# Introduction to simplicial homology

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```							                  Introduction to simplicial homology
Marcus Pivato
December 17, 2003

Topology studies curves, surfaces, networks, and other ‘spaces’, and concerns the properties
of these spaces which remain invariant under deformation. Topology can be roughly divided in
two parts:

Microtopology studies the ‘microscopic’ or ‘local’ structure of a space. For example, the set
R of real numbers and the set Q of rational numbers look superﬁcially the same (they
are both ‘lines’), but they diﬀer at a microscopic level (R is an unbroken continuum,
whereas Q has inﬁnitely many microscopic ‘holes’). Microtopology studies properties like
continuity (of functions) convergence (of sequences), or density (of subsets), and is usually
studied in courses called ‘metric spaces’ or ‘point-set topology’ or ‘general topology’.

Macrotopology studies ‘macroscopic’ or ‘global’ properties of a space. For example, there is a
saying that, ‘A topologist can’t tell the diﬀerence between a coﬀee cup and a donut’ —the
reason being that both are 3-dimensional objects which have a single ‘hole’ or ‘handle’ in
them. In a ‘Salvador Dali’ universe, a coﬀee cup can easily deform into a donut (Fig 1).
Likewise, although a square is geometrically diﬀerent from a circle (the square has four
straight lines which join at sharp corners, while the circle is a single smooth curve), they
are topologically the same (both consist of a single closed curve containing a ‘hole’).
The tools of macrotopology are mainly combinatorial or algebraic, so it is often called
combinatorial topology or algebraic topology.

Macrotopology is not an idle mathematical game. Many important mathematical phenom-
ena are determined entirely by the macrotopological structure of an object. For example:

Figure 1: A homotopy from a coﬀee cup to a donut.

1
• In complex analysis, the path integral of an analytic function along a closed curve
depends only on the global topological properties of the curve.
• In dynamical systems, the existence of ﬁxed points or limit cycles in a smooth dynam-
ical system is sometimes forced by the global topology of the underlying state space.
• In diﬀerential geometry, the existence of certain global geometric structures on a
manifold (eg. orientable frame ﬁelds) is determined by the global topology of the manifold.
• In topological quantum ﬁeld theory, physical phenomena are related to the global
topological properties of particle interaction diagrams.
• In group cohomology theory, the algebraic properties of a group are studied through
the global topology of an associated classifying space.
A basic and important macrotopological tool is simplicial homology. Given a topological
space (eg. a surface) X, we do the following:
1. Represent X combinatorially as an assemblage of simple geometric building blocks called
simplices.
2. Study the macrotopology of X algebraically, by computing the homology groups of this
simplicial representation.

Simplices and Complexes
Simplices are the ‘simplest’ nontrivial geometric objects, which we use as ‘building blocks’ to
construct topological spaces. For example:

Figure 2(A) A 0-simplex is a point.
Figure 2(B) A (convex) 1-simplex is a line segment connecting two points.
Figure 2(C) A (convex) 2-simplex is a triangle (including the interior region).
Figure 2(D) A (convex) 3-simplex is a tetrahedron (including the enclosed volume).

Remember that we are concerned only with topological properties, not geometric ones. Thus,
when considering a 1-simplex, it doesn’t matter what the length of the line segment is (or even
if it is a straight line). Likewise, when considering a 2-simplex, it doesn’t matter whether the
triangle is equilateral or isosceles, (or even if the edges are straight lines).
In general, if n ≤ m, and v0 , v1 , . . . , vn ∈ Rm are (n + 1) distinct points, then the (convex)
n-simplex spanned by v0 , . . . , vn is the set of all convex combinations:
∆{v0 , . . . , vn }   =   {r0 v0 + · · · + rn vn ; r0 , . . . , rn ∈ [0, 1], r0 + · · · + rn = 1}.

Example 1:

2
a                                 a
{a,b}
d
a                   a                       b
{a,b,c}
(A)                             (B)                                                                                               {a,b,c,d}
c                    b             c
b                                                            (C)                                   (D)             b
a
(a,b)

{a,b,c}                                 +a         -a                -a                   +b
(G)                                   (b,a)
(E)                     c
(F)                    b                                          +a                      -b
a                                                                                                                        (H)
a                                           a
a                                    a

d                                          d
(a,
a)

c)

(b,
(a,
(c,

b)

)                                       )
a)

(a,b,c)                             (a,c,b)                                      d,c                                      c,d
)

)
(a,            (a,b,d                    (a,

(a,d,b
(b,c)                                (c,b)
(b,c,d)                                  (b,d,c)
c                      b             c                        b                    (a,c,b                                   (a,b,c
c                       )                      c                )
(I)                                                                      b                                          b
(a,b,c,d)                 (J)            (a,b,d,c)

Figure 2: (A) A (convex) 0-simplex.      (B) A (convex) 1-simplex.    (C) A (convex) 2-simplex.   (D)
A (convex) 3-simplex. (E) A topological 1-simplex.   (F) An topological 2-simplex.   (G) Two oriented
0-simplices. (H) Two oriented 1-simplices. (I) Two oriented 2-simplices. (J) Two oriented 3-simplices.

(A)                               (B)                              (C)                                (D)

Figure 3: Building a simplicial complex:  (A) Begin with the 0-skeleton X(0) . (B) Add the 1-simplices
(1)
which make up with the 1-skeleton X . (C) Add the 2-simplices which make up with the 2-skeleton X(2) .
(D) Add the 3-simplices which make up with the 3-skeleton X(3) .

3
(A)                                          (B)                                   (C)
β                                                                                                                  ε
γβ
d               d
α                    γ                           a
δ
b                               α                                      ε                               δ                           δ
a       δ                                           b                                                                          c                                       c
α                   d
λ                           κ
λ
κ           δ                               ζ                                       γ
η                   φ
a               δ                   b           b                           a
c
γ θφ
δ                          η ζ                                         b
∆                                                                                          b                                                       θ
b
α                       γ           β           α               γ           β

β

β
β
∆
a       α           a
a               δ                   b               a           δ            b                         a

(D)                                                                          (E)                    a                                       a
α                           γ
α           γ

γ                                                      c
α               β                      b
β                           b           β               c
a
(F)                a                                                       a
α
α                   α
α                γ                                                   α
a                       β                                           a
c           γ                                   c
γ                   c
β
b                           a               β                       β
a                   β                       a                                                  α α                                         b
α           ∆           γ                   α                                                                  γ
∆                                                              c       c
a                   β                       a                                                  β β
b

Figure 4: Triangulations of various surfaces (A) A cylinder                                            o
(B) A M¨bius strip. Notice that this object
has only one side.                  (C) A disk with two holes cut in it.                   (D) A torus. (E) A disk. (D) A sphere.

