Relative homology groups are constructed for a pair $(X, A)$, where $A$ is a subspace of the space $X$. The group of relative chains $C_n(X, A)$ is the quotient group $C_n(X)/C_n(A)$. The boundary operator passes to equivalence classes, so the relative chains form a chain complex. The short exact sequence $0 \to C(A) \to C(X) \to C(X, A) \to 0$ induces a long exact sequence of homology groups. For $A = \{x\}$, the relative homology of the pair coincides with the reduced homology of the space $X$.

Chain Groups

Relative homology groups are constructed for a pair of spaces: a topological space $X$ and its subspace $A$.

For the space $X$, chain groups are constructed. For the subspace $A$, chain groups are also constructed, but the basis is chosen in a special way. For each $n$, only those singular simplices whose images lie in $A$ are included in the basis.

Singular Simplices

The chain groups of a space $X$ are formed from the singular simplices of this space. For a subspace $A \subseteq X$, one takes only those singular simplices whose image lies in $A$. The chain group constructed in this way is denoted $C(A)$. Chain groups are defined both for the space $X$ and for its subspace $A$.

Definition of Relative Chains

Let $X$ be a topological space and $A \subset X$. The relative chain group is defined as the quotient group

$C_n(X, A) = C_n(X) / C_n(A).$

The elements of $C_n(X, A)$ are equivalence classes of chains from $C_n(X)$, where two chains are considered equivalent if their difference lies in $C_n(A)$.

Chain Complex of Relative Chains

The groups $C_n(X, A)$ form a chain complex. Its terms are:

$\dots \rightarrow C_{n+1}(X, A) \rightarrow C_n(X, A) \rightarrow C_{n-1}(X, A) \rightarrow \dots$

Boundary Operator

A boundary operator acts on this complex. To determine its action, take an equivalence class from $C_n(X, A)$ and apply the boundary to its representative.

Boundary Operator on Relative Chains

The elements of the relative chain group are equivalence classes. For convenience, a class is denoted by a single letter, which reduces the number of brackets when working with homologies.

The boundary operator ∂ commutes with passage to the equivalence class. Therefore, it acts on relative chains. The composition of boundary operators is zero, and the relative chains form a chain complex.

The operator acting on equivalence classes and the boundary homomorphism of the chain complex for the subspace A are different operators.

The relative homology group of the pair (X, A) is defined as the kernel of the operator ∂ modulo the image of the operator ∂ of the next dimension, that is, ker ∂ₙ / im ∂ₙ₊₁, where ∂ acts on relative chains.

Consider a class α belonging to the n-th homology group.

Relative chain of space A

Let the element $a$ belong to the $n$-th group of chains of the space $X$, and let the relative chain $a$ represent some homology class.

If the boundary $\partial_n a$ is equal to zero, then $a$ is a cycle. Such a chain is trivial modulo $A$. In this case, the boundary $\partial_n a$ itself belongs to the $(n-1)$-th group of chains of the space $A$.

Next, consider the case where the relative chain $a$ represents a trivial class in homology. Then there exists an element $b$ from the group of $(n+1)$-chains of the space $X$ for which the equality $a = \partial_{n+1} b$ holds.

If two equivalence classes coincide, they differ by an element of the group of chains of the space $A$. In other words, the relative chain is represented in the form $a = \partial_{n+1} b + c,$ where $c$ belongs to the $n$-th group of chains of the space $A$.

The next step defines a short exact sequence of complexes.

Short Exact Sequence of Chain Complexes

We consider the sequence of chain complexes

0 → C(A) → C(X) → C(X, A) → 0,

where C(A) is the chain complex of the space A, C(X) is the chain complex of the space X, and C(X, A) is the relative chain complex. The extreme terms of the sequence are zero.

The map i from C(A) to C(X) is the usual inclusion. The map j from C(X) to C(X, A) is the quotient map.

It is assumed that the image of the map i is equal to the kernel of the map j.

A short exact sequence of chain complexes requires more than the equality of the image of one map with the kernel of the next. The additional condition in this fragment is not fully stated: the exposition breaks off at this point.

Commutativity of the diagram

Maps between chain complexes must commute with the boundary operators in each of the complexes. This condition is checked directly, and it holds.

