Algebraic Topology [lecture notes]

81 Pages·2017·0.63 MB·English

by  Birgit Richter

Algebraic Topology, summer term 2017
Birgit Richter
Fachbereich Mathematik der Universita¨t Hamburg, Bundesstraße 55, 20146 Hamburg,
Germany
E-mail address: [email protected]

CHAPTER 1
Homology theory
1. Chain complexes
Definition 1.1. A chain complex is a sequence of abelian groups, (Cn)n∈Z, together with homomor-
phisms d : C →C for n∈Z, such that d ◦d =0.
n n n−1 n−1 n
Let R be an associative ring with unit 1 . A chain complex of R-modules can analoguously be defined
R
as a sequence of R-modules (Cn)n∈Z with R-linear maps dn: Cn →Cn−1 with dn−1◦dn =0.
Definition 1.2.
• The d are differentials or boundary operators.
n
• The x∈C are called n-chains.
n
• Is x∈C and d x=0, then x is an n-cycle.
n n
Z (C):={x∈C |d x=0}.
n n n
• If x∈C is of the form x=d y for some y ∈C , then x is an n-boundary.
n n+1 n+1
B (C):=Im(d )={d y,y ∈C }.
n n+1 n+1 n+1
Note that the cycles and boundaries form subgroups of the chains. As d ◦d =0, we know that the
n n+1
image of d is a subgroup of the kernel of d and thus
n+1 n
B (C)⊂Z (C).
n n
We’ll often drop the subscript n from the boundary maps and we’ll just write C for the chain complex.
∗
Definition 1.3. The abelian group H (C):=Z (C)/B (C) is the nth homology group of the complex
n n n
C .
∗
Notation: We denote by [c] the equivalence class of a c∈Z (C).
n
If c,c(cid:48) ∈C satisfy that c−c(cid:48) is a boundary, then c is homologous to c(cid:48). That’s an equivalence relation.
n
Examples:
1) Consider
(cid:40)
Z n=0,1
C =
n
0 otherwise
and let d be the multiplication with N ∈N, then
1
(cid:40)
Z/NZ n=0
H (C)=
n
0 otherwise.
2) Take C =Z for all n∈Z and
n
(cid:40)
idZ n odd
d =
n
0 n even.
What is the homology of this chain complex?
2’) Consider C =Z for all n∈Z again, but let all boundary maps be trivial. What is the homology of this
n
chain complex?
3
Definition 1.4. Let C and D be two chain complexes. A chain map f: C → D is a sequence of
∗ ∗ ∗ ∗
homomorphisms f : C →D such that dD◦f =f ◦dC for all n, i.e., the diagram
n n n n n n−1 n
C dCn (cid:47)(cid:47)C
n n−1
fn fn−1
(cid:15)(cid:15) (cid:15)(cid:15)
D dDn (cid:47)(cid:47)D
n n−1
commutes for all n.
Such an f sends cycles to cycles and boundaries to boundaries. We therefore obtain an induced map
H (f): H (C)→H (D)
n n n
via H (f) [c]=[f c].
n ∗ n
There is a chain map from the chain complex mentioned in Example 1) to the chain complex D that is
∗
concentrated in degree zero and has D = Z/NZ. Note, that (f ) is an isomorphism on zeroth homology
0 0 ∗
groups.
Are there chain maps between the complexes from Examples 2) and 2’)?
Lemma 1.5. If f: C →D and g: D →E are two chain maps, then H (g)◦H (f)=H (g◦f) for
∗ ∗ ∗ ∗ n n n
all n.
When do two chain maps induce the same map on homology?
Definition 1.6. A chain homotopy H between two chain maps f,g: C → D is a sequence of homo-
∗ ∗
morphisms (Hn)n∈Z with Hn: Cn →Dn+1 such that for all n
dD ◦H +H ◦dC =f −g .
n+1 n n−1 n n n
... dCn+2 (cid:47)(cid:47)C dCn+1 (cid:47)(cid:47)C dCn (cid:47)(cid:47)C dCn−1 (cid:47)(cid:47)...
n+1 n n−1
Hn+1 fn+1 gn+1 Hn fn gn Hn−1 fn−1 gn−1
...(cid:119)(cid:119) dDn+2 (cid:47)(cid:47)D(cid:21)(cid:21) (cid:9)(cid:9) (cid:119)(cid:119) dDn+1 (cid:47)(cid:47)D(cid:21)(cid:21) (cid:9)(cid:9) (cid:119)(cid:119) dDn (cid:47)(cid:47)D(cid:21)(cid:21) (cid:9)(cid:9) dDn−1 (cid:47)(cid:47)...
n+1 n n−1
If such an H exists, then f and g are (chain) homotopic: f ∼g.
