INTRODUCTION TO ALGEBRAIC TOPOLOGY
MARTIN ˇCADEK
3. Simplicial and singular homology
3.1. Exact sequences. A sequence of homomorphisms of Abelian groups or modules
over a ring
. . .
fn+1
−−−→ An
fn
−−→ An−1
fn−1
−−−→ An−2
fn−2
−−−→ . . .
is called an exact sequence if
Im fn = Ker fn−1.
Exactness of the following sequences
O −→ A
f
−−→ B, B
g
−−→ C −→ 0, 0 −→ C
h
−−→ D −→ 0
means that f is a monomorphism, g is an epimorphism and h is an isomorphism,
respectively.
A short exact sequence is an exact sequence
0 −→ A
i
−→ B
j
−→ C −→ 0.
In this case C ∼= B/A. We say that the short exact sequence splits if one of the
following three equivalent conditions is satisfied:
(1) There is a homomorphism p : B → A such that pi = idA.
(2) There is a homomorphism q : C → B such that jq = idC.
(3) There are homomorphisms p : B → A and q : C → B such that ip + qj = idB.
The last condition means that B ∼= A ⊕ C with isomorphism (p, q) : B → A ⊕ C.
Exercise. Prove the equivalence of (1), (2) and (3).
3.2. Chain complexes. The chain complex (C, ∂) is a sequence of Abelian groups
(or modules over a ring) and their homomorphisms indexed by integers
. . .
∂n+2
−−−→ Cn+1
∂n+1
−−−→ Cn
∂n
−−→ Cn−1
∂n−1
−−−→ . . .
such that
∂n−1∂n = 0.
This conditions means that Im ∂n ⊆ Ker ∂n−1. The homomorphism ∂n is called a
boundary operator. A chain homomorphism of chain complexes (C, ∂C
) and (D, ∂D
) is
a sequence of homomorphisms of Abelian groups (or modules over a ring) fn : Cn → Dn
which commute with the boundary operators
∂D
n fn = fn−1∂C
n .
1
2
3.3. Homology of chain complexes. The n-th homology group of the chain complex
(C, ∂) is the group
Hn(C) =
Ker ∂n
Im ∂n+1
.
The elements of Ker ∂n = Zn are called cycles of dimension n and the elements of
Im ∂n+1 = Bn are called boundaries (of dimension n). If a chain complex is exact,
then its homology groups are trivial.
The component fn of the chain homomorphism f : (C, ∂C
) → (D, ∂D
) maps cycles
into cycles and boundaries into boundaries. It enables us to define
Hn(f) : Hn(C) → Hn(D)
by the prescription Hn(f)[c] = [fn(c)] where [c] ∈ Hn(C∗) and [fn(c)] ∈ Hn(D∗
) are
classes represented by the elements c ∈ Zn(C) and fn(c) ∈ Zn(D), respectively.
3.4. Long exact sequence in homology. A sequence of chain homomorphisms
. . . −→ A
f
−−→ B
g
−−→ C −→ . . .
is exact if for every n ∈ Z
. . . −→ An
fn
−−→ Bn
gn
−−→ Cn −→ . . .
is an exact sequence of Abelian groups.
Theorem. Let 0 → A
i
−→ B
j
−→ C → 0 be a short exact sequence of chain complexes.
Then there is a connecting homomorphism ∂∗ : Hn(C) → Hn−1(A) such that the
sequence
. . .
∂∗
−−→ Hn(A)
Hn(i)
−−−→ Hn(B)
Hn(j)
−−−−→ Hn(C)
∂∗
−−→ Hn−1(A)
Hn−1(i)
−−−−−→ . . .
is exact.
Proof. Define the connecting homomorphism ∂∗. Let [c] ∈ Hn(C) where c ∈ Cn is a
cycle. Since j : Bn → Cn is an epimorphism, there is b ∈ Bn such that j(b) = c.
Further, j(∂b) = ∂j(b) = ∂c = 0. From exactness there is a ∈ An−1 such that
i(a) = ∂b. Since i(∂a) = ∂i(a) = ∂∂b = 0 and i is a monomorphism, ∂a = 0 and a is
a cycle in An−1. Put
∂∗[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.
3.5. Chain homotopy. Let f, g : C → D be two chain homomorphisms. We say
that they are chain homotopic if there are homomorphisms sn : Cn → Dn+1 such that
∂D
n+1sn + sn−1∂C
n = fn − gn for all n.
The relation to be chain homotopic is an equivalence. The sequence of maps sn is
called a chain homotopy.
3
Theorem. If two chain homomorphism f, g : C → D are chain homotopic, then
Hn(f) = Hn(g).
Exercise. Prove the previous theorem from the definitions.
3.6. Five Lemma. Consider the diagram
A //
f1 ∼=