Previously, the following statement was proved. If a short exact sequence of chain complexes is given, then there exists a long exact sequence of their homology groups. It has the form

$\dots \to H_n(A) \to H_n(X) \to H_n(X, A) \to H_{n-1}(A) \to \dots$

In this sequence, the homomorphism denoted by the symbol $D$ connects the homology groups of adjacent terms.

Action of D on homology classes

In the previous lecture, we considered a homology class $c \in H_n(X, A)$. Its representative is the chain $C \in C_n(X)$.

It is known that the map $\iota$ is an epimorphism. A further statement about the action of $D$ on homology classes is supposed to follow from this fact, but the argument in this fragment is not complete.

Existence of the element B

Choice of the element B

Consider an element B belonging to the chain group C_n, for which ι(B) = C. Here C is a cycle representing a homology class, and ι is the map that assigns to the element B its equivalence class.

One can take the element C itself as B. Then, under factorization, the element B corresponds to the equivalence class containing C.

Action of the boundary operator

For the chosen element B, the boundary operator ∂ gives ∂B. From the choice of B it follows that ∂B belongs to the kernel of the map ι. This kernel coincides with the image of the inclusion, so ∂B equals the image of some element A from C_{n-1}.

If we take B = C, then ∂B = ∂C = 0, since C represents a homology class. Thus, for B = C, the equality ∂B = 0 holds.

Connection with relative chains

Earlier it was proved that ∂A = 0 for such an element A. The element A can be chosen from those with the help of which the action of the operator ∂ is defined.

The operator ∂ acts on the elements of the group as a boundary. Therefore, the class C is assigned the homology class to which the element ∂C corresponds.

A relative chain is defined by the condition that its boundary falls into the subspace A. The pair (X, A) and the quotient group by A are related precisely through this condition.

Next steps

Two lemmas are required for the further proof: the splitting lemma and the five lemma. The splitting lemma is considered first.

Splitting of Abelian Groups

A short exact sequence of abelian groups is considered. All groups in this sequence are abelian, and the sequence itself is exact.

Second claim

For a short exact sequence of abelian groups or modules $0 \to A \to B \to C \to 0$, the following claim is equivalent to the previous one.

There exists a homomorphism from $B$ to $A$ whose composition with the map from $A$ to $B$ is the identity map of $A$. Such a homomorphism is called a left inverse to the embedding of $A$ into $B$.

Second condition: there exists a homomorphism from $C$ to $B$ whose composition with the map from $B$ to $C$ is the identity map of $C$. This homomorphism is a right inverse to the projection of $B$ onto $C$.

Third condition: there exists an isomorphism between $B$ and the direct sum $A \oplus C$.

Composition of Mappings and the Commutative Diagram

A diagram that must be commutative is considered. It consists of the zero group, group A, group B, group A + C, group C, and again the zero group. The mapping F between the groups is an isomorphism. Commutativity means that the composition of F and I equals H. The composition of K and F equals J.

The mappings act as follows. The homomorphism H maps each element a of group A to the pair (a, 0). The homomorphism K maps the pair (a, c) to the element c.

The notation B is already occupied by another object of the diagram, so the label on the arrow is written incorrectly. The correct statement consists in the existence of an isomorphism.

The stated assertions are equivalent. This equivalence can be proved in different ways.

Constructing the Maps F and I from A into A+C

Consider a short exact sequence of abelian groups $0 \to A \xrightarrow{I} B \xrightarrow{J} C \to 0$, together with a map $P\colon B \to A$ such that $P \circ I = \mathrm{id}_A$. We need to show that three statements about the existence and properties of such maps are equivalent.

The map F

Define the map $F\colon B \to A \oplus C$ by $F(b) = (P(b), J(b))$. An element $b \in B$ first goes to the pair $(P(b), J(b))$, and then the first component of the pair is selected.

The composition of $P$ and $I$ equals the identity map $\mathrm{id}_A$, since $P \circ I = \mathrm{id}_A$. The composition $F \circ I$ equals the map $H\colon A \to A \oplus C$ that sends $a$ to the pair $(a, 0)$. Substituting $a \in A$ into the projection onto the first component gives $a$, that is, the projection onto the first factor is the identity on $A$.

Similarly, one proves that the third statement implies the second.