We will later see geometrically defined examples of chain homotopies.
Proposition 1.7.
(a) Being chain homotopic is an equivalence relation.
(b) If f and g are homotopic, then H (f)=H (g) for all n.
n n
Proof. (a) If H is a homotopy from f to g, then −H is a homotopy from g to f. Each f is homotopic
to itself with H =0. If f is homotopic to g via H and g is homotopic to h via K, then f is homotopic to h
via H +K.
(b) We have for every cycle c∈Z (C ):
n ∗
H (f)[c]−H (g)[c]=[f c−g c]=[dD ◦H (c)]+[H ◦dC(c)]=0.
n n n n n+1 n n−1 n
(cid:3)
Definition 1.8. Let f: C → D be a chain map. We call f a chain homotopy equivalence, if there is
∗ ∗
a chain map g: D →C such that g◦f (cid:39)id and f◦g (cid:39)id . The chain complexes C and D are then
∗ ∗ C∗ D∗ ∗ ∗
chain homotopically equivalent.
Note, that such chain complexes have isomorphic homology. However, chain complexes with isomorphic
homology do not have to be chain homotopically equivalent. (Can you find a counterexample?)
4
Definition 1.9. IfC andC(cid:48) arechaincomplexes, thentheirdirect sum, C ⊕C(cid:48),isthechaincomplex
∗ ∗ ∗ ∗
with
(C ⊕C(cid:48)) =C ⊕C(cid:48) =C ×C(cid:48)
∗ ∗ n n n n n
with differential d=d given by
⊕
d (c,c(cid:48))=(dc,dc(cid:48)).
⊕
Similarly, if(C(j),d(j)) isafamilyofchaincomplexes, thenwecandefinetheirdirectsumasfollows:
∗ j∈J
((cid:77)C(j)) :=(cid:77)C(j)
∗ n n
j∈J j∈J
as abelian groups and the differential d is defined via the property that its restriction to the jth summand
⊕
is d(j).
2. Singular homology
Let v ,...,v be n+1 points in Rn+1. Consider the convex hull
0 n
n n
(cid:88) (cid:88)
K(v ,...,v ):={ t v | t =1,t (cid:62)0}.
0 n i i i i
i=0 i=0
Definition 2.1. If the vectors v −v ,...,v −v are linearly independent, then K(v ,...,v ) is the
1 0 n 0 0 n
simplex generated by v ,...,v . We denote such a simplex by simp(v ,...,v ).
0 n 0 n
Example. The standard topological n-simplex is ∆n :=simp(e ,...,e ). Here, e is the vector in Rn+1 that
0 n i
has a 1 in coordinate i+1 and is zero in all other coordinates. The first examples are: ∆0 is the point e ,
0
∆1 is the line segment between e and e , ∆2 is a triangle in R3 and ∆3 is homeomorphic to a tetrahedron.
0 1
The coordinate description of the n-simplex is
(cid:88)
∆n ={(t ,...,t )∈Rn+1| t =1,t (cid:62)0}.
0 n i i
We consider ∆n as ∆n ⊂Rn+1 ⊂Rn+2 ⊂....
The boundary of ∆1 consists of two copies of ∆0, the boundary of ∆2 consists of three copies of ∆1. In
general, the boundary of ∆n consists of n+1 copies of ∆n−1.
We need the following face maps for 0(cid:54)i(cid:54)n
d =dn−1: ∆n−1 (cid:44)→∆n;(t ,...,t )(cid:55)→(t ,...,t ,0,t ,...,t ).
i i 0 n−1 0 i−1 i n−1
Theimageofdn−1in∆nisthefacethatisoppositetoe . Itisthesimplexgeneratedbye ,...,e ,e ,...,e .
i i 0 i−1 i+1 n
Draw the examples of the faces in ∆1 and ∆2!
Lemma 2.2. Concerning the composition of face maps, the following rule holds:
dn−1◦dn−2 =dn−1◦dn−2, 0(cid:54)j <i(cid:54)n.
i j j i−1
Example: face maps for ∆0 and composition into ∆2: d ◦d =d ◦d .
