Solutions to selected exercises

Jorge Becerra Garrido∗

Last updated: 2nd January 2019

Exercise 1.2. Let M be a set and let i : M −→ Z[M] be the map sending m to the function f : M −→ Z,f (m) = 1 and f (n) = 0 for n 6= m. Let A be an abelian group and let ψ : M −→ A be a map of sets.

Show that there exists a unique group homomorphism ψ : Z[M] −→ M such that ψ i = ψ.

M A

Z[M]

ψ

iψ

Proof. By simplicity, recall the notation (or even better, the isomorphism Z[M] ' ⊕M Z) we use to

denote as k1m1 + · · · + krmr to the function f : M −→ Z sending f (m1) = k1, . . . , f (mr) = kr andeverything else to 0 . Under this description, i(m) = 1 ·m.

Define ψ(k1m1 + · · ·+ krmr) := k1ψ(m1) + · · ·+ krψ(mr), which is clearly a group homomorphism,since it is linear by definition. Moreover, (ψ i)(m) = ψ(1 ·m) = 1 · ψ(m) = ψ(m).

The uniqueness follows since if ψ is another group homomorphism satisfying ψ i = ψ then

ψ(k1m1 + · · ·+ krmr) = ψ(k1m1) + · · ·+ ψ(krmr) = k1ψ(1 ·m1) + ·+ krψ(1 ·mr)

= k1ψ(1 ·m1) + · · ·+ krψ(1 ·mr) = ψ(k1m1 + · · ·+ krmr).

Exercise 1.5. If Ci is a family of chain complexes indexed by a set I, then⊕

i∈I Ci is the chain complex with(⊕

i∈I Ci)n :=⊕

i∈I Cin and differential ∂((ci)i∈I) = (∂ci)i∈I .

Show that there is a canonical isomorphism

Φ :⊕i∈I

Hn(Ci) −→ Hn(⊕i∈I

Ci).

Proof. Define Φ(([ci])i∈I) := [(ci)i∈I ]. This is a well defined group homomorphism, since Φ(([ci +∂di])i∈I) = [(ci + ∂di)i∈I ] = [(ci)i∈I + ∂(di)i∈I ] = [(ci)i∈I ] = Φ(([ci])i∈I). Surjectivity is clear, and forinjectivity, if [(ci)i∈I ] = 0, then (ci)i∈I ∈ Im ∂ so (ci)i∈I = ∂(di)i∈I = (∂di)i∈I by the definition of thedifferential in the direct sum. Therefore [ci] = [∂di] = 0 for all i ∈ I, and ([ci])i∈I = 0.

Exercise 2.1. Let K be a field and let C be a chain of K-vector spaces such that only finitely many of the Cn arenon-trivial vector spaces, and such that each Cn is a finite dimensional vector space.

Show that the following equation holds:

∑k≥0

(−1)k dim Ck = ∑k≥0

(−1)k dim Hk(C)

Proof. Since only finitely many Cn’s are non-trivial, we can suppose that they are below one indexn ∈N, so we have the following situation:

· · · −−−−→ 0∂n+1−−−−→ Cn

∂n−−−−→ Cn−1∂n−1−−−−→ · · · −−−−→ C1

∂1−−−−→ C0

∗These solutions are based on joined work with Guillermo Bijkerk Vila for the course Algebraic Topology in Autumn 2017.

1

Note that the equality makes sense as in both sides the sum is finite, because Ck = 0 and ∂k = 0 (andhence Hk(C) = 0) for k > n. Since the spaces are vector spaces and the differentials are linear maps, wecan use the dimension formula dim Ck = dim Ker ∂k + dim Im ∂k, and hence

n

∑k=0

(−1)k dim Ck = dim C0 +n

∑k=1

(−1)k(dim Ker ∂k + dim Im ∂k)

= dim C0 +n

∑k=1

(−1)k dim Ker ∂k −n

∑k=1

(−1)k−1 dim Im ∂k

= dim C0 +n

∑k=1

(−1)k dim Ker ∂k −n−1

∑k=0

(−1)k dim Im ∂k+1

= dim C0 − dim Im ∂1 +n−1

∑k=1

(−1)k(dim Ker ∂k − dim Im ∂k+1) + (−1)n dim Ker ∂n

= dim C0 − dim Im ∂1 +n−1

∑k=1

(−1)k(dim Ker ∂k − dim Im ∂k+1)

+ (−1)n(dim Ker ∂n − dim Im ∂n+1)

=n

∑k=0

(−1)k dim Hk(C),

where in the penultimate equality we have taken into account that Im ∂n+1 = 0.