The kernel of F

Let $b \in B$ be such that $P(b) = 0$ and $J(b) = 0$. By exactness of the sequence at $B$, $b$ lies in the image of $I$. Hence there exists $a \in A$ with $I(a) = b$.

Apply $P$ to the equality $I(a) = b$. We get $a = P(b) = 0$. Then $b = I(0) = 0$. The kernel of $F$ is trivial, so $F$ is injective.

Surjectivity of F

Let $(a, c) \in A \oplus C$. Since $J$ is surjective, there exists $b \in B$ with $J(b) = c$. Set $b' = b + I\bigl(a - P(b)\bigr).$

Since $J \circ I = 0$, we have $J(b') = J(b) = c$. Furthermore, $P(b') = P(b) + P(I(a - P(b))) = P(b) + (a - P(b)) = a$. Therefore $F(b') = (a, c)$, and $F$ is surjective.

Thus $F$ is an isomorphism. This proves that the third statement implies the first.

Equivalence of the statements

The first statement implies the third. The second statement is equivalent to the third. It follows that the first and second statements are equivalent.

If the sequence is exact, then all three statements hold simultaneously. If at least one of them holds, the others hold as well. All three statements are equivalent.

Notation and Commutativity of the Diagram

Two exact sequences of abelian groups are considered. The groups of the first sequence are denoted A', B', C', D', E'. Homomorphisms between them are given, among which are F, G, H, J, as well as R, S, T, U. Homomorphisms between groups in adjacent positions are denoted L, M, N, P, Q.

All groups are abelian. Both sequences are exact. The diagram composed of these groups and homomorphisms is commutative.

The following is claimed: if the homomorphisms M and P are isomorphisms, then, under the stated conditions, a conclusion is drawn about the homomorphism in the central position. The formulation in this fragment is incomplete, so its conclusion is not given here.

Proof of surjectivity (epimorphism L)

Properties of the mappings

The mapping L is an epimorphism, that is, a surjective mapping. The mapping Q is a monomorphism, that is, an injective mapping. The homomorphisms M and P are isomorphisms.

The kernel of Q is zero, and the image of L is H'.

Isomorphism of the central homomorphism

The mappings L and Q, as well as the isomorphisms M and P, set out the conditions under which the central homomorphism N is an isomorphism. Both of the isomorphisms mentioned above lead to the conclusion that N is an isomorphism.

Proof plan

First, it is proved that N is surjective. The injectivity of N is proved similarly.

To prove the surjectivity of N, it is sufficient to consider only part of the diagram.

Proof of Injectivity

Injectivity is proved by a scheme that differs from the proof of surjectivity. The argument runs in the reverse direction.

Consider an element C' of the chain group C'. Apply the map T to it, and denote the result by T(C').

The map P is surjective. Therefore, there exists an element D of the group D such that P(D) = D'. In other words, P(D) = T(C').

The image of T coincides with the kernel of U. Consequently, U(T(C')) = 0.

Next, the commutativity of the diagram is used. The element D, introduced in the previous step, is related to the maps Q and I. The equality Q(I(D)) = U(P(D)) holds.

Exact Sequence and Connection of Homologies

The element P(D) equals T(C'), and this element is zero. It follows that U(ι(D)) = 0.

The map U has trivial kernel. Therefore ι(D) = 0, and D belongs to the kernel of ι.

By exactness of the sequence, the image of H coincides with the kernel of ι. Hence D lies in the image of H. Consequently, there exists an element C from the group C such that h(C) = D.

Next, consider T(C'). Since T(C') = P(D), and P(D) = P(h(C)), commutativity of the diagram gives T(C') = T(n(C)).

The map T is a homomorphism. Therefore T(C' − n(C)) = 0. In other words, the element C' − n(C) belongs to the kernel of T.

Commutativity and the Isomorphism of Homomorphisms

Image and Kernel

The image of the map s coincides with the kernel of the map t. Therefore, there exists an element b′ from the group B′ for which the equality s(b′) = c′ − n(c) holds.

Using the Isomorphism m

The map m is an isomorphism, and therefore also an epimorphism. Consequently, there exists an element b from the group B such that m(b) = b′.

Next, the commutativity of the diagram is used. It implies the equality s(m(b)) = n(g(b)). Since m(b) = b′, we obtain s(b′) = n(g(b)).

