Simplical Homology Seminar Algorithms

Home , Simplicial homology

Simplical homology Seminar algorithms

Jorrick Sleijster Contents • Chain complexes – Free abelian groups

• Simplicial homology

• Betti numbers • Homotopy invariance Simplicial homology

Recall: Simplex

• 0-simplex

• 1-simplex

• 2-simplex

• 3-simplex

• d-simplex P-chain functions Definition: An oriented simplex. A simplex, σ, together with an orientation. • An orientation is equal to a particular ordering of a vertex set. • Any simplex with a dimension of at least 1, has multiple orientations. P-chain functions Definition: An oriented simplex. A simplex, σ, together with an orientation. • An orientation is equal to a particular ordering of a vertex set. • Any simplex with a dimension of at least 1, has multiple orientations. P-chain functions Definition: A p-chain function, c, goes from a set of oriented p-simplices to integers, such that:

• c(σ) = −c(σ∗) = 1 if σ and σ∗ are opposite orientations of the same simplex.

• c(σ) = 0 for all other oriented p-simplices. P-chain functions Deﬁnition: A p-chain function, c, goes from a set of oriented p-simplices to integers, such that:

• c(σ) = −c(σ∗) = 1 if σ and σ∗ are opposite orientations of the same simplex.

• c(σ) = 0 for all other oriented p-simplices. P Each p-chain is a ﬁnite linear combination c = niσi of the p-simplexes σi. P-chain functions Deﬁnition: A p-chain function, c, goes from a set of oriented p-simplices to integers, such that:

• c(σ) = −c(σ∗) = 1 if σ and σ∗ are opposite orientations of the same simplex.

• c(σ) = 0 for all other oriented p-simplices. P Each p-chain is a ﬁnite linear combination c = niσi of the p-simplexes σ . i v2 A 0-chain example: n0v0 + n1v1 + n2v2.

v0 v1 P-chain functions Deﬁnition: A p-chain function, c, goes from a set of oriented p-simplices to integers, such that:

• c(σ) = −c(σ∗) = 1 if σ and σ∗ are opposite orientations of the same simplex.

• c(σ) = 0 for all other oriented p-simplices. P Each p-chain is a ﬁnite linear combination c = niσi of the p-simplexes σ . i v2 A 0-chain example: n0v0 + n1v1 + n2v2.

v0 v1 P-chain functions Deﬁnition: A p-chain function, c, goes from a set of oriented p-simplices to integers, such that:

• c(σ) = −c(σ∗) = 1 if σ and σ∗ are opposite orientations of the same simplex.

• c(σ) = 0 for all other oriented p-simplices. P Each p-chain is a ﬁnite linear combination c = niσi of the p-simplexes σ . i v2 A 0-chain example: n0v0 + n1v1 + n2v2.

A 1-chain example: v0 v1 n0(v0v1) + n1(v1v2) + n2(v2v0). Abelian groups • The abelian group is a group of multiple elements.

• Operations do not depend on the order of the group Free abelian groups • A free abelian group is an abelian group with a basis. • A generating set consisting of linearly independent elements. • Such that every possible combination of the group can be found by adding/substracting base elements. • An abelian group generated by the elements x1, ..., xk, consists of all combinations of the form n1x1 + ... + nkxk where ni are integers.

v0 v1 Free abelian groups • A free abelian group is an abelian group with a basis. • A generating set consisting of linearly independent elements. • Such that every possible combination of the group can be found by adding/substracting base elements. • An abelian group generated by the elements x1, ..., xk, consists of all combinations of the form n1x1 + ... + nkxk where ni are integers.

• 0-chain ' Z ⊕ Z ⊕ Z with as basis hv0, v1, v2i

v0 v1 Free abelian groups

Example. X = {a, b} and G = Z2 ⊕ Z4 Free abelian groups

Example. X = {a, b} and G = Z2 ⊕ Z4 f : X → G (homomorphism) Free abelian groups

Example. X = {a, b} and G = Z2 ⊕ Z4 f : X → G (homomorphism) • f(a) • f(b) Free abelian groups

Example. X = {a, b} and G = Z2 ⊕ Z4 f : X → G (homomorphism) • f(a) = (1, 2) • f(b) = (0, 3) Free abelian groups

Example. X = {a, b} and G = Z2 ⊕ Z4 f : X → G (homomorphism) • f(a) = (1, 2) • f(b) = (0, 3) Now we deﬁne our free abelian group(FA) on X. FA(X) = ma + nb with m, n ∈ Z Free abelian groups