Exercise 2.2. Let f : A −→ B be a abelian group homomorphism, set Q := B/ Im f and let q : B −→ Q be thecanonical homomorphism to the quotient.

(i) Show that if p : B −→ P is a homomorphism of abelian groups with p f = 0, then there exists a uniquehomomorphism p′ : B −→ P with p′ q = p.

(ii) Suppose that q′ : B −→ Q′ is a homomorphism of abelian groups such that q′ f = 0 and such that forany homomorphism of abelian groups p : B −→ P with p f = 0, there exists a unique homomorphismp′ : Q′ −→ P with p′ q′ = p. Show that there exists a unique isomorphism r : Q −→ Q′ such thatr q = q′.

(iii) Let C and D be chain complexes of abelian groups and let f : C −→ D be a chain map. For n > 0, letEn = Dn/ Im fn and let qn : Dn −→ En be the quotient map. Show that there exists a unique family ofmaps ∂E

n : En −→ En−1 such that

· · ·∂E

3−−−−→ E2∂E

2−−−−→ E1∂E

1−−−−→ E0

is a chain complex and the qn define a chain map q : D −→ E.

Proof. For (i), let us define p′ : Q −→ P in the obvious way: p′(b) := p(b). First of all let us check that itis well-defined: another representative of b is b + f (a) for some a ∈ A, and in this case

p′(b + f (a)) = p(a + f (a)) = p(a) + (p f )(a) = p(a),

hence it does not depend on the choice of the representative of b.

T B P

Q

0

f p

qp′

Defined in this way, it holds p′ q = p obviously (since q is the projection to the quotient), and itis unique, because if ϕ : Q −→ P was another group homomorphism satisfying ϕ q = p, then for allb ∈ Q it holds

p′(b) = p(b) = ϕ(q(b)) = ϕ(b).

2

To see (ii), we consider the following diagram:

Q

r

Af // B

q??

q′ //

q

Q′

r′

Q

Since q′ f = 0 by assumption, following (i) we know that the morphism r : Q −→ Q′ exists andis unique, and the upper triangle commutes. Now, q f = 0, so again by assumption r′ : Q′ −→ Qalso exists and is unique, and the lower triangle commutes. This means that we can just consider thebig, outer triangle and take r′ r : Q −→ Q as the morphism that makes it commute. But the simplerdiagram

Af // B

q //

q

Q

id

Q

does the job as well, and so by unicity of the homomorphism Q −→ Q it must be r′ r = idQ.By exchanging Q and Q′ and the corresponding homomorphisms in the diagram above and doing

an analogous reasoning, we obtain that r r′ = idQ′ , and therefore r : Q −→ Q′ is the desired isomorph-ism, having r′ : Q′ −→ Q as its inverse.

For (iii), we have the following diagram:

· · · // Cn+1∂C

n+1 //

fn+1

Cn∂C

n //

fn

Cn−1 //

fn−1

· · ·

· · · // Dn+1∂D

n+1 //

qn+1

Dn∂D

n //

qn

Dn−1 //

qn−1

· · ·

En+1 En En−1

where the inner squares between C and D commute. Now consider the (sub)diagram

Cnfn // Dn

qn //

∂Dn

En

Dn−1

qn−1 // En−1

Note that qn−1 ∂Dn is a group homomorphism, and

(qn−1 ∂n) fn = qn−1 ( fn−1 ∂Cn ) = 0 ∂C

n = 0,

where we have used the fact that composition is associative and f is a chain map. Thus, by part (i)of this exercise, there exists a unique homomorphism ∂E

n : En −→ En−1 such that the (sub)diagramcommutes. This shows that the desired family of maps ∂E

n exists and is indeed unique. Moreover, wealso directly get that the qn define a chain map q : D −→ E, because by definition the maps ∂E

n are suchthat ∂E

n qn = qn−1 ∂Dn holds.