4
(a) If n = 1, then ∆{v0 , v1 } = {r0 v0 + r1 v1 ; r0 , r1 ∈ [0, 1], r0 + r1 = 1} is the line segment
joining v0 to v1 .

(b) If n = 2, then then ∆{v0 , v1 , v3 } = {r0 v0 + r1 v1 + r3 v3 ; r0 , r1 , r ∈ [0, 1], r0 + r1 + r3 = 1}
is the (ﬁlled) triangle with vertices v0 , v1 and v3 . (Exercise)

The boundary of the n-simplex ∆{v0 , . . . , vn } is the set of all possible (n − 1)-simplices
obtained by removing one vertex:

∂∆{v0 , . . . , vn }
=      ∆{v1 , . . . , vn }, ∆{v0 , v2 , . . . , vn }, ∆{v0 , v1 , v3 , . . . , vn }, . . . , ∆{v1 , . . . , vn−1

Example 2:

(a) If n = 0, then ∆{v0 } = v0 is just a point, and ∂v0 = ∅ is an empty set.

(b) If n = 1, then ∂∆{v0 , v1 } = {v0 , v1 } is just the two end-points of the line segment.

(c) If n = 2, then then ∂∆{v0 , v1 , v2 } =                   ∆{v0 , v1 }, ∆{v1 , v2 }, ∆{v2 , v0 }          is just the
three sides of the triangle.

A topological n-simplex is any subset S ⊂ Rm which can be obtained from a convex n-
simplex ∆ through a continuous deformation (see Figures 2(E) and (F)). We say S is homotopic
to ∆. The boundary of S is just the image of boundary of ∆ under this deformation.
A simplicial complex is a geometric structure obtained by joining together topological
simplices of various dimensions, so that their boundaries match up. Technically, we construct
a simplicial complex as follows:

Figure 3(A) Begin with a collection X(0) of 0-simplices (ie. points). We call X(0) the 0-
skeleton of the complex.

Figure 3(B) Attach a set X(1) of topological 1-simplices (ie. curves), so that for every simplex
∆ ∈ X(1) , the two boundary points in ∂∆ are both elements of X(0)
We call X(1) the 1-skeleton of the complex.

Figure 3(C) Next, attach a set X(2) of topological 2-simplices, so that for every simplex
∆ ∈ X(2) , the three boundary lines in ∂∆ are each elements of X(1) .
We call X(2) the 2-skeleton of the complex.

Figure 3(D) Inductively, suppose you have assembled the skeletons X(0) , X(1) , . . . , X(k−1) .
The k-skeleton is a set X(k) of topological k-simplices, so that for every ∆ ∈ X(k) ,
the (k + 1) distinct boundary simplices in ∂∆ are all elements of X(k−1) .

5
Suppose n is the maximum dimension of the simplices you add. The resulting n-dimensional
simplicial complex is just the union
X     =     X(0) ∪ X(1) ∪ X(2) ∪ · · · ∪ X(n) .
Many familiar geometric objects can be represented as simplicial complexes, by cutting them
up into simplices of various dimensions. For example, we can represent a surface as a simplicial
complex by dividing it up into (topologically deformed) triangles which are joined seamlesslyy
along their edges, so that their vertices always coincide (ie. the vertex of one triangle cannot
be in the middle of the edge of another triangle). This is called triangulation.
Figure 4 shows triangulations of some simple surfaces. In each picture, we ﬁrst show the
original surface with the triangulation drawn upon it; then we show the same surface ‘peeled
open’ along the boundaries of the triangulation. In the ‘peeled open’ version, the same vertex or
edge may appear more than once. We have labelled the vertices (with Roman letters), and both
coloured and labelled the edges (with Greek letters) so that you can see which vertices/edges
are actually ‘the same’. The arrows along the edges indicate the way in which the two ‘copies’
of the edge must be ‘zipped together’ to recover the original surface.

Orientation
Topological objects do not come with an intrinsic ‘orientation’, but we’ll assign them an orien-
tation in order to set up our algebraic machinery1 . This is not a totally artiﬁcial or arbitrary
process; often orientation has an important physical or geometric interpretation in applications
(eg. electrical charge; the direction of electric current; the polarity of a magnetic ﬁeld; the
chirality of a molecule, etc.)

Figure 2(G) An oriented 0-simplex is a 0-simplex with a sign: either positive (+) or neg-
ative (−).
Figure 2(H) An oriented 1-simplex is a topological 1-simplex with a direction. If the
boundary points of the simplex are a and b, then the simplex has two orientations: a→b,
and b→a. We indicate these with the ordered pairs (a, b) and (b, a), respectively.
Remark: Each orientation of the 1-simplex corresponds to an orientation of the 0-
simplices which are its boundaries: a→b corresponds to {−a, +b}, whereas a→b cor-
responds to {+a, −b}.
Notation: It will be convenient to use ‘negation’ to represent orientation reversal. Thus,
‘−(a, b) means the same thing as ‘(b, a)’
Figure 2(I) An oriented 2-simplex is a topological 2-simplex with a uniform choice of ori-
entations along its boundary edges. By this we mean that we orient the edges so that the
‘to’ vertex of each edge is the ‘from’ vertex’ of the next edge.
1
Loosely speaking, we assign each simplex a ‘positive’ or ‘negative’ orientation so that objects in our algebraic
structure have inverses —ie. so that we get a group.