Example. X = {a, b} and G = Z2 ⊕ Z4 f : X → G (homomorphism) • f(a) = (1, 2) • f(b) = (0, 3) Now we deﬁne our free abelian group(FA) on X. FA(X) = ma + nb with m, n ∈ Z F : FA(X) → G Free abelian groups

Example. X = {a, b} and G = Z2 ⊕ Z4 f : X → G (homomorphism) • f(a) = (1, 2) • f(b) = (0, 3) Now we define our free abelian group(FA) on X. FA(X) = ma + nb with m, n ∈ Z F : FA(X) → G F (ma + nb) = mf(a) + nf(b) = m(1, 2) + n(0, 3) = (m, 2m + 3n) Free abelian groups Definition of kernel ••Given a homomorphism f : G → H. • The kernel of G, ker(f), is {g ∈ G|f(g) = 0} Free abelian groups Definition of kernel ••Given a homomorphism f : G → H. • The kernel of G, ker(f), is {g ∈ G|f(g) = 0}

Example: Let’s calculate the kernel for the homomorphism F. Free abelian groups Deﬁnition of kernel ••Given a homomorphism f : G → H. • The kernel of G, ker(f), is {g ∈ G|f(g) = 0}

Example: Let’s calculate the kernel for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n) Free abelian groups Deﬁnition of kernel ••Given a homomorphism f : G → H. • The kernel of G, ker(f), is {g ∈ G|f(g) = 0}

Example: Let’s calculate the kernel for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n)

• m? • n? Free abelian groups Deﬁnition of kernel ••Given a homomorphism f : G → H. • The kernel of G, ker(f), is {g ∈ G|f(g) = 0}

Example: Let’s calculate the kernel for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n)

• m should be even • n? Free abelian groups Deﬁnition of kernel ••Given a homomorphism f : G → H. • The kernel of G, ker(f), is {g ∈ G|f(g) = 0}

Example: Let’s calculate the kernel for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n)

• m should be even • n should be a multiple of 4. Free abelian groups Deﬁnition of image ••Given a homomorphism f : G → H. • The image of f, im(f), given a g ∈ G is {f(g) ∈ H} Free abelian groups Deﬁnition of image ••Given a homomorphism f : G → H. • The image of f, im(f), given a g ∈ G is {f(g) ∈ H}

Now consider the image for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n) Free abelian groups Deﬁnition of image ••Given a homomorphism f : G → H. • The image of f, im(f), given a g ∈ G is {f(g) ∈ H}

Now consider the image for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n)

Examples are: Free abelian groups Deﬁnition of image ••Given a homomorphism f : G → H. • The image of f, im(f), given a g ∈ G is {f(g) ∈ H}

Now consider the image for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n)

Examples are: m = 0, n = 0 ⇒ (0, 0) Free abelian groups Deﬁnition of image ••Given a homomorphism f : G → H. • The image of f, im(f), given a g ∈ G is {f(g) ∈ H}

Now consider the image for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n)

Examples are: m = 0, n = 0 ⇒ (0, 0) m = 1, n = 0 ⇒ (1, 2) Free abelian groups Deﬁnition of image ••Given a homomorphism f : G → H. • The image of f, im(f), given a g ∈ G is {f(g) ∈ H}

Now consider the image for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n)

Examples are: m = 0, n = 0 ⇒ (0, 0) m = 1, n = 0 ⇒ (1, 2) m = 1, n = 1 ⇒ (1, 1) Free abelian groups Deﬁnition of image ••Given a homomorphism f : G → H. • The image of f, im(f), given a g ∈ G is {f(g) ∈ H}

Now consider the image for the homomorphism F.

F : FA(X) → Z2 ⊕ Z4 F (ma + nb) = m(1, 2) + n(0, 3) = (m, 2m + 3n)

Examples are: m = 0, n = 0 ⇒ (0, 0) m = 1, n = 0 ⇒ (1, 2) m = 1, n = 1 ⇒ (1, 1) m = 2, n = 2 ⇒ (0, 2) Boundary operator Let’s now deﬁne the boundary operator

The boundary operator is deﬁned as given a dimension p, report the set of all boundaries for the given dimension.

∂p : Cp(K) → Cp−1(K) Boundary operator Let’s now deﬁne the boundary operator

The boundary operator is deﬁned as given a dimension p, report the set of all boundaries for the given dimension.