· · · // Cn+1∂C

n+1 //

fn+1

Cn∂C

n //

fn

Cn−1 //

fn−1

· · ·

· · · // Dn+1∂D

n+1 //

qn+1

Dn∂D

n //

qn

Dn−1 //

qn−1

· · ·

· · · // En+1∂E

n+1 // En∂E

n // En−1 // · · ·

3

Now, to see that this really is a chain complex, we need to show that ∂En ∂E

n+1 = 0. Indeed, by com-mutativity of the inner squares,

(∂En ∂E

n+1)(x) = (∂En ∂E

n+1 qn+1)(x) = (∂En qn ∂D

n+1)(x) = qn−1 ∂Dn ∂D

n+1)(x) = qn−1(0) = 0,

where x denotes the equivalence class of x in the quotient group En+1.

(Thanks Mireia Martínez i Sellarès for this proof).

Exercise 3.4. (i) Let 0 −→ A′j−→ A

q−→ A −→ 0 be a short exact sequence of abelian groups. Show thatthe following statements are equivalent (such a sequence is called split):

(a) q admits a section s.

(b) j admits a retraction r.

(c) There is a commutative diagram

0 A′ A A 0

0 A′ A′ ⊕ A A 0

j q

f

i p

where f is an isomorphism and i(a′) = (a′, 0) and p(a′, a) = a.

(ii) Give an example of a short exact sequence which is not split.

(iii) Let X be a space and X′ ⊂ X a subspace, and let r : X −→ X′ be a retraction. Use (i) to construct anisomorphism

Hn(X; A) −→ Hn(X′; A)⊕ Hn(X, X′; A).

Proof. (i). Clearly, (c) =⇒ (a) and (c) =⇒ (b) by defining r := π f and s := f−1 ι, where π(a′, a) = a′

and ι(a) = (0, a). By the commutativity of the diagram of (c), we get that r j = IdA′ and q s = IdA.(b) =⇒ (c). Define f := (r, q), that is, f (a) := (r(a), q(a)). It gives a commutative diagram as

desired, so it remains to check that it is isomorphism: for injectivity, if 0 = f (a) = (r(a), q(a)), the,r(a) = 0 and q(a) = 0, so a ∈ Ker q = Im j by exactness, and a = j(a′) for some a′ ∈ A′, and thereforea′ = rj(a′) = r(a) = 0, thus a = 0. For surjectivity, given (a′, a) with a = q(a) for some a ∈ A,considerx := a′− r(a). Then a+ j(x) is the desired element, since f (a+ j(x)) = (r(a) + x, q(a) + qj(x)) = (a′, a).

(a) =⇒ (c). Here it is more convenient to define the isomorphism in the other direction: considerg : A′ ⊕ A −→ A, g(a′, a) := j(a′) + s(a). For injectivity, if g(a′, a) = 0, then s(a) = −j(a′) ∈ Im j =Ker q, so 0 = qs(a) = a, and also a′ = 0 as j(a′) = 0 and j is injective. For surjectivity, given a ∈ A,set a = q(a). We see that a− s(a) ∈ Ker q = Im j, so a− s(a) = j(a′) for some a′ ∈ A′ and we are inbusiness.

(ii). Consider the short exact sequence 0 −→ Z·n−→ Z

q−→ Z/nZ −→ 0, with n ≥ 2. Then q cannothave a section s: indeed, in such a case, if s([1]) = k, then 0 = s([n · 1]) = nk, so k = 0 and s = 0, whatcannot happen if s is a section.

Another more direct way to argue is: Z is free torsion, whereas Z⊕Z/nZ is not.

(iii). For every n ∈ N, the retraction r induces a retraction r∗ : Cn(X; A) −→ Cn(X′; A) by func-tioriality, which is a chain map. By (i), there are isomorphisms (r∗, q) : Cn(X; A) −→ Cn(X′; A) ⊕Cn(X, X′; A), where q : Cn(X; A) −→ Cn(X, X′; A). Being r∗ and q chain maps, so is

(r∗, q) : C(X; A) −→ C(X′; A)⊕ C(X, X′; A).

We conclude by applying exercise 1.5.

Exercise 4.2. Let X be a topological space and let X′′ ⊂ X′ ⊂ X be subspaces, and let A be an abelian group. Let

Hn(X′, X′′; A) −→ Hn(X, X′′; A)

be the map induced by the inclusion X′ −→ X, let

Hn(X, X′′; A) −→ Hn(X, X′; A)

4

be the map induced by the identity of X and let

Hn(X, X′; A) −→ Hn−1(X′, X′′; A)

be the composite of the connecting homomorphism Hn(X, X′; A) −→ Hn−1(X′; A) from the long exact sequenceof the pair (X, X′) and the map Hn−1(X′; A) −→ Hn−1(X′, X′′; A) from the long exact sequence of the pair(X′, X′′).