B //
f2 ∼=

C //
f3

D //
f4 ∼=

E
f5 ∼=

¯A // ¯B // ¯C // ¯D // ¯E
If the horizontal sequences are exact and f1, f2, f4 and f5 are isomorphisms, then f3
is also an isomorphism.
Exercise. Prove 5-lemma.
3.7. 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 ∆-complex, which is a special case of CW-complex.
Let v0, v1, . . . , vn be points in Rm
such that v1 − v0, v2 − v0, vn − v0 are linearly independent.
The n-simplex [v0, v1, . . . , vn] with the vertices v0, v1, . . . , vn is the subspace
of Rm
{
n
i=0
tivi;
n
i=1
ti = 1, ti ≥ 0}
with a given ordering of vertices. A face of this simplex is any simplex determined by
a proper subset of vertices in the given ordering.
Let ∆α, α ∈ J be a collection of simplices. Subdivide all their faces of dimension
i into sets Fi
β. A ∆-complex is a quotient space of disjoint union α∈J ∆α obtained
by identifying simplices from every Fi
β into one single simplex via affine maps which
preserve the ordering of vertices. Thus every ∆-complex is determined only by combinatorial
data.
A special case of ∆-complex is a finite simplicial complex. It is a union of simplices
the vertices of which lie in a given finite set of points {v0, v1, . . . , vn} in Rm
such that
v1 − v0, v2 − v0, . . . , vn − v0 are linearly independent.
Example. Torus, real projective space of dimension 2 and Klein bottle are ∆-complexes
as one can see from the following pictures.
In all the cases we have two sets F2
whose elements are triangles, three sets F1
every with two segments and one set F0
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 ∆-complexes.
To every ∆-complex X we can assign the chain complex (C, ∂) where Cn(X) is a free
Abelian group generated by n-simplices of X (i. e. the rank of Cn(X) is the number
4
b
b
b
b
b
b
a a a a a a
c c c
Figure 3.1. Torus, RP2
and Klein bottle as ∆-complexes
of the sets Fn
and the boundary operator on generators is given by
∂[v0, v1, . . . , vn] =
n
i=0
(−1)i
[v0, . . . , ˆvi . . . , vn].
Here the symbol ˆvi means that the vertex vi is omitted. Prove that ∂∂ = 0.
The simplicial homology groups of ∆-complex X are the homology groups of the
chain complex defined above. Later, we will show that these groups are independent
of ∆-complex structure.
Exercise. Compute simplicial homology of S2
(find a ∆-complex structure), RP2
,
torus and Klein bottle (with ∆-complex structures given in example above).
Let X and Y be two ∆-complexes and f : X → 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 fn : Cn(X) → Cn(Y ) induced by
the map f. This chain map enables us to define homomorphism of simplicial homology
groups induced by f.
Having a ∆-subcomplex A of a ∆-complex X (i. e. subspace of X formed by some of
the simplices of X) we can define simplicial homology groups Hn(X, A). The definition
is the same as for singular homology in paragraph 3.9. These groups fit into the long
exact sequence
· · · → Hn(A) → Hn(X) → Hn(X, A) → Hn−1(A) → . . .
See again 3.9.
3.8. Singular homology. The standard n-simplex is the n-simplex
∆n
= {(t0, t1, . . . , tn) ∈ Rn+1
;
n
i=0
ti = 1; ti ≥ 0}.
The j-th face of this standard simplex is the (n−1)-dimensional simplex [e0, . . . , ˆej, . . . , en]
where ej is the vertex with all coordinates 0 with the exception of the j-th one which
is 1. Define
εj
n : ∆n−1
→ ∆n
5
as the affine map εj
n(t0, t1, . . . , tn−1) = (t0, . . . , tj−1, 0, tj, . . . , tn−1) which maps
e0 → e0, . . . , ej−1 → ej−1, ej → ej+1, . . . , en−1 → en.
It is not difficult to prove
Lemma. εk
n+1εj
n = εj+1
n+1εk
n for k < j.
A singular n-simplex in a space X is a continuous map σ : ∆n
→ X. Denote the
free Abelian group generated by all the singular n-simplices by Cn(X) and define the
boundary operator ∂n : Cn(X) → Cn−1(X) by
∂n(σ) =
n
i=0
(−1)i
σεi
n
for n ≥ 0. Put Cn(X) = 0 for n < 0. Using the lemma above one can show that
∂n+1∂n = 0.
The chain complex (Cn, ∂n) is called the singular chain complex of the space X. The
singular homology groups Hn(X) of the space X are the homology groups of the chain
complex (Cn(X), ∂n), i. e.
Hn(X) =
Ker ∂n
Im ∂n+1
.
Next consider a map f : X → Y . Define the chain homomorhism Cn(f) : Cn(X) →
Cn(Y ) on singular n-simplices as the composition
Cn(f)(σ) = fσ.
From definitions it is easy to show that these homomorphisms commute with boundary
operators. Hence this chain homomorphism induces homomorphisms
f∗ = Hn(f) : Hn(X) → Hn(Y ).
Moreover, Hn(idX) = idHn(X) and Hn(fg) = Hn(f)Hn(g). It means that Hn 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.
Prove the lemma above and ∂n+1∂n = 0.
Show directly from the definition that the singular homology groups of a point are
H0(∗) = Z and Hn(∗) = 0 for n = 0.
3.9. Singular homology groups of a pair. Consider a pair of topological spaces
(X, A). Then the Cn(A) is a subgroup of Cn(X). Hence we get this short exact
sequence
0 → Cn(A)
i
−→ Cn(X)
j
−→
Cn(X)
Cn(A)
→ 0.
Since the boundary operators in Cn(A) are restrictions of boundary operators in
Cn(X), we can define boundary operators
∂n :
Cn(X)
Cn(A)
→
Cn−1(X)
Cn−1(A)
.
6
We will denote this chain complex as (C(X, A), ∂) and its homology groups as Hn(X, A).
Notice that the factor Cn(X)/Cn(A) is a free Abelian group generated by singular simplices
σ : ∆n
→ X such that σ(∆n
) A. We will need it later.
A map f : (X, A) → (Y, B) induces the chain homomorphism Cn(f) : Cn(X) →
Cn(Y ) which restricts to a chain homomorphism Cn(A) → Cn(B) since f(A) ⊆ B.
Hence we can define the chain homomorphism
Cn(f) : Cn(X, A) → Cn(Y, B)
which in homology induces the homomorphism
f∗ = Hn(f) : Hn(X, A) → Hn(Y, B).
We can again conclude that Hn is a functor from the category Top2
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 → Y in Top can be considered as the
object (X, ∅) and the morphism ˆf = f : (X, ∅) → (Y, ∅) in the category Top2
and
Hn(X, ∅) = Hn(X), Hn( ˆf) = Hn(f).
3.10. Long exact sequence for singular homology. Consider inclusions of spaces
i : A → X, i : B → Y and maps j : (X, ∅) → (X, A), j : (Y, ∅) → (Y, B) induced
by idX and idY , respectively. Let f : (X, A) → (Y, B) be a map. Then there are
connecting homomorphisms ∂X
∗ and ∂Y
∗ such that the following diagram
∂X
∗
// Hn(A)
i∗
//
(f/A)∗