∂p : Cp(K) → Cp−1(K)

Where Cp(K) is a p-chain. The basis of this group are the p-simplices found in simplex K. Boundary operator Let’s now deﬁne the boundary operator

The boundary operator is deﬁned as given a dimension p, report the set of all boundaries for the given dimension.

∂p : Cp(K) → Cp−1(K)

Now let σ = [v0, ..., vp] be an oriented p-simplex with p > 0. We can now deﬁne

p X i d ∂pσ = ∂p[vo, ..., vp] = (−1) [v0, ..., vi, ..., vp] i=0 d Where vi is the item not contained in the simplex. Boundary operator Recall:

p X i d ∂pσ = ∂p[vo, ..., vp] = (−1) [v0, ..., vi, ..., vn] i=0

Now let’s look at some examples. Boundary operator Recall:

p X i d ∂pσ = ∂p[vo, ..., vp] = (−1) [v0, ..., vi, ..., vn] i=0

Now let’s look at some examples.

∂0[v0] ∂1[v0v1] ∂2[v0v1v2] ∂3[v0v1v2v3] Boundary operator Recall:

p X i d ∂pσ = ∂p[vo, ..., vp] = (−1) [v0, ..., vi, ..., vn] i=0

Now let’s look at some examples.

∂0[v0] = 0 ∂1[v0v1] ∂2[v0v1v2] v0 ∂3[v0v1v2v3]

v1 Boundary operator Recall:

p X i d ∂pσ = ∂p[vo, ..., vp] = (−1) [v0, ..., vi, ..., vn] i=0

Now let’s look at some examples.

∂0[v0] = 0 ∂1[v0v1] = v1 − v0 ∂2[v0v1v2] ∂3[v0v1v2v3] v0 v2

v1 Boundary operator Recall:

p X i d ∂pσ = ∂p[vo, ..., vp] = (−1) [v0, ..., vi, ..., vn] i=0

Now let’s look at some examples.

∂0[v0] = 0 ∂1[v0v1] = v1 − v0 ∂2[v0v1v2] = v1v2 − v0v2 + v0v1 v0 v2 ∂3[v0v1v2v3]

v1 Boundary operator

∂0[v0] = 0 ∂1[v0v1] = v1 − v0 ∂2[v0v1v2] = [v1v2] − [v0v2] + [v0v1] ∂3[v0v1v2v3] = [v1v2v3] − [v0v2v3] + [v0v1v3] − [v0v1v2]

v0 v2

v1 Boundary operator Now that we have seen the rule for simplexes. What about complexes? Boundary operator Now that we have seen the rule for simplexes. What about complexes?

Given a complex A, how do we calculate the boundary? Boundary operator Now that we have seen the rule for simplexes. What about complexes?

Given a complex A, how do we calculate the boundary?

Let A consists of multiple x-simpleces K0, ..Km. Then the ∂x of this complex is deﬁned by n0∂x(K0) + .. + nm∂x(Km) Boundary operator Now that we have seen the rule for simplexes. What about complexes?

Given a complex A, how do we calculate the boundary?

Let A consists of multiple x-simpleces K0, ..Km. Then the ∂x of this complex is deﬁned by n0∂x(K0) + .. + nm∂x(Km)

v0 v2

v1 Boundary operator Now that we have seen the rule for simplexes. What about complexes?

Given a complex A, how do we calculate the boundary?

Let A consists of multiple x-simpleces K0, ..Km. Then the ∂x of this complex is deﬁned by n0∂x(K0) + .. + nm∂x(Km)

∂1(A) = n0∂1(v0v1) + n1∂1(v1v2) + n2∂1(v2v3) + v n3∂1(v3v0) + n4∂1(v2v0) 3

v0 v2

v1 Boundary operator Now that we have seen the rule for simplexes. What about complexes?

Given a complex A, how do we calculate the boundary?