Show that the sequence

· · · Hn(X′, X′′; A) Hn(X, X′′; A) Hn(X, X′; A)

Hn−1(X′, X′′; A) Hn−1(X, X′′; A) Hn−1(X, X′; A)

Hn−2(X′, X′′; A) · · ·

∂n

∂n−1

is exact.

Proof. The first step is to make sure that the long exact sequences of the pairs (X′, X′′), (X, X′′) and(X, X′) can be interlaced to form the following “braid diagram” (I will omit the coefficient from thenotation):

Hn(X) Hn(X, X′′) Hn−1(X′, X′′)

Hn(X′) Hn(X, X′′) Hn−1(X′) Hn−1(X, X′′)

Hn(X′, X′′) Hn−1(X′′) Hn−1(X)

f4

g3 h1

h3 i1f3

f2

g1

g4

i2

i3f1

g2 h4

h2 i4

The braid commutes by the naturality of the long exact sequence of a pair (induced by the mapsof pairs (X′, X′′) −→ (X, X′′) and (X, X′′) −→ (X, X′)). Now the claim is that if the green, yellowand blue sequences are exact, so is the red one, which is the desired one (note that h1 here is preciselydefined as i2 h3, as required). The exercise here boils down to a diagram chasing problem of abeliangroups. It needs to be check all possible inclusions one by one. I will just make one of them explicit:

Let’s see for example that Ker g1 ⊂ Im f1: if a ∈ Hn(X, X′′), as 0 = h3g1a = h2g4a,there is b ∈Hn(X′, X′′) such that g2b = g4a. Note that g4( f1b − a) = 0, so f1b − a ∈ Ker g4 = Im f4, so there isd ∈ Hn(X) such that f4d = f1b− a. As g3 = g1 f4, d ∈ Ker g3 = f3, so there is e ∈ Hn(x′) such thatf3e = d. Finally, b− f2e is the desired element: indeed,

f1(b− f2e) = f1b− f1 f2e = f1b− f4 f3e = f1b− f4d = f1b− f1b + a = a.

Alternative proof. Another approach is to come up with a short exact sequence of chain complexes indu-cing the previous long exact sequence in homology. Indeed,

0 −→ C(X′, X′′; A) −→ C(X, X′′; A) −→ C(X, X′; A) −→ 0

is the desired sequence (the homomorphisms are induced by the inclusion C(X′; A) −→ C(X; A) andthe identity of C(X; A), respectively). It remains to the reader to check that it is exact, that the arrowsare chain maps, and that the induced morphisms in homology are the one described before. For thehomomorphism corresponding to the composite, recall that elements of Hn(X, X′; A) are representedby chains c ∈ Cn(X; A) such that ∂nc ∈ Cn−1(X′; A).

5

Exercise 5.1. Let X be a topological space and A an abelian group. Let U, V ⊂ X satisfying X =U ∪

V, and let

U

U ∩V X

V

jU

iV

iU

jV

be the natural inclusion maps. Define groups homomorphisms

∂n : Hn(X; A) −→ Hn−1(U ∩V; A)

such that the sequence of groups homomorphisms

· · · Hn(U ∩V; A) Hn(U; A)⊕ Hn(V; A) Hn(X; A)

Hn−1(U ∩V; A) Hn−1(U; A)⊕ Hn−1(V; A) Hn−1(X; A)

Hn−2(U ∩V; A) · · ·

(iU∗ ,iV∗ )

(jU∗−jV∗

)

∂n

∂n−1

is exact. This is called the Mayer-Vietoris sequence of the cover U, V

Proof. Consider the sequence of chain maps that levelwise is

0 Cn(U ∩V; A) Cn(U; A)⊕ Cn(V; A) Cn(SO ; A) 0(iU∗ ,iV∗ )

(jU∗−jV∗

)(∗)

and note that it is enough to check that it is exact, because since by hypothesis the coverO := U, V isadmissible, the inclusion SO(X) in S(X) induces an isomorphism (small simplices theorem)

Hn(C(SO(X); A))∼−→ Hn(X; A),

and therefore the long exact sequence of the previous sequence (∗) will give the requested diagram(and the morphisms ∂n are precisely the composite of the connecting homomorphisms coming from thecorresponding long exact sequence with the previous isomorphism coming from the small simplicestheorem). Note that here we must also take into account that Hn(C⊕D; A) = Hn(C; A)⊕ Hn(D; A) forchain complexes C, D (exercise 5, sheet 1).

