarrow{\ \ }\dots
\end{multline*}
is exact.

\begin{proof} The covering $\mathcal{U}=\{A,B\}$ satisfies conditions of Lemma
\ref{HOet}. The sequence of chain complexes
$$0\longrightarrow \frac{C(A\cap
B)}{C(C)}\overset{i}\longrightarrow\frac{C(A)}{C(C)}\oplus\frac{C(B)}{C(c)}
\overset{j}\longrightarrow\frac{C^{\mathcal{U}}(X)}{C(C)}\longrightarrow 0$$
where $i(x)=(x,x)$ and $j(x,y)=x-y$ is exact. Consequently, it
induces a long exact sequence. Using Lemma \ref{HOet} we get
that $H_n(C^{\mathcal{U}}(X),C(C))=H_n(X,C)$, which completes the
proof.
\end{proof}
\end{cislo}

\begin{cislo}\label{HOeq}{\bf Equality of simplicial and singular
homology.} Let $(X,A)$ be a pair of $\Delta$-complexes. Then the natural
inclusion of simplicial and singular chain complexes
$C^{\Delta}(X,A)\hookrightarrow C(X,A)$ induces the isomorphism of simplicial
and singular homology groups
$$H_n^{\Delta}(X,A)\cong H_{n}(X,A).$$
\end{cislo}

\begin{proof}[Outline of the proof] Consider the long exact
sequences for the pair $(X^k,X^{k-1})$ of skeletons of $X$.
We get
$$
\xymatrix{
H_{n+1}^{\Delta}(X^k,X^{k-1}) \ar[r] \ar[d]  & H_n^{\Delta}(X^{k-1}) \ar[r]
\ar[d]& H_n^{\Delta}(X^k) \ar[r] \ar[d]&H_n^{\Delta}(X^k,X^{k-1})
\ar[r] \ar[d]& H_{n-1}^{\Delta}(X^{k-1})  \ar[d] \\
H_{n+1}(X^k,X^{k-1}) \ar[r]  & H_n(X^{k-1}) \ar[r] &
H_n(X^k) \ar[r] &H_n(X^k,X^{k-1})
\ar[r] & H_{n-1}(X^{k-1})
}
$$
Using induction by $k$ we have  $H_{i}^{\Delta}(X^{k-1})=
H_{i}(X^{k-1})$ for all $i$. Further, $C_{i}^{\Delta}(X^k,X^{k-1})$
is according to definition zero if $i\ne k$ and free Abelian of
rank equal the number of $i$-simplices $\Delta^i_{\alpha}$ if $i=k$.
The homology
groups $H_{i}^{\Delta}(X^k,X^{k-1})$ have the same description.

Since
$$\coprod_{\alpha}\Delta^k_{\alpha}/\coprod_{\alpha}\partial
\Delta_{\alpha}^k= X^k/X^{k-1}$$
we get the isomorphism
$$H^{\Delta}_i(X^k/X^{k-1})\to H_i(\coprod_{\alpha}\Delta^k_{\alpha}/
\coprod_{\alpha}\partial\Delta_{\alpha}^k)= H_i(X^k/X^{k-1}).$$
Applying 5-lemma (see \ref{HO5lemma}) in the diagram above, we
get that $H^{\Delta}_n(X^k)\to  H_n(X^k)$ is an isomorphism.

If $X$ is finite $\Delta$-complex,
we are ready. If it is not,
we have to prove that $H_n^{\Delta}(X)=H_n(X)$. See [Hatcher], %\cite{Ha},
page 130.
\end{proof}