Let A consists of multiple x-simpleces K0, ..Km. Then the ∂x of this complex is deﬁned by n0∂x(K0) + .. + nm∂x(Km)

∂1(A) = n0∂1(v0v1) + n1∂1(v1v2) + n2∂1(v2v3) + v n3∂1(v3v0) + n4∂1(v2v0) 3

= n0(v1 − v0) + n1(v2 − v1) + n2(v3 − v2) +n3(v0 − v3) + n4(v0 − v2) v0 v2

v1 Boundary operator

Now consider the 1-chain ∂2[v0v1v2]. What happens when we call the operator ∂1 to this 1-chain?

v0 v2

v1 Boundary operator

Now consider the 1-chain ∂2[v0v1v2]. What happens when we call the operator ∂1 to this 1-chain?

v0 v2 ∂2[v0v1v2] = [v1v2] − [v0v2] + [v0v1]

v1 Boundary operator

Now consider the 1-chain ∂2[v0v1v2]. What happens when we call the operator ∂1 to this 1-chain?

v0 v2 ∂2[v0v1v2] = [v1v2] − [v0v2] + [v0v1]

v1 Now let’s apply the ∂1 operation to this. ∂1[[v1v2] − [v0v2] + [v0v1]] = Boundary operator

Now consider the 1-chain ∂2[v0v1v2]. What happens when we call the operator ∂1 to this 1-chain?

v0 v2 ∂2[v0v1v2] = [v1v2] − [v0v2] + [v0v1]

v1 Now let’s apply the ∂1 operation to this. ∂1[[v1v2] − [v0v2] + [v0v1]] =

v2 − v1 − (v2 − v0) + v1 − v0 = v2 − v1 − v2 + v0 + v1 − v0 = Boundary operator

Now consider the 1-chain ∂2[v0v1v2]. What happens when we call the operator ∂1 to this 1-chain?

v0 v2 ∂2[v0v1v2] = [v1v2] − [v0v2] + [v0v1]

v1 Now let’s apply the ∂1 operation to this. ∂1[[v1v2] − [v0v2] + [v0v1]] =

v2 − v1 − (v2 − v0) + v1 − v0 = v2 − v1 − v2 + v0 + v1 − v0 = v2 − v2 + v1 − v1 + v0 − v0 = 0 Chain complexes A chain complex is a sequence of homomorphisms, if for each n, fnfn+1 = 0.

∂n+1 ∂n ∂n−1 ∂1 ∂0 .. −→ Cn+1 −−−→ Cn −→ Cn−1 −−−→ .. −→ C1 −→ C0 −→ 0 Chain complexes A chain complex is a sequence of homomorphisms, if for each n, fnfn+1 = 0.

∂n+1 ∂n ∂n−1 ∂1 ∂0 .. −→ Cn+1 −−−→ Cn −→ Cn−1 −−−→ .. −→ C1 −→ C0 −→ 0

Remember the last slide? Chain complexes A chain complex is a sequence of homomorphisms, if for each n, fnfn+1 = 0.

∂n+1 ∂n ∂n−1 ∂1 ∂0 .. −→ Cn+1 −−−→ Cn −→ Cn−1 −−−→ .. −→ C1 −→ C0 −→ 0

Now take as simple homomorphism f : G → H. Chain complexes A chain complex is a sequence of homomorphisms, if for each n, fnfn+1 = 0.

∂n+1 ∂n ∂n−1 ∂1 ∂0 .. −→ Cn+1 −−−→ Cn −→ Cn−1 −−−→ .. −→ C1 −→ C0 −→ 0

Now take as simple homomorphism f : G → H. • The kernel of f is the subgroup f −1(0) of G. Chain complexes A chain complex is a sequence of homomorphisms, if for each n, fnfn+1 = 0.

∂n+1 ∂n ∂n−1 ∂1 ∂0 .. −→ Cn+1 −−−→ Cn −→ Cn−1 −−−→ .. −→ C1 −→ C0 −→ 0

Now take as simple homomorphism f : G → H. • The kernel of f is the subgroup f −1(0) of G. • The image of f is the subgroup f(G) of H. Chain complexes A chain complex is a sequence of homomorphisms, if for each n, fnfn+1 = 0.

∂n+1 ∂n ∂n−1 ∂1 ∂0 .. −→ Cn+1 −−−→ Cn −→ Cn−1 −−−→ .. −→ C1 −→ C0 −→ 0

Now take as simple homomorphism f : G → H. • The kernel of f is the subgroup f −1(0) of G. • The image of f is the subgroup f(G) of H. Simplicial homology groups

• Group of p-cycles: Zp(K). The kernel of ∂p. Simplicial homology groups

• Group of p-cycles: Zp(K). The kernel of ∂p.

• Group of p-boundaries: Bp(K). The image of ∂p+1. Simplicial homology groups

• Group of p-cycles: Zp(K). The kernel of ∂p.

• Group of p-boundaries: Bp(K). The image of ∂p+1. Simplicial homology groups

• Group of p-cycles: Zp(K). The kernel of ∂p.