6
If the three vertices are {a, b, c}, then the three edges are ∆{a, b}, ∆{b, c}, and ∆{c, a}.
The 2-simplex has two possible orientations, corresponding to the clockwise path a→b→c→a
amd the counterclockwise path a→c→b→a. We indicate these with the ordered triples
(a, b, c) and (a, c, b), respectively.
Note: There is no reason the vertex a has to come ‘ﬁrst’. Thus the triples (b, c, a) and
(c, a, b) both represent the same orientation as (a, b, c). Likewise, the triples (c, b, a) and
(b, a, c) both represent the same orientation as (a, c, b).
Remark: Each orientation of the 2-simplex corresponds to an orientation of its 1-simplex
boundaries. The orientation (a, b, c) corresponds to the orientations (a, b), (b, c), and (c, a),
whereas the orientation (a, c, b) corresponds to the orientations (b, a), (c, b), and (a, c).
Observe that, in each case the boundary edges are oriented so that the ‘signs’ of their
boundary points cancel out. For example, (a, b, c) corresponds to (a, b), (b, c), and (c, a),
which yields {+a, −b}, {+b, −c}, and {+c, −a}.
Notation: It will be convenient to use ‘negation’ to represent orientation reversal. Thus,
‘−(a, b, c) means the same thing as ‘(a, c, b)’

Figure 2(H) An oriented 3-simplex is a topological 3-simplex with a uniform choice of
orientations along its boundary faces. By this we mean that we orient the faces so that,
whenever two faces have a common edge, they assign it opposite orientation.
The two orientations of the 3-simplex ∆{a, b, c, d} are indicated by the ordered 4-tuples
(a, b, c, d) and (a, b, d, c). We interpret these 4-tuples as follows:

• The faces of (a, b, c, d) have orientation: (b, c, d), -(a, c, d), (a, b, d), and −(a, b, c).
• The faces of (a, b, d, c) have orientation: (b, d, c), -(a, d, c), (a, b, c), and −(a, b, d).

In both cases, we obtain the orientation of the four faces by systematically deleting the
ﬁrst, second, third, or fourth entry from the 4-tuple, and then assigning alternating signs:
+, -, +, -, etc.
Remark: This rule is equivalent to the following rule: To orient (a, b, c, d) (for example),
assign the ‘bottom’ face the orientation (b, c, d). Now each other face has exactly one
edge in common with the bottom face; orient the other faces so that each assigns opposite
orientation to the edge it shares with (b, c, d). (Exercise: Check that this yields an
equivalent orientation.)
Remark: Again, there is no reason the vertex a has to come ‘ﬁrst’. We can permute
the order of the vertices. Even permutations preserve sign, whereas odd permutations
reverse it (exercise). Thus, for example, (b, a, d, c) and (a, d, b, c) both represent the same
orientation as (a, b, c, d), but (b, a, c, d) has the opposite orientation.

At this point, the general pattern is clear. We extend the concept of orientation inductively
to higher dimensional simplices as follows:

7
An oriented n-simplex is a topological n-simplex ∆ whose boundary (n−1)-
simplices are oriented such that, whenever two (n−1)-simplices of ∂∆ share a com-
mon (n−2)-simplex, they assign it opposite orientation.

The two orientations of the n-simplex ∆{a0 , a1 , . . . , an } are indicated by the ordered (n+1)-
tuples (a0 , a1 , . . . , an ) and (an , . . . , a1 , a0 ). We interpret these (n + 1)-tuples as follows:

• The faces of (a0 , a1 , . . . , an ) have orientation: (a1 , . . . , an ), - (a0 , a2 , . . . , an ), (a0 , a1 , a3 , . . . , an ),
...

• The faces of (an , . . . , a1 , a0 ) have the opposite orientation

In both cases, we obtain the orientation of the n faces by systematically deleting the ﬁrst,
second, third, etc. entry from the n-tuple, and then assigning alternating signs: +, -, +, -, etc.
We indicate this by a formal ‘sum’:

∂(a0 , a1 , . . . , an ) = (a1 , . . . , an ) − (a0 , a2 , . . . , an ) + (a0 , a1 , a3 , . . . , an ) − . . .
n
=           (−1)k (a0 , a1 , . . . , ak , . . . , an )                                    (1)
k=0

Here, “ ak ” means we are removing ak . For example,

If (a, b) is an oriented 1-simplex, then ∂(a, b) = b − a
If (a, b, c) is an oriented 2-simplex, then ∂(a, b, c) = (a, b) + (b, c) + (c, a)
= (b, c) − (a, c) + (a, b).
If (a, b, c, d) is an oriented 3-simplex, then ∂(a, b, c, d) = (b, c, d) − (a, c, d) + (a, b, d) − (a, b, c).

These formal ‘sums’ are examples of (n−1)-chains (see below).
Remark: Again, there is no reason the vertex a1 has to come ‘ﬁrst’. We can permute the
order of the vertices. Even permutations preserve sign, whereas odd permutations reverse it
(exercise).

Chains
A multiset is a set in which the ‘same’ element can appear multiple times. For example, the
prime factorization of the number 360 is 360 = 5 × 2 × 2 × 2 × 3 × 3. Thus, the multiset of
prime factors of 370 is {2, 2, 2, 3, 3, 5}.
Suppose X is a simplicial complex, and n ∈ N. Then

(C1) An n-chain in X is a multiset of oriented n-simplices of X, with the convention that
simplices with opposite orientation cancel out.

8
For example, suppose X is the triangulated cylinder from Figure 4(A) on page 4. Then a
2-chain is a multiset of oriented 2-simplices of X —in other words, a multiset drawn from the
set {±α, ±β, ±γ, ±δ}. The ‘cancellation rule’ means that the multiset

{α, −γ, α, −δ, α, β, −α, −α, β, γ, γ, }

is equivalent to the multiset {α, β, β, γ, −δ}. We can write this multiset as a formal Z-linear
combination of α, β, γ and δ:
α + 2β + γ − δ.
Thus, we come to our second (equivalent) deﬁnition of an n-chain:

(C2) An n-chain in X is a formal sum z1 ∆1 + z2 ∆2 + · · · + zn ∆n , where ∆1 , . . . , ∆n ∈ X(n)
are oriented n-simplices of X, and z1 , . . . , zn ∈ Z

For example, equation (1) expressed the (oriented) boundary of an (oriented) n-simplex as
a formal ‘sum’ of (oriented) (n−1) simplices. Now we see that

The boundary of an n-simplex is an (n−1)-chain.

If we write an n-chain as a formal sums of n-simplices, then we can ‘add’ two chains together
in the obvious way. For example:

α + 2β + γ − δ        +   2α − 3β + γ − 2δ           =       3α − β + 2γ − 3δ .