Hn(X1
)
j∗
//
f∗

Hn(X, A)
∂X
∗
//
f∗

Hn−1(A)
i∗
//
(f/A)∗
∂Y
∗
// Hn(B)
i∗
// Hn(Y )
j∗
// Hn(Y, B)
∂Y
∗
// Hn−1(B)
i∗
//
commutes and its horizontal sequences are exact.
An analogous theorem holds also for simplicial homology.
Remark. Consider the functor I : Top2
→ Top2
which assigns to every pair (X, A)
the pair (A, ∅). The commutativity of the last square in the diagram above means
that ∂∗ is a natural transformation of functors Hn and Hn−1 ◦ I defined on Top2
.
Proof. We have the following commutative diagram of chain complexes
0 // C(A)
C(i)
//
C(f/A)

C(X)
C(j)
//
C(f)

C(X, A) //
C(f)

0
0 // C(B)
C(i )
// C(Y )
C(j )
// C(Y, B) // 0
with exact horizontal rows. Then Theorem 3.4 and the construction of connecting
homomorphism ∂∗ imply the required statement.
Remark. It is useful to realize how ∂∗ : Hn(X, A) → Hn−1(A) is defined. Every
element of Hn(X, A) is represented by a chain x ∈ Cn(X) with a boundary ∂x ∈
7
Cn−1(A). This is a cycle in Cn(A) and from the definition in 3.4 we have
∂∗[x] = [∂x].
3.11. Homotopy invariance. If two maps f, g : (X, A) → (Y, B) are homotopic,
then they induce the same homomorphisms
f∗ = g∗ : Hn(X, A) → Hn(Y, B).
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], Theorem 2.10 and Proposition
2.19 or [Spanier], Chapter 4, Section 4.
Corollary. If X and Y are homotopy equivalent spaces, then
Hn(X) ∼= Hn(Y ).
3.12. Excision Theorem. There are two equivalent versions of this theorem.
Theorem (Excision Theorem, 1st version). Consider spaces C ⊆ A ⊆ X and suppose
that ¯C ⊆ int A. Then the inclusion
i : (X − C, A − C) → (X, A)
induces the isomorphism
i∗ : Hn(X − C, A − C)
∼=
−→ Hn(X, A).
Theorem (Excision Theorem, 2nd version). Consider two subspaces A and B of a
space X. Suppose that X = int A ∪ int B. Then the inclusion
i : (B, A ∩ B) → (X, A)
induces the isomorphism
i∗ : Hn(B, A ∩ B)
∼=
−→ Hn(X, A).
The second version of Excision Theorem holds also for simplicial homology if we
suppose that A and B are ∆-subcomplexes of a ∆-complex X and X = A∪B. In this
case the proof is easy since the inclusion
Cn(i) : Cn(B, A ∩ B) → Cn(A ∪ B, A)
is an isomorphism, namely the both chain complexes are generated by the same n-
simplices.
Exercise. Show that the theorems above are equivalent.
The proof of Excision Theorem for singular homology can be found in [Hatcher],
pages 119 – 124, or in [Spanier], 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.
8
Lemma. Let U = {Uα; α ∈ J} be a collection of subsets of X such that X =
α∈J int Uα. Denote the free chain complex generated by singular simplices σ with
σ(∆n
) ∈ Uα for some α as CU
n (X). Then
CU
n (X)) → Cn(X)
induces isomorphism in homology.
Proof of Excision Theorem. Consider U = {A, B}. Then the inclusion
Cn(i) : Cn(B, A ∩ B) →
CU
n (X)
Cn(A)
is an isomorphism and, moreover, according to the previous lemma, the homology of
the second chain complex is Hn(X, A).
3.13. Homology of disjoint union. Let X = α∈J Xα be a disjoint union. Then
Hn(X) =
α∈J
Hn(Xα).
The proof follows from the definition and connectivity of σ(∆n
) in X for every
singular n-simplex σ.
3.14. Reduced homology groups. For every space X = ∅ we define the augmented
chain complex ( ˜C(X), ˜∂) as follows
˜Cn(X) =
Cn(X) for n = −1,
Z for n = −1.
with ˜∂n = ∂n for n = 0 and ∂0( j
i=1 niσi) = j
i=1 ni. The reduced homology groups