• Group of p-boundaries: Bp(K). The image of ∂p+1.

• p-th homology group: Hp(K) = Zp(K)/Bp(K). Simplicial homology groups Why do simplicial homology groups exists? Simplicial homology groups Why do simplicial homology groups exists?

• How many holes does this empty donut have? Simplicial homology groups Now let’s take a ball and a disk. Are they homeomorphic, meaning, are they geometrically speaking the same thing? Simplicial homology groups Now let’s take a ball and a disk. Are they homeomorphic, meaning, are they geometrically speaking the same thing?

Can I take the ﬁrst, and deform it into the second without tearing or squishing things too much? And afterwards also transform it back? Simplicial homology groups Now let’s take a ball and a disk. Are they homeomorphic, meaning, are they geometrically speaking the same thing?

Can I take the ﬁrst, and deform it into the second without tearing or squishing things too much? And afterwards also transform it back? Homology groups allows us to answer both of these questions by using mathematics. Ex. Circle homology groups Ex. Circle homology groups e2 v0 v2

e0 e1

v1 Ex. Circle homology groups e2 • 0-simplices = {v0, v1, v2} v0 v2

e0 e1

v1 Ex. Circle homology groups e2 • 0-simplices = {v0, v1, v2} v0 v2

e0 e1 • 1-simplices = {v0v1, v1v2, v2v0}

v1 Ex. Circle homology groups e2 • 0-simplices = {v0, v1, v2} v0 v2

e0 e1 • 1-simplices = {v0v1, v1v2, v2v0} v ∂2 ∂1 ∂0 1 C2 −→ C1 −→ C0 −→ 0 Ex. Circle homology groups e2 • 0-simplices = {v0, v1, v2} v0 v2

e0 e1 • 1-simplices = {v0v1, v1v2, v2v0} v ∂2 ∂1 ∂0 1 C2 −→ C1 −→ C0 −→ 0

C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i Ex. Circle homology groups e2 • 0-simplices = {v0, v1, v2} v0 v2

e0 e1 • 1-simplices = {v0v1, v1v2, v2v0} v ∂2 ∂1 ∂0 1 C2 −→ C1 −→ C0 −→ 0

C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

C1 ' Z ⊕ Z ⊕ Z hv0v1 v1v2 v2v0i Ex. Circle homology groups e2 • 0-simplices = {v0, v1, v2} v0 v2

e0 e1 • 1-simplices = {v0v1, v1v2, v2v0} v ∂2 ∂1 ∂0 1 C2 −→ C1 −→ C0 −→ 0

C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

C1 ' Z ⊕ Z ⊕ Z hv0v1 v1v2 v2v0i

C2 ' 0 Ex. Circle homology groups e2 • 0-simplices = {v0, v1, v2} v0 v2

e0 e1 • 1-simplices = {v0v1, v1v2, v2v0} v ∂2 ∂1 ∂0 1 C2 −→ C1 −→ C0 −→ 0

C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

C1 ' Z ⊕ Z ⊕ Z hv0v1 v1v2 v2v0i

C2 ' 0

∂0 = 0 v0, v1, v2 → 0

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2 Ex. Circle homology groups e2 • 0-simplices = {v0, v1, v2} v0 v2

e0 e1 • 1-simplices = {v0v1, v1v2, v2v0} v ∂2 ∂1 ∂0 1 C2 −→ C1 −→ C0 −→ 0

C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

C1 ' Z ⊕ Z ⊕ Z hv0v1 v1v2 v2v0i

C2 ' 0

∂0 = 0 v0, v1, v2 → 0

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

∂2 = 0 0 → 0 Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1

v1 Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

v1 Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1 Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2 Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

B0 = im∂1 ' Z ⊕ Z ⊕ Z hv1 − v0, v2 − v1, v0 − v2i Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

B0 = im∂1 ' Z ⊕ Z ⊕ Z hv1 − v0, v2 − v1, v0 − v2i

Now what is H0? Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

B0 = im∂1 ' Z ⊕ Z ⊕ Z hv1 − v0, v2 − v1, v0 − v2i

Now what is H0?

First option: Setting all elements of B0 to 0. Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

B0 = im∂1 ' Z ⊕ Z ⊕ Z hv1 − v0, v2 − v1, v0 − v2i

Now what is H0?

First option: Setting all elements of B0 to 0.

v1 − v0 = 0 v2 − v1 = 0 v0 − v2 = 0 Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

B0 = im∂1 ' Z ⊕ Z ⊕ Z hv1 − v0, v2 − v1, v0 − v2i

Now what is H0?