2 0 0 1
Proof. Both expressions yield
dn−1◦dn−2(t ,...,t )=(t ,...,t ,0,...,0,t ,...,t )=dn−1dn−2(t ,...,t ).
i j 0 n−2 0 j−1 i−1 n−2 j i−1 0 n−2
(cid:3)
Let X be an arbitrary topological space, X (cid:54)=∅.
Definition 2.3. A singular n-simplex in X is a continuous map α: ∆n →X.
Note, that α just has to be continuous, not smooth or anything!
Definition 2.4. Let S (X) be the free abelian group generated by all singular n-simplices in X. We
n
call S (X) the nth singular chain module of X.
n
5
Elements of S (X) are finite sums (cid:80) λ α with λ =0 for almost all i∈I and α : ∆n →X.
n i∈I i i i i
For all n(cid:62)0 there are non-trivial elements in S (X), because we assumed that X (cid:54)=∅: we can always
n
takeanx ∈X andtheconstantmapκ : ∆n →X asα. Byconvention,wedefineS (∅)=0foralln(cid:62)0.
0 x0 n
If we want to define maps from S (X) to some abelian group then it suffices to define such a map on
n
generators.
Example. What is S (X)? A continuous α: ∆0 →X is determined by its value α(e )=:x ∈X, which is a
0 0 α
(cid:80) (cid:80)
pointinX. Asingular0-simplex λ α canthusbeidentifiedwiththeformalsumofpoints λ x .
i∈I i i i∈I i αi
For instance if you count the zeroes and poles of a meromorphic function with multiplicities then this gives
an element in S (X). In algebraic geometry a divisor is an element in S (X).
0 0
Definition 2.5. We define ∂ : S (X)→S (X) on generators
i n n−1
∂ (α)=α◦dn−1
i i
and call it the ith face of α.
On S (X) we therefore get ∂ ((cid:80) λ α )=(cid:80) λ (α ◦dn−1).
n i j j j j j j i
Lemma 2.6. The face maps on S (X) satisfy
n
∂ ◦∂ =∂ ◦∂ , 0(cid:54)j <i(cid:54)n.
j i i−1 j
Proof. The proof follows from the one of Lemma 2.2. (cid:3)
Definition 2.7. We define the boundary operator on singular chains as ∂: S (X) → S (X), ∂ =
n n−1
(cid:80)n (−1)i∂ .
i=0 i
Lemma 2.8. The map ∂ is a boundary operator, i.e., ∂◦∂ =0.
Proof. We calculate
n−1 n
(cid:88) (cid:88) (cid:88)(cid:88)
∂◦∂ =( (−1)j∂ )◦( (−1)i∂ )= (−1)i+j∂ ◦∂
j i j i
j=0 i=0
(cid:88) (cid:88)
= (−1)i+j∂ ◦∂ + (−1)i+j∂ ◦∂
j i j i
0(cid:54)j<i(cid:54)n 0(cid:54)i(cid:54)j(cid:54)n−1
(cid:88) (cid:88)
= (−1)i+j∂ ◦∂ + (−1)i+j∂ ◦∂ =0.
i−1 j j i
0(cid:54)j<i(cid:54)n 0(cid:54)i(cid:54)j(cid:54)n−1
(cid:3)
We therefore obtain the singular chain complex, S (X),
∗
∂ ∂ ∂ ∂
...→S (X)−→S (X)−→...−→S (X)−→S (X)→0.
n n−1 1 0
We abbreviate Z (S (X)) by Z (X), B (S (X)) by B (X) and H (S (X)) by H (X).
n ∗ n n ∗ n n ∗ n
Definition 2.9. For a space X, H (X) is the nth singular homology group of X.
n
Note that Z (X)=S (X).
0 0
As an example of a 1-cycle consider a 1-chain c=α+β+γ where α,β,γ: ∆1 →X such that α(e )=
1
β(e ), β(e )=γ(e ) and γ(e )=α(e ) and calculate that ∂c=0.
0 1 0 1 0
We need to understand how continuous maps of topological spaces interact with singular chains and
singular homology.
Let f: X →Y be a continuous map.