It is clear that that the set of all n-chains forms an abelian group under this addition. This is
called the nth chain group of X, and written Cn (X). Observe that every n-simplex in X can
be treated as an n-chain (eg. the n-simplex ∆ corresponds to the chain 1 · ∆.). It is easy to see
that the n-simplices of X generate the chain group Cn (X). Indeed, it is easy to prove:

Lemma 3        Suppose X is a simplicial complex with n-skeleton X(n) . Then Cn (X) is a free
abelian group, generated by X(n) .                                                             2

Thus, a more ‘abstract’ deﬁnition of an n-chain is as follows:

(C3) The nth chain group Cn (X) is the free abelian group generated by the n-skeleton X(n) .
An n-chain is any element of Cn (X).

It is an easy exercise to verify that deﬁnitions (C1), (C2), and (C3) are equivalent.

9
Boundaries and Cycles
Recall that the boundary of an n-simplex is an (n − 1)-chain. We can deﬁne the boundary
of an n-chain to be the sum of the boundaries of its constituent n-simplices. Formally, if
χ = z1 ∆1 + z2 ∆2 + · · · + zn ∆n is an n-chain, then we deﬁne:

∂χ    =    z1 · ∂∆1 + z2 · ∂∆2 + · · · + zn · ∂∆n .                 (2)

For example, suppose X is the triangulated cylinder from Figure 4(A) on page 4. Observe that

∂α = a − a = 0,          ∂δ = b − a,     and     ∂γ = b − a.

Hence, if χ = α + 2δ + γ, then

∂χ    =    ∂α + 2∂δ + γ        =   0 + 2(b − a) + (b − a)      =     3b − 3a.

By deﬁning boundaries in this way, we transform any n-chain into an (n−1)-chain; hence we
get boundary map ∂n+1 : Cn (X)−→Cn−1 (X).

Lemma 4       ∂n : Cn (X)−→Cn−1 (X) is a group homomorphism.                               2

Proof: We have deﬁned the map ∂n+1 in eqn (2), so that the ‘the boundary of a sum is the
sum of the boundaries’. It is easy to check that this yields a homomorphism.         2

An n-boundary is an n-chain which is the boundary of some (n + 1)-chain. For example,
suppose X is the triangulated cylinder from Figure 4(A) on page 4, where the 1-simplices
{α, β, γ, δ} have the orientations indicated by the arrows, and the 2-simplices ∆ and have
the orientations indicated by the clockwise turning arrows. Then
• (b − a) is a 1-boundary, because (b − a) = ∂γ.

• α + δ − γ is a 2-boundary, because ∂∆ = α + δ − γ.

• α − β is a 2-boundary, because ∂(∆ +      ) = (α + δ − γ) + (γ − β − δ) = α − β.

Equivalently, an n-boundary is an element of the image subgroup ∂n+1 Cn+1 (X) . The set of
all n-boundaries is denoted Bn (X). We summarize:

Bn (X)   =    ∂n+1 Cn+1 (X)     is a subgroup of Cn (X).

An n-cycle is an n-chain χ whose boundary is trivial, meaning that all the oriented (n−1)-
simplices in ∂χ cancel out, leaving 0. For example, suppose X is the triangulated torus from
Figure 4(D) on page 4, with the simplices oriented as shown. Then
• α is a 1-cycle, because ∂α = (a − a) = 0.

10
-a
(A) -a
{} = 0
(a,b)

1                +b                         0
+b

(B)                                                    +a          -a                                                   +a          -a
a

(a,
a)
(c,

b)
(a,b,c)                                                                                                                                  +b
2                                           +b                           -c
-c                                                      1
c                  b                                                                                          +c
+c        (b,c)           -b                                                             -b
(a,b) + (b,c) + (c,a)                                     (b-a) + (c-b) + (a-c) = 0
(C)                                                                                                                                       a (a,d a
a        a                                                               )                (d,
a                                    a)
a                                                                                                 d

)
(c,a
a
c)         d                                                            c)
d
,d,

(a,b
d                                           (d,
)

)
(a

(a,c
(a,b,d

)
d

)
(b,d
c

(b,a
c

)
(a,c,b                                                       c          (c,b)                                 d
b
c          )                             db
b (c
,d )
c                                         3                                                             2

(d,b)
b
(a,b,c,d)          b
c       (b,c,d)                                                    c          (b,c)
b
b                             (a,c)+(c,b)+(b,a)                   (a,c)-(b,c)-(a,b)
(a,c,b) + (a,d,c) + (a,b,d) + (b,c,d)                     + (a,d)+(d,c)+(c,a)
=         + (a,d)-(c,d)-(a,c)
= 0
+ (a,b)+(b,d)+(d,a)                 + (a,b)-(d,b)-(a,d)
+ (b,c)+(c,d)+(d,b)                 + (b,c)+(c,d)+(d,b)

Figure 5: (A) ∂0 ◦ ∂1 (ab) = ∂0 (b − a) = 0.                 (B) ∂1 ◦ ∂2 (abc) = ∂1 [(ab) + (bc) + (ca)] = (b − a) + (c −
b) + (a − c) = 0.               (C) ∂2 ◦ ∂3 (abcd) = ∂2 [(bcd) − (acd) + (abd) − (abc)] = 0.

• ∆+               is a 2-cycle, because ∂(∆ +                          ) = (α + β − γ) + (γ − α − β) = 0.

Equivalently, an n-cycle is an element of the kernel subgroup ‘ker(∂n ) ⊂ Cn (X). The set of
all n-cycles is denoted Zn (X). We summarize:

Zn (X)        =            ker(∂n ) is a subgroup of Cn (X).

Remark: Recall that, if v0 is a (unoriented) 0-simplex, then ∂v0 = ∅ is the empty set. It
is therefore reasonable to deﬁne the boundary of any 0-chain to be 0. In other words, ∂0 ≡ 0.
It follows that Z0 (X) = C0 (X).
We now come to what might be called the ‘Fundamental Theorem of Homology’.

Theorem 5                      Let X be a simplicial complex. Then

11
(a) For any β ∈ Bn (X),       ∂(β) = 0. In other words,                  ∂n ◦ ∂n+1           ≡     0.

(b) Thus, Bn (X) is a subgroup of Zn (X).

Proof:       For part (a), it suﬃces to show:
Claim 1:         If ∆ is any (n + 1)-simplex, then ∂n (∂n+1 (∆)) = 0.