First option: Setting all elements of B0 to 0.

v1 − v0 = 0 v2 − v1 = 0 v0 − v2 = 0

v0 = v1 = v2 Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

B0 = im∂1 ' Z ⊕ Z ⊕ Z hv1 − v0, v2 − v1, v0 − v2i

Now what is H0?

First option: Setting all elements of B0 to 0.

v0 = v1 = v2

H0 ' Z as we only have multiples of v0 + v1 + v2 Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

B0 = im∂1 ' Z ⊕ Z ⊕ Z hv1 − v0, v2 − v1, v0 − v2i

Second option: Remove all redundant pieces of B0. Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

B0 = im∂1 ' Z ⊕ Z ⊕ Z hv1 − v0, v2 − v1, v0 − v2i

Second option: Remove all redundant pieces of B0.

v0 − v2 can be calculated based on v1 − v0 and v2 − v1. Making B0 ' Z ⊕ Z. Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

∂1 = e0 → v1 − v0 e1 → v2 − v1 e2 → v0 − v2

B0 = im∂1 ' Z ⊕ Z ⊕ Z hv1 − v0, v2 − v1, v0 − v2i

Second option: Remove all redundant pieces of B0.

v0 − v2 can be calculated based on v1 − v0 and v2 − v1. Making B0 ' Z ⊕ Z.

H0 = Z0/B0 ' Z ⊕ Z ⊕ Z/Z ⊕ Z ' Z Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

Either way we conclude the following:

H0 ' Z because we can only have a multiple of v0 + v1 + v2 Ex. Circle homology groups e2 H0 = Z0/B0 v0 v2

e0 e1 Z0 =ker∂0 = C0 ' Z ⊕ Z ⊕ Z hv0, v1, v2i

B0 =im∂1 v1

Either way we conclude the following:

H0 ' Z because we can only have a multiple of v0 + v1 + v2

Any element of this H0 looks like n(v0 + v1 + v2) where n ∈ Z Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1

v1 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1

v1 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 v ∂1(n0e0 + n1e1 + n2e2) 1 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 v ∂1(n0e0 + n1e1 + n2e2) 1

= n0(v1 − v0) + n1(v2 − v1) + n2(v0 − v2) Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 v ∂1(n0e0 + n1e1 + n2e2) 1

= n0(v1 − v0) + n1(v2 − v1) + n2(v0 − v2)

= v0(−n0 + n2) + v1(n0 − n1) + v2(n1 − n2) Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 v ∂1(n0e0 + n1e1 + n2e2) 1

= n0(v1 − v0) + n1(v2 − v1) + n2(v0 − v2)

= v0(−n0 + n2) + v1(n0 − n1) + v2(n1 − n2) = 0 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 v ∂1(n0e0 + n1e1 + n2e2) 1

= n0(v1 − v0) + n1(v2 − v1) + n2(v0 − v2)

= v0(−n0 + n2) + v1(n0 − n1) + v2(n1 − n2) = 0

⇒ −n0 + n2 = 0 n0 − n1 = 0 n1 − n2 = 0 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 v ∂1(n0e0 + n1e1 + n2e2) 1

= n0(v1 − v0) + n1(v2 − v1) + n2(v0 − v2)

= v0(−n0 + n2) + v1(n0 − n1) + v2(n1 − n2) = 0

⇒ −n0 + n2 = 0 n0 − n1 = 0 n1 − n2 = 0

⇒ n0 = n1 = n2 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 v ∂1(n0e0 + n1e1 + n2e2) 1

= n0(v1 − v0) + n1(v2 − v1) + n2(v0 − v2)

= v0(−n0 + n2) + v1(n0 − n1) + v2(n1 − n2) = 0

⇒ −n0 + n2 = 0 n0 − n1 = 0 n1 − n2 = 0

⇒ n0 = n1 = n2

Z1 ' Z he0 + e1 + e2i Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 ' Z he0 + e1 + e2i

v1 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 ' Z he0 + e1 + e2i

B1 =im∂2 v1 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 ' Z he0 + e1 + e2i

B1 =im∂2 v1

C2 = 0 ⇒ B1 = 0 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 ' Z he0 + e1 + e2i