Definition 2.10. The map f =S (f): S (X)→S (Y) is defined on generators α: ∆n →X as
n n n n
f (α)=f ◦α:∆n −α→X −f→Y.
n
6
Lemma 2.11. For any continuous f: X →Y we have
S (X) fn (cid:47)(cid:47)S (Y)
n n
∂X ∂Y
(cid:15)(cid:15) (cid:15)(cid:15)
S (X) fn−1 (cid:47)(cid:47)S (Y),
n−1 n−1
i.e., (f ) is a chain map and hence induces a map H (f): H (X)→H (Y).
n n n n n
Proof. By definition
n n
(cid:88) (cid:88)
∂Y(f (α))= (−1)i(f ◦α)◦d = (−1)if ◦(α◦d )=f (∂Xα).
n i i n−1
i=0 i=0
(cid:3)
Of course, the identity map on X induces the identity map on H (X) for all n (cid:62) 0 and if we have a
n
composition of continuous maps
f g
X −→Y −→Z,
then S (g◦f)=S (g)◦S (f) and H (g◦f)=H (g)◦H (f). In categorical language, this says precisely
n n n n n n
that S (−) and H (−) are functors from the category of topological spaces and continuous maps into the
n n
category of abelian groups. Taking all S (−) together turns S (−) into a functor from topological spaces
n ∗
and continuous maps into the category of chain complexes with chain maps as morphisms.
One implication of Lemma 2.11 is that homeomorphic spaces have isomorphic homology groups:
X ∼=Y ⇒H (X)∼=H (Y) for all n(cid:62)0.
n n
Our first (not too exciting) calculation is the following:
Proposition 2.12. The homology groups of a one-point space pt are trivial but in degree zero,
(cid:40)
0, if n>0,
H (pt)∼=
n Z, if n=0.
Proof. For every n(cid:62)0 there is precisely one continuous map α: ∆n →pt, namely the constant map.
We denote this map by κ . Then the boundary of κ is
n n
n n (cid:40)
∂κ =(cid:88)(−1)iκ ◦d =(cid:88)(−1)iκ = κn−1, n even,
n n i n−1
0, n odd.
i=0 i=0
For all n we have S (pt)∼=Z generated by κ and therefore the singular chain complex looks as follows:
n n
... ∂=0 (cid:47)(cid:47)Z∂=idZ(cid:47)(cid:47)Z ∂=0 (cid:47)(cid:47)Z.
(cid:3)
3. H and H
0 1
Before we calculate anything, we define a map.
Proposition 3.1. For any topological space X there is a homomorphism ε: H (X)→Z with ε(cid:54)=0 for
0
X (cid:54)=∅.
Proof. IfX (cid:54)=∅,thenwedefineε(α)=1foranyα: ∆0 →X,thusε((cid:80) λ α )=(cid:80) λ onS (X).
i∈I i i i∈I i 0
As only finitely many λ are non-trivial, this is in fact a finite sum.
i
We have to show that this map is well-defined on homology, i.e., that it vanishes on boundaries. One
possibilityistoseethatεcanbeinterpretedasthemaponsingularchainsthatisinducedbytheprojection
map of X to a one-point space.
7
(cid:80)
One can also show the claim directly: Let S (X)(cid:51)c=∂b be a boundary and write b= ν β with
0 i∈I i i
β : ∆1 →X. Then we get
i
(cid:88) (cid:88) (cid:88) (cid:88)
∂b=∂ ν β = ν (β ◦d −β ◦d )= ν β ◦d − ν β ◦d
i i i i 0 i 1 i i 0 i i 1
i∈I i∈I i∈I i∈I
and hence
(cid:88) (cid:88)
ε(c)=ε(∂b)= ν − ν =0.
i i
i∈I i∈I
(cid:3)
We said that S (∅) is zero, so H (∅)=0 and in this case we define ε to be the zero map.
0 0
If X (cid:54)= ∅, then any α: ∆0 → X can be identified with its image point, so the map ε on S (X) counts
0
points in X with multiplicities.
Proposition 3.2. If X is a path-connected, non-empty space, then ε: H (X)∼=Z.
0
Proof. As X is non-empty, there is a point x∈X and the constant map κ with value x is an element
x
in S (X) with ε(κ ) = 1. Therefore ε is surjective. For any other point y ∈ X there is a continuous path
0 x
ω: [0,1]→X with ω(0)=x and ω(1)=y. We define α : ∆1 →X as
ω
α (t ,t )=ω(1−t ).