˜Hn(X) are the homology groups of the augmented chain complex. From the definition
it is clear that
˜Hn(X) = Hn(X) for n = 0
and
˜Hn(∗) = 0 for all n.
For pairs of spaces we define ˜Hn(X, A) = Hn(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
˜Hn(X) = ˜Hn(X, ∗) = Hn(X, ∗).
Using this equality and the long exact sequence for unreduced homology we get that
H0(X) ∼= H0(X, ∗) ⊕ H0(∗) ∼= ˜H0(X) ⊕ Z.
9
Lemma. Let (X, A) be a pair of CW-complexes, X = ∅. Then
˜Hn(X/A) = Hn(X, A)
and we have the long exact sequence
· · · → ˜Hn(A) → ˜Hn(X) → ˜Hn(X/A) → ˜Hn−1(A) → . . .
Proof. According to example in Section 2
(X, A) → (X ∪ CA, CA) → (X ∪ CA/CA, ∗) = (X/A, ∗)
is the composition of an excision and a homotopy equivalence. Hence ˜Hn(X/A) =
Hn(X, A). The rest folows from the long exact sequence of the pair (X, A).
Exercise. Prove that ˜Hn( Xα) ∼= ⊕ ˜Hn(Xα).
˜Hn can be considered as a functor from Top∗ to Abelian groups.
3.15. The long exact sequence of a triple. Three spaces (X, B, A) with the
property A ⊆ B ⊆ X are called a triple. Denote i : (B, A) → (X, A) and j : (X, A) →
(X, B) maps induced by the inclusion B → X and idX, respectively. Analogously as
for pairs one can derive the following long exact sequence:
. . .
∂∗
−→ Hn(B, A)
i∗
−→ Hn(X, A)
j∗
−→ Hn(X, B)
∂∗
−→ Hn−1(B, A)
i∗
−→ . . .
3.16. Singular homology groups of spheres. Consider the long exact sequence
of the triple (∆n
, ∂∆n
, Λn−1
= ∂∆n
− ∆n−1
):
· · · → Hi(∆n
, Λn−1
) → Hi(∆n
, ∂∆n
)
∂∗
−→ Hi−1(∂∆n
, Λn−1
) → Hi−1(∆n
, Λn−1
) → . . .
The pair (∆n
, Λn−1
) is homotopy equivalent to (∗, ∗) and hence its homology groups
are zeroes. Next using Excision Theorem and homotopy invariance we get that
Hi(∆n
, Λn−1
) ∼= Hi(∆n−1
, ∂∆n−1
). Consequently, we get an isomorphism
Hi(∆n
, ∂∆n
) ∼= Hi−1(∆n−1
, ∂∆n−1
).
Using induction and computing Hi(∆1
, ∂∆1
) = Hi([0, 1], {0, 1}) ∼= Hi−1({0, 1}, {0})
we get that
Hi(∆n
, ∂∆n
) =
Z for i = n,
0 for i = n.
Doing the induction carefully we can find that the generator of the group Hn(∆n
, ∂∆n
) =
Z is determined by the singular n-simplex id∆n .
The pair (Dn
, Sn−1
) is homeomorphic to (∆n
, ∂∆n
). Hence it has the same homology
groups. Using the long exact sequence for this pair we obtain
˜Hi−1(Sn−1) = Hi(Dn
, Sn−1
) =
0 for i = n,
Z for i = n.
10
3.17. Mayer-Vietoris exact sequence. Denote inclusions A∩B → A, A∩B → B,
A → X, B → X by iA, iB, jA, jB, respectively. Let C → A ∩ B and suppose that
X = int A ∪ int B. Then the following sequence
. . .
∂∗
−−→ Hn(A ∩ B, C)
(iA∗,iB∗)
−−−−−→ Hn(A, C) ⊕ Hn(B, C)
jA∗−jB∗
−−−−−→ Hn(X, C)
∂∗
−−→ Hn−1(A ∩ B, C) −→ . . .
is exact.
Proof. The covering U = {A, B} satisfies conditions of Lemma 3.12. The sequence of
chain complexes
0 −→
C(A ∩ B)
C(C)
i
−→
C(A)
C(C)
⊕
C(B)
C(c)
j
−→
CU
(X)
C(C)
−→ 0
where i(x) = (x, x) and j(x, y) = x − y is exact. Consequently, it induces a long
exact sequence. Using Lemma 3.12 we get that Hn(CU
(X), C(C)) = Hn(X, C), which
completes the proof.
3.18. Equality of simplicial and singular homology. Let (X, A) be a pair of
∆-complexes. Then the natural inclusion of simplicial and singular chain complexes
C∆
(X, A) → C(X, A) induces the isomorphism of simplicial and singular homology
groups
H∆
n (X, A) ∼= Hn(X, A).
Outline of the proof. Consider the long exact sequences for the pair (Xk
, Xk−1
) of
skeletons of X. We get
H∆
n+1(Xk
, Xk−1
) //