B1 =im∂2 v1

C2 = 0 ⇒ B1 = 0

H1 = Z1/B1 ⇒ Z1/0 ⇒ Z1 Ex. Circle homology groups e2 H1 = Z1/B1 v0 v2

e0 e1 Z1 =ker∂1 ' Z he0 + e1 + e2i

B1 =im∂2 v1

C2 = 0 ⇒ B1 = 0

H1 = Z1/B1 ⇒ Z1/0 ⇒ Z1 ' Z Ex. Circle homology groups e2 • H0 ' Z v0 v2

e0 e1 • H1 ' Z

v1 Ex. Circle homology groups e2 • H0 ' Z v0 v2

e0 e1 • H1 ' Z

• Hn = 0 for n ≥ 2 v1 Ex. Circle homology groups e2 • H0 ' Z v0 v2

e0 e1 • H1 ' Z

• Hn = 0 for n ≥ 2 v1

So what do these values mean? Ex. Circle homology groups e2 • H0 ' Z v0 v2

e0 e1 • H1 ' Z

• Hn = 0 for n ≥ 2 v1

So what do these values mean? Homology groups enable for distiguishing shapes by examining their holes. Example: A circle vs a disk and a sphere vs a circle. Ex. Circle homology groups e2 • H0 ' Z v0 v2

e0 e1 • H1 ' Z

• Hn = 0 for n ≥ 2 v1

So what do these values mean? Homology groups enable for distiguishing shapes by examining their holes. Example: A circle vs a disk and a sphere vs a circle.

A circle has a 1-dimensional hole, were a disk does not. And a sphere has a 2-dimensional hole while a circle has 1. Ex. Circle homology groups e2 • H0 ' Z v0 v2

e0 e1 • H1 ' Z

• Hn = 0 for n ≥ 2 v1

So what do these values mean? Homology groups enable for distiguishing shapes by examining their holes. Example: A circle vs a disk and a sphere vs a circle.

A circle has a 1-dimensional hole, were a disk does not. And a sphere has a 2-dimensional hole while a circle has 1.

As they have diﬀerent homology groups, they are not the same thing/homeomorphic. Ex. Circle homology groups e2 • H0 ' Z v0 v2

e0 e1 • H1 ' Z

• Hn = 0 for n ≥ 2 v1

So what do these values mean? Homology groups enable for distiguishing shapes by examining their holes. Example: A circle vs a disk and a sphere vs a circle.

A circle has a 1-dimensional hole, were a disk does not. And a sphere has a 2-dimensional hole while a circle has 1.

But what about H0 then? Special example H0

v1 v3

e0 e1

v0 v2 Special example H0

∂2 ∂1 ∂0 v1 v3 C2 −→ C1 −→ C0 −→ 0 e0 e1 C0 ' Z ⊕ Z ⊕ Z ⊕ Z ⇒ v0 v1 v2 v3 v0 v2 C1 ' Z ⊕ Z ⇒ e0 e1

C2 ' 0 Special example H0

∂2 ∂1 ∂0 v1 v3 C2 −→ C1 −→ C0 −→ 0 e0 e1 C0 ' Z ⊕ Z ⊕ Z ⊕ Z ⇒ v0 v1 v2 v3 v0 v2 C1 ' Z ⊕ Z ⇒ e0 e1

C2 ' 0

H0 = Z0/B0 Special example H0

∂2 ∂1 ∂0 v1 v3 C2 −→ C1 −→ C0 −→ 0 e0 e1 C0 ' Z ⊕ Z ⊕ Z ⊕ Z ⇒ v0 v1 v2 v3 v0 v2 C1 ' Z ⊕ Z ⇒ e0 e1

C2 ' 0

H0 = Z0/B0

Z0 =ker∂0 ' C0 Special example H0

∂2 ∂1 ∂0 v1 v3 C2 −→ C1 −→ C0 −→ 0 e0 e1 C0 ' Z ⊕ Z ⊕ Z ⊕ Z ⇒ v0 v1 v2 v3 v0 v2 C1 ' Z ⊕ Z ⇒ e0 e1

C2 ' 0

H0 = Z0/B0

Z0 =ker∂0 ' C0

B0 =im∂1 = ∂1(e0) + ∂1(e1) ' Z ⊕ Z hv1 − v0, v3 − v2i Special example H0

∂2 ∂1 ∂0 v1 v3 C2 −→ C1 −→ C0 −→ 0 e0 e1 C0 ' Z ⊕ Z ⊕ Z ⊕ Z ⇒ v0 v1 v2 v3 v0 v2 C1 ' Z ⊕ Z ⇒ e0 e1