For the sake of simplicity, call i = (iU∗ , iV∗ ) and j =(

jU∗−jV∗

)in (∗), and let us check its exactness:

• i injective:

Note that i injective iff iU∗ injective and iV∗ injective, but both are because they precisely are theinduced morphisms by the inclusions.

• j surjective:

If aσ ∈ Cn(SO(X); A), then σ ∈ S(U)n or σ ∈ S(V)n; so aσ is the image of either aσ or −aσ,respectively. For a finite sum of elements of Cn(SO(X); A) we just consider the correspondingsum of elements in the direct sum.

• Im i = Ker j:

If aσ ∈ Cn(U ∩V; A), i sends it to (aσ, aσ) ∈ Im i, and by j it goes to 0, and linearity proves it fora sum. Conversely, if an element (aσ, bσ′) is in the kernel of j, then aσ− bσ = 0 ∈ Cn(SO(X); A).This implies that σ = σ′ and a = b; hence σ ∈ S(U ∩V)n.

6

Remark. The Mayer-Vietoris sequence is also natural, that is, if f : X −→ Y is a map of spaces and X =U1∪V1 and Y = U2∪V2 with f (U1) ⊂ U2 and f (V1) ⊂ V2, then there is a commutative diagram relatingboth sequences, as for the long exact sequence of a pair. This follows from the fact that the Mayer-Vietoris sequence comes up from the long exact sequence induced by a certain short exact sequence ofchain maps. The conditions on the covers implies that there is a commutative diagram of chain mapswith exact rows

0 C(U1 ∩V1; A) C(U1; A)⊕ C(V1; A) C(SO(X); A) 0

0 C(U2 ∩V2; A) C(U2; A)⊕ C(V2; A) C(SO(Y); A) 0

which induces the desired commutative diagram relating the Mayer-Vietoris sequences.

Remark. The Mayer-Vietoris sequence also holds for de Rham cohomology: for a smooth manifold Mof dimension n, the differential of forms d induces a cochain complex

0 −→ C∞(M)d−→ Ω1(M)

d−→ Ω2(M)d−→ · · · d−→ Ωn(M)

d−→ 0

(the “co” means that the degree goes up), because d2 = 0, and one can take cohomology groupsHk

dR(M) := Ker d/ Im d. For two open sets U, V ⊂ M such that U ∪V = M, the inclusions

U ∩V ⇒ U ä V −→ M

induce exact sequences

0 −→ Ωk(M) −→ Ωk(U)⊕Ωk(V) −→ Ωk(U ∩V) −→ 0

which induce the Mayer-Vietoris sequence in de Rham cohomology.

Exercise 5.2. Compute the homology groups Hn(Sm; A) using the Mayer-Vietoris sequence.

Proof. This is a very easy exercise and I will not spell it out. The proof goes by induction: for m = 0 itboils down to the computation of the homology groups of the one-point space. For m > 0, one take theopen sets

U = (x0, . . . , xm) ∈ Sm : xm > −1/3and

V := (x0, . . . , xm) ∈ Sm : xm < 1/3,which are contractible; and one has that U ∩V is homotopy equivalent to Sm−1.

Exercise 5.3. Let X be a topological space, A an abelian group and n ≥ 0 an integer. Let X′ ⊆ X be a non-emptyclosed subspace of X that is a neighbourhood deformation retract. As usual, we write X/X′ for the quotient of Xthat identifies all points in X′.

(i) Show that the quotient map X −→ X/X′ induces an isomorphism of relative homology groups

Hn(X, X′; A)∼−→ Hn(X/X′, X′/X′; A).

(ii) Show that the relative homology groups Hn(X, X′; A) are isomorphic to the homology groups Hn(X/X′; A),if n ≥ 1, and to H0(X/X′; A)/A if n = 0.