H∆
n (Xk−1
) //

H∆
n (Xk
) //

H∆
n (Xk
, Xk−1
) //

H∆
n−1(Xk−1
)

Hn+1(Xk
, Xk−1
) // Hn(Xk−1
) // Hn(Xk
) // Hn(Xk
, Xk−1
) // Hn−1(Xk−1
)
Using induction by k we have H∆
i (Xk−1
) = Hi(Xk−1
) for all i. Further, C∆
i (Xk
, Xk−1
)
is according to definition zero if i = k and free Abelian of rank equal the number of isimplices
∆i
α if i = k. The homology groups H∆
i (Xk
, Xk−1
) have the same description.
Since
α
∆k
α/
α
∂∆k
α = Xk
/Xk−1
we get the isomorphism
H∆
i (Xk
/Xk−1
) → Hi(
α
∆k
α/
α
∂∆k
α) = Hi(Xk
/Xk−1
).
Applying 5-lemma (see 3.6) in the diagram above, we get that H∆
n (Xk
) → Hn(Xk
) is
an isomorphism.
If X is finite ∆-complex, we are ready. If it is not, we have to prove that H∆
n (X) =
Hn(X). See [Hatcher], page 130.
11
CZ.1.07/2.2.00/28.0041
Centrum interaktivních a multimediálních studijních opor pro inovaci výuky a efektivní učení