Proof:       Recall that the deﬁnition of ‘oriented (n + 1)-simplex’ on page 7 required:
whenever two n-simplices of ∂∆ share a common (n−1)-simplex, they assign
it opposite orientation.
This condition is precisely what we need for ∂n (∂n+1 (∆)) = 0. Indeed, for n = 1, 2, or
3, the equation can be checked by inspecting Figure 5. For higher dimensions, we use the
deﬁning equation (1). Suppose ∆ = (a0 , a1 , . . . , an , an+1 ) is an (n+1)-simplex. Then
n+1
∂n+1 (∆) =          (−1)k (a0 , a1 , . . . , ak , . . . , an+1 ).
k=0
n+1
Hence, ∂n ∂n+1 (∆)        =         (−1)k · ∂n (a0 , a1 , . . . , ak , . . . , an+1 )
k=0
n+1 n+1
=                 (−1)k (−1)j · (a0 , a1 , . . . , aj , . . . , ak , . . . , an+1 ).
k=0     j=0
j=k

It is an exercise to verify that all terms cancel out, leaving 0.                 .......... 2 [Claim 1]

Thus, if χ = z1 ∆1 + z2 ∆2 + · · · + zn ∆n is any (n + 1)-chain, and β = ∂n+1 (χ), then

∂n (β)     =    ∂n ∂n+1 (χ)    = ∂n (z1 · ∂n+1 ∆1 + z2 · ∂n+1 ∆2 + · · · + zn · ∂n+1 ∆n )
= z1 · ∂n ∂n+1 ∆1 + z2 · ∂n ∂n+1 ∆2 + · · · + zn · ∂n ∂n+1 ∆n
= z1 · 0 + z2 · 0 + · · · + zn · 0 = 0,

as desired. Part (b) follows immediately from (a).                                                                      2

Thus, the group Zn (X) can be partitioned into cosets of Bn (X). If two cycles in Zn (X)
belong to the same coset of Bn (X), we say they are homologous. Let Bn = Bn (X). Then

α and β are homologous          ⇐⇒          (α + Bn ) = (β + Bn )                 ⇐⇒             (α − β) ∈ Bn

⇐⇒          (α − β) = ∂χ, for some χ ∈ Cn+1 (X)                         .

12
(A)                               (B)                             (C)         b       c             (D)
x                                   δ                                                   d                   b       c
χ
(b,c)   (c,d)

b)
χ               d
γ                             x

(a,
(c,d)

b)
(a,
γ                        a
y
y                 X                             X                        a
X                                                            a and d are                         a and d are
homologous                        not homologous
x and y are                      x and y are                     χ = (a,b) + (b,c) + (c,d)             χ = (a,b) + (c,d)
connected                       not connected                   χ = (b-a) + (c-b) + (d-c)             χ = (b - a) + (d - c)
= (d - a)

Figure 6: (A) The points x and y are connected by the path γ.      (B) The points x and y are not connected.
The path γ is inadmissible because it leaves X. The path δ is inadmissible because it is broken (discontinuous).
(C) The 0-simplices a and d are homologous via the 1-chain χ, because ∂γ = (d − a). (D) The 0-simplices a
and d are not homologous. The 1-chain γ is not a homology, because ∂γ = (d − c) + (b − a) = (d − a).

(A)                                           (B)                                        (C)
α
α
β                                           β
γ                                                                                                 δ
δ

α is homotopic to β, but not to γ or δ.          A homotopy from α to β.                        A nullhomotopy of δ.
β is homotopic to α, but not to γ or δ.
γ is not homotopic to α, β, or δ. Likewise, δ is not homotopic to α, β, or γ.
δ is nullhomotopic, but α, β, and γ are not nullhomotopic.

Figure 7: Homotopy of loops.

α
β                        α is homologous to β, but not to γ or δ.
β is homologous to α, but not to γ or δ.
γ is not homologous to α, β, or δ.
δ is not homologous to α, β, or γ.

δ is nullhomologous; ie δ is a boundary.
γ                          δ         α, β, and γ are not nullhomologous.

Figure 8: Homology of 1-chains.

13
(A)                                                         (B)             ∆2         ∆4         ∆6
α                                                ∆1
∆2
∆3
∆4
∆5
∆6
∆7                              ∆1         ∆3         ∆5          ∆7
∆17                            ∆8                              ∆17 ∆                             ∆8
β                             ∆16
∆15
∆9
∆10
16

∆15
χ                      ∆9
∆10
∆14                                                                  ∆14
∆13 ∆11
∆12                                                          ∆13         ∆11
∆12
The region between −α and β is
Let χ = ∆1 + ∆2 + ... + ∆17 be the 2-chain
a union of 2-simplices, ∆1,...,∆17
representing the region between −α and β
To obtain a homology from                                         ∆2             ∆4          ∆6                  (C)
∆1         ∆3              ∆5            ∆7
α to β, we must find a
∆17 ∆                                        ∆8
2-chain χ whose boundary
χ is (β−α).
16
χ                     ∆9
=                          (β−α)
∆15                                      ∆10
∆14

∆13
It then follows that                                                                        ∆12 ∆11
(β−α) ε B1(X); hence                                                   Thus, χ = ∆1 + ∆2 + ... + ∆17         = ( β − α)
β+ B1(X) = α+ B1(X).                                                   (because all the purple arrows cancel off in pairs)

Figure 9: The 1-cycles α and β are homologoous, because χ is a 2-chain such that ∂χ = β − α.

a                                                                                        a
A nullhomology of δ.                                                                 Let ∆1 = (a,b,f).
∆1
f               b     First observe that                                                f              b Then ∆1 = (a,b) + (b,f) + (f,a).
e
δ   c           δ = (a,b) + (b,c) + (c,d)                                               δ1           Let δ1 = (b,c) + (c,d) + (d,e) +
(e,f) + (f,b).
+ (d,e) + (e,f) + (f,a).
d                                                                                                    Then δ = δ1 + ∆1

Let ∆2 = (b,c,f).
Then ∆2 = (b,c) + (c,f) + (f,b).                                                               Let ∆3 = (f,c,e).
f                                                                                        f
∆2 b                                                                                                Then ∆3 = (f,c) + (c,e) + (e,f).
δ2      Let δ2 = (c,d) + (d,e) +                                                            ∆3
e                   Let δ3 = (c,d) + (d,e) + (e,c).
c            (e,f) + (f,c).
δ3    c
Then δ2 = δ3 + ∆3
Then δ1 = δ2 + ∆2
a          We conclude that
∆                         + ∆4
δ = ∆1 + ∆2 + ∆3
Let ∆4 = (c,d,e).                                                          f ∆1 b
∆3 2     = (∆1 + ∆2 + ∆3 + ∆4)
e    ∆4      c Then ∆4 = (c,d) + (d,e) + (e,c)                                          e       c
∆4     Hence, δ ε B1(X), so δ
d                        = δ 3.                                                          is nullhomologous
d

Figure 10: The 1-cycle δ is nullhomologous. If χ = ∆1 +∆2 +∆3 +∆4 , then ∂χ = ∂∆1 +∂∆2 +∂∆3 +∂∆4 = δ.