Proof. (i). Let us consider the following commutative diagram, where U is the neighbourhood of X′

such that X′ is a deformation retract of U (and therefore X′ is homotopy equivalent to U):

Hn(X, X′; A) Hn(X, U; A) Hn(X− X′, U − X′; A)

Hn(X/X′, X′/X′; A) Hn(X/X′, U/X′; A) Hn(X/X′ − X′/X′, U/X′ − X′/X′; A)

f

π∗

g

π∗

h i

7

We will show that π∗ is an isomorphism by showing that the rest of labelled arrows in the diagramsare too. By the “2 out of 3” property in the following diagram, we see that f is an isomorphism,

Hn(X′; A) Hn(X; A) Hn(X, X′; A)

Hn(U; A) Hn(X; A) Hn(X, U; A)

∼ ∼ f

because the inclusion X′ −→ U is a homotopy equivalence. Moreover, the homotopy H : U × I −→ Udefines another homotopy H : U/X′ × I −→ U/X′, continuous and well defined by the universalproperty of the quotient topology,

U × I U U/X′

U/X′ × I

H

H

since for x′1, x′2 ∈ X′, we have H(x′1, t) ∈ X′, H(x′2, t) ∈ X′, and hence they are the same point in thequotient U/X′. In particular, the previous diagram says that X′/X′ is a deformation retract of U/X′

with homotopy H. One concludes that h is isomorphism.

But we also have, by excision theorem, that g and i are also isomorphisms. For that just note that

one can apply the theorem since X′ = X′ ⊂ O =O ⊆

U, where O is the open set such that X′ ⊆ O ⊆ U;

and the same analogously with the quotient space (because the definition of the quotient topology).

Finally, it turns out that π∗ is also isomorphism trivially, since

π|X−X′ : X− X′ −→ (X− X′)/X′ = X/X′ − X′/X′,

π|U−X′ : U − X′ −→ (U − X′)/X′ = U/X′ − X′/X′,

are homeomorphisms (there’s nothing to identify!).

Altogether, one has that π∗ is a group isomorphism because it is composite of isomorphisms.

(ii). Just note that the previous isomorphism tells us that

Hn(X, X′; A) = Hn(X/X′, X′/X′; A) = Hn(X/X′, ∗; A) =

Hn(X/X′; A) n > 0H0(X/X′; A)/H0(∗; A) n = 0

,

where in the last equality we have used the long exact sequence of the pair (X/X′, ∗).

Exercise 6.3. Let X be a topological space, A an abelian group and n ≥ 0.

(i) Construct a natural isomorphism Hn(X; A) ' Hn+1(ΣX; A) relating the n-th reduced homology group ofX to the (n + 1)-st reduced homology group of its suspension.

(ii) What is ΣSn?

Proof. (i). Consider the cones

C•X :=X× [0, 1/2 + ε]

X× 0 , C•X :=X× [1/2− ε, 1]

X× 1 ,

and note that they form a cover of the suspension ΣX.

8

Analogously to the unreduced case, there are Mayer-Vietoris sequences and long exact sequences inreduced homology. By applying the reduced Mayer-Vietoris sequence to this cover, we obtain that thefollowing sequence is exact:

Hn+1(X× (1/2− ε, 1/2 + ε); A) Hn+1(C•X; A)⊕ Hn+1(C•X; A) Hn+1(ΣX; A)

Hn(X× (1/2− ε, 1/2 + ε) Hn(C•X; A)⊕ Hn(C•X; A) Hn(ΣX; A)

Hn−1(X× (1/2− ε, 1/2 + ε); A) · · ·

Now, X × (1/2− ε, 1/2 + ε) is homotopy equivalent to X via a deformation retraction, and bothC•X and C•X are contractible1, hence Hn(C•X; A) ⊕ Hn(C•X; A) = 0 for all n ≥ 0, so the previousexact sequence becomes

· · · −→ Hn+1(X; A) −→ 0 −→ Hn+1(ΣX; A) −→ Hn(X; A) −→ 0 −→ · · · ,

and therefore we obtain an isomorphism (given by the connecting homomorphism)

Hn+1(ΣX; A) ' Hn(X; A).

The statement about naturality means that if Y is another topological space and f : X −→ Y is acontinuous map, then we get a commutative diagram

Hn+1(ΣX; A) Hn(X; A)

Hn+1(ΣY; A) Hn(Y; A)

∼

F∗ f∗

∼

where F∗ is induced by the map F : ΣX −→ ΣY, F([(x, t)]) = [( f (x), t)] induced by f in the suspensionof the spaces. This follows from the naturality of the Mayer-Vietoris sequence for the cover C•Y∪C•Y =ΣY and the corresponding induced morphisms (see remark after 5.1).