Comparing the two expressions for s(b′), we get:

c′ − n(c) = n(g(b)).

Hence c′ is expressed in terms of the elements b and c:

c′ = n(g(b)) + c.

Since n is a homomorphism, the sum n(g(b)) + c is written in the form indicated above.

The Converse Statement

The converse statement is proved analogously. Suppose that some element c is mapped to zero. From this assumption it follows that the element c itself is equal to zero. The argument in the other direction follows the same scheme.

Reduced Homologies

Next, reduced homologies are considered. They were discussed at the previous class. This section is declared important for understanding the material, so it is suggested to return to it and restore the definitions.

Reduced Homology and Chain Complexes

Definition via the Augmented Complex

Reduced homology is defined as the homology of the augmented chain complex. The augmentation is given by a homomorphism $\varepsilon$, which maps zero-dimensional chains to $\mathbb{Z}$.

Short Exact Sequence of Augmented Complexes

We consider a short exact sequence of augmented chain complexes. For a pair $(X, A)$, the quotient complex $C(X)/C(A)$, with augmentation added, gives $\mathbb{Z}/\mathbb{Z}$ in degree zero, which is zero. The homomorphism induced by the augmentation is the quotient map.

The sequence has the following properties:

• it remains a chain complex;
• it is exact;
• the diagram commutes with the boundary homomorphisms.

A short exact sequence of chain complexes induces a long exact sequence of homology groups. For reduced homology, this sequence has the form:

$\dots \to \tilde H_n(A) \to \tilde H_n(X) \to H_n(X, A) \to \tilde H_{n-1}(A) \to \dots$

Reduced Homology of a Point

Let $A$ consist of a single point $x \in X$. The ordinary singular homology of a point is $\mathbb{Z}$ when $n = 0$ and zero when $n > 0$. If we add augmentation, the reduced homology of a point is trivial for all $n \geq 0$.

Since the sequence is exact and the reduced homology of a point is zero, the isomorphism

$\tilde H_n(X) \cong H_n(X, A)$

holds for every $n \geq 0$.

Thus, when $A = \{x\}$, the relative homology coincides with the reduced homology of the space $X$.

Alternative Definition of Relative Homology

Relative homology is sometimes defined differently, namely by modifying the original chain complex.

Reduced Homology of a Point and the Long Sequence

Reduced Homology as Relative Homology

The reduced homology of a space X coincides with the relative homology of the pair (X, x₀), where x₀ is a point of X.

Relation Between Ordinary and Reduced Homology

For n > 0, the reduced homology coincides with the ordinary homology: H̃ₙ(X) ≅ Hₙ(X).

For n = 0, the relation H₀(X) ≅ H̃₀(X) ⊕ ℤ holds.

The Long Sequence of the Pair (X, x₀)

Consider the long exact sequence of the pair (X, x₀). The homology of a point is zero for n ≥ 1, so for n ≥ 2 the homology of the point vanishes. Then Hₙ(X) ≅ Hₙ(X, x₀), and these groups are isomorphic to the reduced homology.

The case n = 1 requires a separate argument. Let r: X → {x₀} be a retraction, and i: {x₀} → X the inclusion. The restriction of r to the point is the identity map, so r ∘ i = id. Consequently, the composition of the induced maps r∗ ∘ i∗ is the identity on H₀(point) = ℤ. Hence i∗ is injective.

Since the map i∗ is injective, the image of the boundary homomorphism ∂ equals the kernel of i∗, that is, zero. From the exact sequence we obtain H₁(X) ≅ H₁(X, x₀).

A general fact is used here twice. If the composition f ∘ g is injective, then g is injective. The proof is by contradiction: if g is not injective, then two distinct points map to one, and their images under f also coincide, so the composition is not injective.

For n = 0, we write out the short exact sequence 0 → H₀(point) → H₀(X) → H₀(X, x₀) → 0. By the splitting lemma there exists r∗ such that r∗ ∘ i∗ = id. Then H₀(X) ≅ H₀(point) ⊕ H₀(X, x₀). By definition H₀(X, x₀) = H̃₀(X), and H₀(point) = ℤ. Hence H₀(X) ≅ H̃₀(X) ⊕ ℤ.