14
Homology is an equivalence relation, and the equivalence classes (ie. the cosets of Bn (X)) are
called homology classes. The set of all homology classes (ie. cosets of Bn (X)) forms a quotient
group, called the nth homology group of X:

Hn (X)    =    Zn (X)/Bn (X).

Before computing some homology groups, we’ll look at the geometric meaning of homology.

Connectivity and 0-Homology
Let X be a topological space (eg. a surface). If x, y ∈ X are two points, then we say that
x is connected to y if there is a continuous (ie. unbroken) path from x to y which does not
leave X, as shown in Figures 6(A,B) on page 13. We then write “x ↔ y”. This deﬁnes
an equivalence relation on X (exercise); the equivalence classes of this relation are called the
connected components of X. Observe that ‘connectivity’ is a macrotopological phenomenon.
In particular

The number of connected components of X is a global topological property of X.

This is exactly the sort of property we can detect with our algebraic-topological tools.
Suppose we represent the space X with a simplicial complex X. The 0-simplices of X are
points; and every 0-simplex can be treated as a 0-chain in C0 (X). Recall that every 0-chain is a
0-cycle (because the boundary of any point, and hence, any 0-chain, is zero). Thus, we regard
0-simplices of X as 0-cycles.
So, suppose a and d are 0-simplices, seen as 0-cycles. Recall that a is homologous to d if
there is some 1-chain χ such that ∂χ = (d − a), as in Figure 6(C) on page 13. We can regard
a 1-chain as an oriented path (or a collection of oriented paths). To say that ∂χ = (d − a) is
to say that χ acts as an oriented path from a to d. From this observation we get the following

Proposition 6      Let X be a topological space, represented by simplicial complex X.

(a) If a and d are two 0-simplices in X, then

a and d are connected (as points in X)      ⇐⇒       a and d are homologous    .

(b) There is a bijective correspondence between the set of connected components of X
and the set of homology classes of 0-simplices in X.

(c) The homology group H0 (X) is a free abelian group, generated by the homology classes
of 0-simplices in X.

(d) Thus, H0 (X) ∼ ZR , where R is the number of connected components in X.
=

15
Proof: (a) follows from the previous argument: any path γ connecting a and d is equivalent
to a 1-chain χ with ∂χ = (d − a).
(b)       First check: Any point in X is connected to a 0-simplex.
It follows that the set of ↔-equivalence classes of points in X is in bijective correspondence
with the set of ↔-equivalence classes of 0-simplices in C0 (X). But part (a) says that these
equivalence classes are exactly the homology classes of 0-simplices in C0 (X).
(c) is an exercise. (d) follows from (b) and (c).                                           2

Example 7:

(a) If X is the space in Figure 6(A) on page 13 (or, equivalently, the simplicial complex in
Figure 6(C)), then H0 (X) = Z.

(b) If X is the space in Figure 6(B) on page 13 (or, equivalently, the simplicial complex in
Figure 6(D)), then H0 (X) = Z2 .

Homotopy and 1-Homology
Let X be a topological space. A loop is a closed, continuous path in X. If α and β are two
loops in X, then we say that α is homotopic to β if α can be continuously deformed into β
(Figure 7 on page 13). In other words, as shown in Figure 7(B), you can make a ‘movie’2 where
the loop α ‘morphs’ into β, without every splitting open or leaving X. Homotopy is clearly an
equivalence relation, and the equivalence classes are called homotopy classes.
Intuitively, the ‘obstructions’ to homotopy are the ‘holes’ in X. For example, in Figure 7(A)
on page 13, the loops α and γ are not homotopic because they do not go around the same holes
the same way. However, α is homotopic to β, because they both go around the same hole the
same way.
A trivial loop is a loop concentrated at a single point —ie. a path that goes nowhere. We
say that a path is nullhomotopic if it can be homotopically ‘shrunk’ down to a trivial loop, as
shown in Figure 7(C) on page 13. Intuitively, the ‘obstruction’ to a nullhomotopy is a ‘hole’ in
the middle of the loop. For example, in Figure 7(A), the loop δ is nullhomotopic, but α is not
nullhomotopic (because α goes around a hole). We have the following intuitive observation:

A loop is nullhomotopic if it doesn’t go around any holes.

The homotopy class of nullhomotopic loops is called the trivial homotopy class; other homotopy
classes are nontrivial.
Observe that homotopy, nullhomotopy, and the presence of ‘holes’ are all macrotopological
phenomena. In particular
2
There is a precise deﬁnition of this, but we won’t bother.

16
The number (and structure) of nontrivial homotopy classes of loops in X is a
global topological property of X.

Recall that a 1-cycle is a 1-chain whose boundary is trivial. Recall that two 1-cycles α and
β are homologous if (α + B1 (X)) = (β + B1 (X)). Equivalently, α and β are homologous if
there is a 2-chain χ such that ∂χ = (β − α), as in Figure 9 on page 14.
In particular, a 1-cycle δ is nullhomologous if δ is homologous to 0. This means there is
a 2-chain χ such that ∂χ = δ, as in as in Figure 10 on page 14.
We can interpret a 1-cycle as a closed loop or a collection of closed loops. Under this
interpretation, a homology between two 1-chains corresponds (loosely) to a homotopy between
two loops. Likewise, a 1-chain is nullhomologous if the corresponding loop is nullhomotopic.

Proposition 8       Let X be a topological space, represented by simplicial complex X. Let
α, β ∈ Z1 (X) be two 1-cycles, and which correpond to (topological) loops α and β.

(a) If α is homotopic to β, then α is homologous to β.

(b) If α is nullhomotopic, then α is nullhomologous                                        2

Notice that the statements in this proposition are not ‘if and only if’. Homology is not quite
equivalent to homotopy, but the two concepts are very similar. In particular, it is convenient,
when computing the homology group H1 (X), to recognize that two 1-cycles are automatically
homologous if we see that they are homotopic.
We thus expect that the existence of ‘holes’ in X should cause H1 (X) to be nontrivial, and
indeed this is often the case...
Example 9:

(a) Let X be the cylinder in Figure 4(A) on page 4. Then H1 (X) = Z.

o
(b) Let X be the M¨bius strip in Figure 4(B). Then H1 (X) = Z.