(ii). We will prove that ΣSn = Sn+1. For the sake of simplicity, we will take the suspension of atopological space X as the quotient (X × [−1, 1])/ ∼ where we identify the points in X × −1 and inX× 1.

Consider the continuous map

Sn × [−1, 1] Sn+1((x0, . . . , xn), t

)(√

1− t2x0, . . . ,√

1− t2xn, t)

It satisfies that equivalent points via ∼ are mapped to the same point (all the points in Sn × −1and Sn × 1 , that are equivalent, are mapped onto the south and north pole, respectively). Therefore,by the Universal Property of the quotient topology, we get a continuous map

ϕ :Sn × [−1, 1]

∼ = ΣSn −→ Sn+1.

We see that it is a bijection, as its inverse morphism is

Sn×[−1,1]∼ = ΣSn Sn+1[(

( y0√1−y2

n+1

, . . . , yn√1−y2

n+1

), yn+1

)](y0, . . . , yn, yn+1)

1A cone is always contractible: in general, given a topological space X, define CX := X×[0,1]X×0 . It turns out that the map

H : CX × [0, 1] −→ CX, H([(x, t)], s) = [(x, (1− s)t)] is a homotopy in CX from the identity on CX to the point X × 0, henceCX is contractible.

9

Since the source is a compact space, and the target is a Hausdorff space, we conclude that ϕ is homeo-morphism, as required.

Exercise 7.2. Let X be a topological space and let f , g : ∂Dn −→ X be two continuous maps. Let X f⋃

∂Dn Dn

be the space obtained by attaching an n−cell to X with attaching map f and let Xg⋃

∂Dn Dn be the space obtainedby attaching an n-cell with attaching map g.

Show that X f⋃

∂Dn Dn and Xg⋃

∂Dn Dn are homotopy equivalent if f and g are homotopic.

Proof. Let H : ∂Dn × I −→ X be the homotopy between f and g. The claim is that X f⋃

∂Dn Dn

is homotopy equivalent to XH⋃

∂Dn×I Dn × I, and similarly Xg⋃

∂Dn Dn is homotopy equivalent toXH

⋃∂Dn×I Dn × I. As the homotopy equivalence is an equivalence relation, we are done with the

claim.We will show that X f

⋃∂Dn Dn is homotopy equivalent to XH

⋃∂Dn×I Dn × I (with g is analogous).

The key property for that is that the cylinder Dn × I deformation retracts to ∂Dn × I ∪Dn × I, as weused in proposition 11.10 in the lecture notes. Denote as F : (Dn × I)× I −→ Dn × I the homotopyrelating F0 = IdDn×I with the retraction F1 = r.

We will now prove that XH⋃

∂Dn×I Dn × I deformation retracts to X f⋃

∂Dn Dn = XH0

⋃∂Dn×I Dn ×

0 ⊂ XH⋃

∂Dn×I Dn × I. By lemma 7.12,

(XH ∪∂Dn×I Dn × I)× I ∼= X× IH×Id ∪(∂Dn×I)×I (Dn × I)× I,

so to give a homotopy from XH⋃

∂Dn×I Dn× I to another space is the same thing as to give homotopiesfrom X× I and Dn × I to the space which “coincide” in ∂Dn × I.

Denote as π1 and π2 the projections from X and Dn × I to the pushout XH⋃

∂Dn×I Dn × I, respect-ively, and define φ1 as the composite

X× Ipr1−→ X

π1−→ XH ∪∂Dn×I Dn × I

and φ2 as the composite

(Dn × I)× I F×Id−→ Dn × Iπ2−→ XH ∪∂Dn×I Dn × I,

so we get the diagram of below with solid arrows (and the square is a pushout diagram):

(∂Dn × I)× I X× I

(Dn × I)× I (XH ∪∂Dn×I Dn × I)× I

XH ∪∂Dn×I Dn × I

H×Id

i×Idφ1

φ2

G

Moreover, for ((y, t), s) ∈ (∂Dn × I)× I,

(φ1 (H × Id))((y, t), s) = φ1(H(y, t), s) = [H(y, t)] = [(y, t)] = φ2((y, t), s)= (φ2 (i× Id))((y, t), s)

as F|∂Dn×I = Id. Therefore, the outer diagram commutes and by the universal property of the pushoutwe get the dashed arrow G. This is the desired homotopy which deformation retracts XH ∪∂Dn×I Dn× Iinto X f

⋃∂Dn Dn. Indeed, G0 = Id and G1 maps to XH

⋃∂Dn×I(∂Dn × I ∪Dn × I) ∼= XH0

⋃∂Dn×I Dn ×

0 = X f⋃

∂Dn Dn since all points in ∂Dn× I are already identified with points of X in the pushout.