ω 0 1 0
Then
∂(α )=∂ (α )−∂ (α )=α (e )−α (e )=α (0,1)−α (1,0)=κ −κ ,
ω 0 ω 1 ω ω 1 ω 0 ω ω y x
and the two generators κ ,κ are homologous. This shows that ε is injective. (cid:3)
x y
From now on we will identify paths w and their associated 1-simplices α .
w
Corollary 3.3. If X is of the form X =(cid:70) X such that the X are non-empty and path-connected,
i∈I i i
then
H (X)∼=(cid:77)Z.
0
i∈I
Inthiscase,thezerothhomologygroupofX isthefreeabeliangroupgeneratedbythepath-components.
Proof. The singular chain complex of X splits as the direct sum of chain complexes of the X :
i
S (X)∼=(cid:77)S (X )
n n i
i∈I
for all n. Boundary summands ∂ stay in a component, in particular,
i
∂: S (X)∼=(cid:77)S (X )→(cid:77)S (X )∼=S (X)
1 1 i 0 i 0
i∈I i∈I
is the direct sum of the boundary operators ∂: S (X )→S (X ) and the claim follows. (cid:3)
1 i 0 i
Next, we want to relate H to the fundamental group. Let X be path-connected and x∈X.
1
Lemma 3.4. Let ω ,ω ,ω be paths in X.
1 2
(a) Constant paths are null-homologous.
(b) If ω (1) = ω (0), then ω ∗ω −ω −ω is a boundary. Here ω ∗ω is the concatenation of ω
1 2 1 2 1 2 1 2 1
followed by ω .
2
(c) If ω (0)=ω (0),ω (1)=ω (1) and if ω is homotopic to ω relative to {0,1}, then ω and ω are
1 2 1 2 1 2 1 2
homologous as singular 1-chains.
(d) Any 1-chain of the form ω¯ ∗ω is a boundary. Here, ω¯(t):=ω(1−t).
8
Proof. For a), consider the constant singular 2-simplex α(t ,t ,t ) = x and c , the constant path on
0 1 2 x
x. Then ∂α=c −c +c =c .
x x x x
For b), we define a singular 2-simplex β: ∆2 →X as follows.
e
2
(cid:1)(cid:21)(cid:65)(cid:75)
(cid:1) (cid:65)
(cid:81)(cid:81)
ω ∗ω (cid:1) (cid:81)(cid:65) ω
1 2(cid:1)(cid:81)(cid:81) (cid:65) 2
(cid:81)
(cid:1)(cid:81)(cid:81) (cid:81)(cid:65)
(cid:1) (cid:81) (cid:45)(cid:81)(cid:65)
e ω e
0 1 1
We define β on the boundary components of ∆2 as indicated and prolong it constantly along the sloped
inner lines. Then
∂β =β◦d −β◦d +β◦d =ω −ω ∗ω +ω .
0 1 2 2 1 2 1
For c): Let H: [0,1]×[0,1]→X a homotopy from ω to ω . As we have that H(0,t)=ω (0)=ω (0),
1 2 1 2
we can factor H over the quotient [0,1]×[0,1]/{0}×[0,1]∼=∆2 with induced map h: ∆2 →X. Then
∂h=h◦d −h◦d +h◦d .
0 1 2
The first summand is null-homologous, because it’s constant (with value ω (1) = ω (1)), the second one is
1 2
ω and the last is ω , thus ω −ω is null-homologous.
2 1 1 2
For d): Consider γ: ∆2 →X as indicated below.
e
2
(cid:1)(cid:21)(cid:65)(cid:75)
(cid:1) (cid:65)
(cid:1) (cid:1)(cid:1)(cid:65)
ω(1) ω
(cid:1) (cid:1) (cid:1)(cid:65)
(cid:1) (cid:1) (cid:1) (cid:65)
(cid:1)
(cid:1) (cid:1) (cid:1) (cid:1)(cid:45)(cid:65)
e ω¯ e
0 1
(cid:3)
Definition 3.5. Let h: π (X,x)→H (X) be the map, that sends the homotopy class of a closed path
1 1
ω, [ω] , to its homology class [ω]=[ω] . This map is called the Hurewicz-homomorphism.
π1 H1
Witold Hurewicz: 1904–1956.
Lemma 3.4 ensures that h is well-defined and
h([ω ][ω ])=h([ω ∗ω ])=[ω ]+[ω ]=h([ω ])+h([ω ])
1 2 1 2 1 2 1 2
thus h is a homomorphism.
Note that for a closed path ω we have that [ω¯]=−[ω] in H (X).