(c) Let X be the 2-holed disk in Figure 4(C). Then H1 (X) = Z2 .

(d) Let X be the torus in Figure 4(D). Then H1 (X) = Z2 .

(e) Let X be the sphere in Figure 4(E). Then H1 (X) = {0}. In this case, there is a ‘hole’ (the
hollow core of the sphere), but this hole is a ‘2-dimensional’ hole, not a ‘1-dimensional’
hole, so we cannot detect it with H1 (X) (we must instead look at H2 (X)).
Intuitively, you can see why 1-chain homology (or loop homotopy) cannot detect the ‘hole’
in the sphere. Any rubber band which you stretch across the surface of the sphere can
be easily slipped oﬀ. In other words, every loop is nullhomotopic.

We will verify some of the claims of Example 9 in the next section. The rest are exercises.

17
a

γ

Figure 11: The circle as a simplicial complex. Here, X(0) = {a} and X(1) = {γ}

Computation of Homology Groups
Proposition 10         Let X the (hollow) circle. Then H0 (X) = Z and H1 (X) = Z, and Hn (X) =
{0} for all n ≥ 2.

Proof: Represent the circle with the simplicial complex in Figure 11. Here, the 0-skeleton
has one point, and the 1-skeleton has one edge, while all the higher dimensional skeletons are
trivial:
X(0) = {a};        X(1) = {γ};       X(n) = ∅, for n ≥ 2.
Thus, C0 (X) is the free abelian group generated by a (ie. the inﬁnite cyclic group generated
by a), and C1 (X) is the free abelian group generated by γ, and Cn (X) is trivial for n ≥ 2.
We write this:
C0 (X) = Z{a};          C1 (X) = Z{γ};        Cn (X) = {0}, for n ≥ 2.
Now, ∂0 (a) = 0 (because all points have trivial boundary). Hence, ∂0 (χ) = 0, for any
χ ∈ C0 (X); hence Z0 (X) = C0 (X) ∼ Z.
=
Also, ∂1 (γ) = (a − a) = 0. Hence, ∂1 (χ) = 0 for any χ ∈ C1 (X); hence Z1 (X) = C1 (X) ∼
=
Z. It also follows from this argument that B0 (X) = ∂1 C1 (X)          = {0}.

Finally, there are   no 2-simplices; hence C2 (X) = {0}; hence B1 (X) = ∂2 C2 (X)         = {0}.
We summarize:
Z0 (X) ∼ Z
=          and B0 (X) = {0}, so H0 (X) = Z0 (X)/B0 (X)               ∼ Z/{0}
=            ∼ Z.
=
Z1 (X) ∼ Z
=          and B1 (X) = {0}, so H1 (X) = Z1 (X)/B1 (X)               ∼ Z/{0}
=            ∼ Z.
=
Zn (X) = {0}        and Bn (X) = {0}, so Hn (X) = Zn (X)/Bn (X)               ∼ {0}/{0}
=            ∼ {0},
=
for all n ≥ 2.                                                                                2

Proposition 11         Let X the (hollow) torus. Then

H0 (X) = Z;     H1 (X) = Z2 ;    and     H2 (X) = Z;
while Hn (X) = {0} for all n ≥ 2.

18
Proof:    We triangulate the torus as in Figure 4(D) on page 4. Thus,
X(0) = {a};                 X(1) = {α, β, γ};               and      X(2) = {∆, }
It follows:
C0 (X) = Z{a} ∼ Z;
=           C1 (X) = Z{α, β, γ} ∼ Z3 ; and C2 (X) = Z{∆, } ∼ Z2 .
=                          =

Claim 1:     B0 (X) = {0}.
Proof: Since α, β, and γ generate C1 (X), it follows that ∂(α), ∂(β), and ∂(γ) generate
B0 (X) = ∂ C1 (X) . But ∂(α) = (a − a) = 0, ∂(β) = (a − a) = 0, and
∂(γ) = (a−a) = 0. Hence, B0 (X) is generated by 0; hence B0 (X) = {0}. 2 [Claim 1]
Now, recall that every 0-chain is a 0-cycle. That is, Z0 (X) = C0 (X) = Z{a}. It follows:
Z0 (X)             Z{a}    ∼
H0 (X)      =               =               =    Z/{0}    =       Z.
B0 (X)              {0}

Claim 2:     Z1 (X) = C1 (X).
Proof: As we saw in Claim 1, ∂(α) = ∂(β) = ∂(γ) = 0. Hence if χ = z1 α+z2 β +z3 γ is
any 1-chain in C1 (X), then ∂χ = z1 · 0 + z2 · 0 + z3 · 0 = 0, so χ ∈ Z1 (X). 2 [Claim 2]
Claim 3:     B1 (X) is the free abelian group generated by (α + β − γ).
Proof:   Since ∆ and        generate C2 (X), it follows that ∂∆ and ∂             generate B1 (X) =
∂ C2 (X) . But

∂∆ = (α + β − γ)         and        ∂     = (γ − α − β) = −(α + β − γ).
Hence, (α + β − γ) generates B1 (X). .................................. 2 [Claim 3]

Z1 (X)              Z{α, β, γ}             ∼
It follows that H1 (X)     =                  =                              =   Z2 ,
B1 (X)           Z (α + β − γ)

where the last step is an exercise.
Claim 4:     Z2 (X) is the free group generated by (∆ +          ).
Proof:     Let χ = z∆ ∆ + z         be a 2-chain. Then
∂χ = z∆ · ∂∆ + z · ∂       =               z∆ · ∂(α + β − γ) − z · (α + β − γ)
= (z∆ − z ) · (α + β − γ).

Hence,    χ ∈ Z2 (X)    ⇐⇒        ∂χ = 0       ⇐⇒         z∆ = −z

⇐⇒      χ = z · (∆ +      ), for some z ∈ Z         ⇐⇒      χ ∈ Z (∆ +      )        .   2 [Claim 4]

19
Now, X(3) = ∅, so C3 (X) = {0}, so B2 (X) = ∂ C3 (X)                = {0}. Thus,

Z2 (X)         Z (∆ +        )
H2 (X)    =              =                         ∼
=    Z/{0}      ∼
=    Z.
B2 (X)                 {0}

The higher homology groups are automatically trivial, because all the higher chain groups
are trivial.                                                                           2

Proposition 12       Let X the (hollow) cylinder. Then H0 (X) = Z and H1 (X) = Z, and
H0 (X) = {0} for all n ≥ 2.