The Excision Theorem

First formulation. Let Z ⊂ A ⊂ X, where the closure of Z is contained in the interior of A. Then the inclusion (X − Z, A − Z) → (X, A) induces an isomorphism of homology groups Hₙ(X − Z, A − Z) ≅ Hₙ(X, A).

Second formulation. Let A and B be subspaces of X such that the interiors of A and B cover X. Then the inclusion (B, A ∩ B) → (X, A) induces an isomorphism Hₙ(B, A ∩ B) ≅ Hₙ(X, A) for all n ≥ 0.

The two formulations are equivalent. In the second version, it suffices to set B = X − Z.

The excision theorem is used here without proof. We consider its application.

Reduced Homology of a Union with a Cone

Let CA be the cone over A, and let X ∪ CA be obtained by gluing X and CA along A. Then the reduced homology of X ∪ CA is isomorphic to the homology of the pair (X, A):

H̃ₙ(X ∪ CA) ≅ Hₙ(X, A).

The proof proceeds in two steps.

1. Consider the long exact sequence of the pair (X ∪ CA, CA). The cone is contractible and homotopy equivalent to a point, so its homology is trivial. Hence H̃ₙ(X ∪ CA) ≅ Hₙ(X ∪ CA, CA).

2. Denote the apex of the cone by v. By the first formulation of the excision theorem,

Hₙ(X ∪ CA, CA) ≅ Hₙ(X ∪ CA − v, CA − v).

Removing the apex does not change the homotopy type. The space X ∪ CA − v deformation retracts onto X, and CA − v deformation retracts onto the base A. Therefore Hₙ(X ∪ CA − v, CA − v) ≅ Hₙ(X, A).

The isomorphism between all these groups follows from the five lemma. The corresponding groups on the left and right are isomorphic, so the middle isomorphism holds as well.

The Homotopy Extension Property

A pair (X, A) has the homotopy extension property if for any continuous map F: X → Z and any homotopy f_t: A → Z such that f₀ coincides with F|A, there exists a homotopy F_t: X → Z such that F₀ = F and F_t|A = f_t for all t.

This property simplifies the computation of homology of pairs. Pairs with this property are called good. Examples include:

• cellular pairs (CW pairs);
• NDR pairs.

An NDR pair (A ⊂ X) is defined as follows. A is closed in X, and there exists an open neighborhood U of A such that A is a strong deformation retract of U. A strong deformation retract is a deformation retract in which the points of A remain fixed throughout the entire homotopy.

If the pair (X, A) has this property, then the quotient space X/A is homotopy equivalent to X ∪ CA.

Pairs with this property occur often, since in practice one usually works with cellular pairs. For pairs without this property, for example a bouquet, the homology is also computed. The homotopy extension property merely simplifies the proofs and yields additional results.

Cofibrations

Let (X, A) be a cofibration. Then the quotient map (X, A) → (X/A, point) induces an isomorphism of homology groups:

Hₙ(X, A) ≅ H̃ₙ(X/A).

This follows from the previous theorem. Since X ∪ CA is homotopy equivalent to X/A, we get Hₙ(X, A) ≅ H̃ₙ(X ∪ CA) ≅ H̃ₙ(X/A).

The Suspension Isomorphism

The suspension of X is the space SX = X × I with the upper and lower bases collapsed. Sometimes the interval is taken as [−1, 1] so that the work reduces to cones.

The reduced homology of the suspension is isomorphic to the reduced homology of the original space shifted by one:

H̃ₙ(SX) ≅ H̃ₙ₋₁(X).

The proof goes as follows. The cone CX is contractible, so its homology vanishes. From the long exact sequence of the pair (CX, X) we obtain Hₙ(CX, X) ≅ H̃ₙ₋₁(X). The pair (CX, X) is a cofibration, so Hₙ(CX, X) ≅ H̃ₙ(CX/X). The quotient space CX/X is homotopy equivalent to the suspension SX.

Homology of Spheres

From the suspension isomorphism it follows that the sphere Sⁿ is the n-fold suspension of S⁰. The homology of S⁰ is computed directly. Then the homology of the spheres Sⁿ is found by induction using the relation H̃ₙ(SX) ≅ H̃ₙ₋₁(X).