1
Definition 3.6. Let G be an arbitrary group, then its abelianization, G , is G/[G,G].
ab
Recall that [G,G] is the commutator subgroup of G. That is the smallest subgroup of G containing all
commutatorsghg−1h−1,g,h∈G. ItisanormalsubgroupofG: Ifc∈[G,G],thenforanyg ∈Gtheelement
gcg−1c−1 is a commutator and also by the closure property of subgroups the element gcg−1c−1c=gcg−1 is
in the commutator subgroup.
Proposition 3.7. The Hurewicz homomorphism factors over the abelianization of π (X,x) and induces
1
an isomorphism
π (X,x) ∼=H (X)
1 ab 1
9
for all path-connected X.
π (X,x) h (cid:47)(cid:47)(cid:51)(cid:51)H (X)
1 1
∼=
p
(cid:15)(cid:15) hab
π (X,x) =π (X,x)/[π (X,x),π (X,x)]
1 ab 1 1 1
Proof. Wewillconstructaninversetoh . Foranyy ∈X wechooseapathu fromxtoy. Fory =x
ab y
we take u to be the constant path on x. Let α be an arbitrary singular 1-simplex and y = α(e ). Define
x i i
φ: S (X)→π (X,x) ongeneratorsasφ(α)=[u ∗α∗u¯ ]andextendφlinearlytoallofS (X), keeping
1 1 ab y0 y1 1
in mind that the composition in π is written multiplicatively.
1
We have to show that φ is trivial on boundaries, so let β: ∆2 →X. Then
φ(∂β)=φ(β◦d −β◦d +β◦d )=φ(β◦d )φ(β◦d )−1φ(β◦d ).
0 1 2 0 1 2
Abbreviating β◦d with α we get as a result
i i
[u ∗α ∗u¯ ][u ∗α ∗u¯ ]−1[u ∗α ∗u¯ ]=[u ∗α ∗u¯ ∗u ∗α ∗u¯ ∗u ∗α¯ ∗u¯ ].
y1 0 y2 y0 1 y2 y0 2 y1 y0 2 y1 y1 0 y2 y2 1 y0
Here, we’ve used that the image of φ is abelian. We can reduce u¯ ∗u and u¯ ∗u and are left with
y1 y1 y2 y2
[u ∗α ∗α ∗α¯ ∗u¯ ] but α ∗α ∗α¯ is the closed path tracing the boundary of β and therefore it is
y0 2 0 1 y0 2 0 1
null-homotopic in X. Thus φ(∂β)=0 and φ passes to a map
φ: H (X)→π (X,x) .
1 1 ab
The composition φ◦h evaluated on the class of a closed path ω gives
ab
φ◦h [ω] =φ[ω] =[u ∗ω∗u¯ ] .
ab π1 H1 x x π1
But we chose u to be constant, thus φ◦h =id.
x ab
(cid:80)
If c = λ α is a cycle, then h ◦φ(c) is of the form [c+D ] where the D -part comes from the
i i ab ∂c ∂c
contributions of the u . The fact that ∂(c) = 0 implies that the summands in D cancel off and thus
yi ∂c
h ◦φ=id . (cid:3)
ab H1(X)
Note, that abelianization doesn’t change anything for abelian groups, i.e., whenever we have an abelian
fundamental group, we know that H (X)∼=π (X,x).
1 1
Corollary 3.8. Knowledge of π gives
1
H (Sn)=0, for n>1, H (S1)∼=Z,
1 1
H (S1×...×S1)∼=Zn,
1
(cid:124) (cid:123)(cid:122) (cid:125)
n
H (S1∨S1)∼=(Z∗Z) ∼=Z⊕Z,
1 ab
(cid:40)
Z, if n=1,
H (RPn)∼=
1 Z/2Z, for n>1.
4. Homotopy invariance
We want to show that two continuous maps that are homotopic induce identical maps on the level of
homology groups.
Heuristics: If α: ∆n → X is a singular n-simplex and if f,g are homotopic maps from X to Y, then
the homotopy from f ◦α to g◦α starts on ∆n×[0,1]. We want to translate this geometric homotopy into
a chain homotopy on the singular chain complex. To that end we have to cut the prism ∆n ×[0,1] into
(n+1)-simplices. In low dimensions this is easy:
∆0×[0,1] is homeomorphic to ∆1, ∆1×[0,1] ∼= [0,1]2 and this can be cut into two copies of ∆2 and
∆2×[0,1] is a 3-dimensional prism and that can be glued together from three tetrahedrons, e.g., like
10