Proof:    We triangulate the cylinder as in Figure 4(A) on page 4. Thus,

X(0) = {a, b};                X(1) = {α, β, γ, δ};               and     X(2) = {∆, }.
It follows:
C0 (X) = Z{a, b} ∼ Z2 ;
=            C1 (X) = Z{α, β, γ, δ} ∼ Z4 ; and C2 (X) = Z{∆, } ∼ Z2 .
=                          =

Claim 1:     B0 (X) is the cyclic subgroup generated by (b − a). That is, B0 (X) = Z{(b − a)}.

Proof: C1 (X) is generated by α, β, γ and δ. Thus, B0 (X) = ∂1 C1 (X) is generated by
∂α, ∂β, ∂γ, and ∂δ. But

∂(α) = (a−a) = 0;            ∂(β) = (b−b) = 0;             ∂(γ) = (b−a); and ∂(δ) = (b−a).

Thus, B0 (X) is generated by (b − a).
To see this explicitly, suppose that χ = z1 α + z2 β + z3 γ + z4 δ is a 1-chain. Then

∂χ = z1 · ∂α + z2 · ∂β + z3 · ∂γ + z4 · ∂δ
= z1 · 0 + z2 · 0 + z3 · (b − a) + z4 · (b − a)            =    (z3 + z4 ) · (b − a).   (3)

Hence, every 1-boundary has the form z · (b − a) for some z ∈ Z. ....... 2 [Claim 1]

Now, recall that every 0-chain is a 0-cycle. That is, Z0 (X) = C0 (X) = Z{a, b}. Hence,

Z0 (X)           Z{a, b}         ∼
H0 (X)    =             =                       =    Z,
B0 (X)          Z{(b − a)}

where the last step is an easy exercise.
Claim 2:     Z1 (X) is the free abelian group generated by α, β, and (γ − δ). That is,
Z1 (X) = Z{α, β, (γ − δ)}.

20
Proof: If χ = z1 α + z2 β + z3 γ + z4 δ is a 1-chain, then eqn.(3) says ∂χ = (z3 + z4 ) · (b − a).
Hence

χ ∈ Z1 (X)      ⇐⇒        ∂χ = 0           ⇐⇒          z4 = −z3 , while z1 and z2 are arbitrary

⇐⇒        χ = z1 α + z2 β + z3 (γ − δ), for some z1 , z2 , z3 ∈ Z

⇐⇒        χ ∈ Z{α, β, (γ − δ)}     .

It is easy to see that α, β and (γ − δ) are Z-linearly independent. ...... 2 [Claim 2]

Claim 3:     B1 (X) is the free abelian group generated by (α + δ − γ) and (γ − β − δ).

Proof:   Since ∆ and         generate C2 (X), it follows that ∂∆ and ∂          generate B1 (X) =
∂ C2 (X) . But

∂∆ = (α + δ − γ)        and     ∂       = (γ − β − δ).

Hence, these elements are 1-boundaries, and generate B1 (X). It is an exercise to check
that ∂∆ and ∂ are Z-linearly independent. .......................... 2 [Claim 3]

Z1 (X)                Z α, β, (γ − δ)
It follows that H1 (X)     =               =                                        ∼
=    Z,
B1 (X)         Z (α + δ − γ), (γ − β − δ)

where the last step is an exercise.
Claim 4:     Z2 (X) = {0}.

Proof:     Let χ = z∆ ∆ + z        be a 2-chain. Then

∂χ = z∆ · ∂∆ + z · ∂    = z∆ · ∂(α + δ − γ) + z · (γ − β − δ)
= z∆ · α − z · β − (z∆ − z ) · γ + (z∆ − z ) · δ.

Hence, χ ∈ Z2 (X)          ⇐⇒       ∂χ = 0    ⇐⇒          z∆ = 0 = z          ⇐⇒      χ = 0 .
2 [Claim 4]

Since B2 (X) ⊂ Z2 (X), it follows that B2 (X) = {0} also. Thus, H2 (X) = Z2 (X)/B2 (X) =
{0}/{0} = {0}. The higher homology groups are automatically trivial, because all the
higher chain groups are trivial.                                                       2

21
Retraction
Observe that the homology groups of the cylinder (Proposition 12) are exactly the same as
those of a circle (Proposition 10). This shouldn’t be surprising, because a cylinder is really just
a circle which has been ‘extruded’ in one direction. Or, conversely, a circle is just a cylinder
which has been ‘compressed’ in one dimension. This is an example of a retraction.
If X is a topological space and Y ⊂ X is a subspace, then a retraction from X into Y
is a continuous deformation3 of X so that every point in X moves continously through X and
eventually ends up inside of Y. We require, furthermore:

• No point of X ever leaves X during the deformation.

• X is not torn or cut during the deformation. In other words, if x, y ∈ X are two points
which are very close together, then x and y remain close, throughout the deformation.

We then say that X is retractable into Y. In particular, if X is retractable down to a single
point, we say that X is contractable.
Example 13:

(a) Let X be a cylinder and let Y be one of its (circular) rims. Then X is retractable into Y
(by continuously shrinking the length of the cylinder down to zero).

(b) A torus is not retractable into a circle. Neither is a sphere.

(c) The unit disk D is contractable down to a single point. So is the unit ball B.

Clearly, retraction is a kind of homotopy. Given the previously discussed relationship be-
tween homology and homotopy, the following theorem is unsurprising

Theorem 14            Let X be a topological space and let Y ⊂ X be a subspace.

(a) If X is retractable into Y, then Hn (X) = Hn (Y) for all n ∈ N.

(b) In particular, if X is contractible, then H0 (X) = Z, and Hn (X) = {0} for all n ≥ 1.2

Example 15:

(a) The cylinder is retractable into a circle. Hence, the cylinder and the circle have the same
homology groups, as seen by comparing Propositions 10 and 12.

(b) The disk D is contractible. Hence, we know automatically that H0 (D) = Z, and Hn (D) =
{0} for all n ≥ 1.

3
There is a precise formal deﬁnition of retraction, but we won’t bother.

22

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