Exercise 9.2. Construct a CW structure on S∞ and show that it is contractible.

10

Proof. Consider the flitration

S0 −→ S1 −→ S2 −→ · · · −→ Sn −→ Sn+1 −→ · · · −→ S∞,

where Sn is the n-skeleton and as we saw Sn+1 arises from Sn by attaching two (n + 1)-cells. Observethat by definition, S∞ =

⋃n≥0 Sn is topologized with the subspace topology of R∞, so it remains to

check that the weak topology in S∞ induced by the filtration is the same as the subspace topology ofR∞. This can be seen, for instance, using the commutativity of the diagram

· · · Sn Sn+1 · · · S∞

· · · Rn Rn+1 · · · R∞

To see that S∞ is contractible, we use the map f : S∞ −→ S∞, f (x0, x1, x2, . . .) := (0, x0, x1, . . .). Theclaim is that f is homotopic both to the identity and to constp, p = (1, 0, 0, . . .), so S∞ is contractible.Indeed, we use the homotopies

H : S∞ × I −→ S∞ , H(x, t) :=(1− t)x + t f (x)|(1− t)x + t f (x)|

and

F : S∞ × I −→ S∞ , F(x, t) :=(1− t) f (x) + t constp(x)|(1− t) f (x) + t constp(x)| .

It is readily verified that in both cases the denominator does not vanish, so the maps are continuous.

Remark. A good way to remember the definition of the Homotopy Extension Property (HEP) of a pair(X, A) is that given the below commutative diagram with solid arrows, there exits a dashed arrow suchthat both triangles commute,

A× 0 X× 0

A× I X× I

Z

f

F

H

Exercise 11.1. Let (X, A) be a pair of spaces with the homology extension property and assume that the inclusionmap i : A −→ X is homotopic to a constant map. Let p : X −→ X/A be the quotient map to the space obtainedfrom X by collapsing A to a point. Show that there exists a continuous map r : X/A −→ X such that r p ishomotopic to the identity on X

Proof. Let F : A × [0, 1] −→ X be the homotopy between the inclusion i = F0 and the constant mapF1 = constx0 , and set f = IdX in the diagram of the remark. By the HEP of the pair (X, A), F extendsto a homotpy H : X × I −→ X. Consider H1 : X −→ X. By the universal property of the quotienttopology, there exists a unique continuous map r : X/A −→ X such that H1 = r p, since H1(a) = x0for all a ∈ A. We conclude as H1 ' H0 = IdX .

Exercise 11.2. Let (X, A) be a pair of spaces with the homotopy extension property and assume that A is acontractible space.

(i) Show that the quotient map p : X −→ X/A is a homotopy equivalence.

(ii) Give an example of a pair of spaces that satisfy the assumptions of Exercise 11.1 but not the assumptions ofthe present exercise.

Proof. (i). Being A contractible, the identity on such space is homotopic to a constant map. Composingthe inclusion i with such a homotopy yields a homotopy F : A × [0, 1] −→ X for which i ' constx0 .Then the previous exercise provides a homotopy H and a map r such that r p ' IdX , so it remains toshow that p r ' IdX/A.

11

The homotopy H induces a homotopy H : X/A × [0, 1] −→ X/A arising in the following form:consider the composite p H : X × I −→ X/A and the map p × Id : X × I −→ X/A × I, whichis a quotient map as I is locally compact (proposition 7.15). For a point (a, t) ∈ X × I with a ∈ A,(p H)(a, t) = [a] ∈ X/A since H|A×I = F and F has image in A. Therefore, by the universal propertyof the quotient topology, there exists H : X/A× [0, 1] −→ X/A. This is the desired homotopy betweenIdX/A and p r, as

H([x], 0) = [H(x, 0)] = [x]

andH([x], 1) = [H(x, 1)] = p(H(x, 1)) = (p r p)(x) = (p r)[x].

(ii). Take (D2, ∂D2 = S1). This pair has the HEP (proposition 11.10), and the inclusion i : S1 −→ D2

is homotopic to a constant map (π1(D2) = 0), but S1 is not contractible as space

12