C2 ' 0

H0 = Z0/B0

Z0 =ker∂0 ' C0

B0 =im∂1 = ∂1(e0) + ∂1(e1) ' Z ⊕ Z hv1 − v0, v3 − v2i

v0 = v1 v2 = v3 Special example H0

∂2 ∂1 ∂0 v1 v3 C2 −→ C1 −→ C0 −→ 0 e0 e1 C0 ' Z ⊕ Z ⊕ Z ⊕ Z ⇒ v0 v1 v2 v3 v0 v2 C1 ' Z ⊕ Z ⇒ e0 e1

C2 ' 0

H0 = Z0/B0

Z0 =ker∂0 ' C0

B0 =im∂1 = ∂1(e0) + ∂1(e1) ' Z ⊕ Z hv1 − v0, v3 − v2i

v0 = v1 v2 = v3

H0 ' Z ⊕ Z Example formula: n1(v0 + v1) + n2(v2 + v3) Special example H0

H0 ' Z ⊕ Z Example formula: n1(v0 + v1) + n2(v2 + v3)

H0 represent the amount of connected components in this chain. Ex. Disk homology groups e2 v0 v2 T e0 e1

v1 Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1 Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1 = e1 + e2 + e0

H1 = Z1/B1 Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1 = e1 + e2 + e0

H1 = Z1/B1

Z1 =ker∂1 ' Z he0 + e1 + e2i Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1 = e1 + e2 + e0

H1 = Z1/B1

Z1 =ker∂1 ' Z he0 + e1 + e2i

B1 =im∂2 ' Z he0 + e1 + e2i Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1 = e1 + e2 + e0

H1 = Z1/B1

Z1 =ker∂1 ' Z he0 + e1 + e2i

B1 =im∂2 ' Z he0 + e1 + e2i

H1 = Z1/B1 = 0 Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1

H2 = Z2/B2 Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1

H2 = Z2/B2

Z2 =ker∂2 = ∂2(n1T ) Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1

H2 = Z2/B2

Z2 =ker∂2 = ∂2(n1T )

= n1(e0 + e1 + e2) = 0 Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1

H2 = Z2/B2

Z2 =ker∂2 = ∂2(n1T )

= n1(e0 + e1 + e2) = 0

n1 = 0 Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1

H2 = Z2/B2

Z2 =ker∂2 = ∂2(n1T )

= n1(e0 + e1 + e2) = 0

n1 = 0

Z2 = 0 Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1

H2 = Z2/B2

Z2 =ker∂2 = ∂2(n1T )

= n1(e0 + e1 + e2) = 0

n1 = 0

Z2 = 0 → B2 =im∂3 = 0 Ex. Disk homology groups e2 C0 and C1 are the same as previously v0 v2 T e0 e1 C2 ' Z hT i

∂2(A) = (v1v2) − (v0v2) + (v0v1) v1

H2 = Z2/B2

Z2 =ker∂2 = ∂2(n1T )

= n1(e0 + e1 + e2) = 0

n1 = 0

Z2 = 0 → B2 =im∂3 = 0

H2 = 0 Ex. Disk homology groups e2 H0 ' Z v0 v2 T e0 e1 H1 = 0

H2 = 0 v1 Ex. Disk homology groups e2 H0 ' Z v0 v2 T e0 e1 H1 = 0

H2 = 0 v1

Hn = 0 where n > 2 Ex. A homology group with a torsion coeﬃcient

v1 e1 v0

T1 e0 T2 e0

e2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 e0 T2 e0

e2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 C0 ' Z ⊕ Z hv0, v1i e0 T2 e0

e2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 C0 ' Z ⊕ Z hv0, v1i e0 T2 e0

C1 ' Z ⊕ Z ⊕ Z he0, e1, e2i e2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 C0 ' Z ⊕ Z hv0, v1i e0 T2 e0

C1 ' Z ⊕ Z ⊕ Z he0, e1, e2i e2 v0 e1 v1 C2 ' Z ⊕ Z hT1,T2i Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 C0 ' Z ⊕ Z hv0, v1i e0 T2 e0

C1 ' Z ⊕ Z ⊕ Z he0, e1, e2i e2 v0 e1 v1 C2 ' Z ⊕ Z hT1,T2i

∂1 = e0 → v1 − v0 e1 → v1 − v0 e2 → v0 − v0 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 C0 ' Z ⊕ Z hv0, v1i e0 T2 e0

C1 ' Z ⊕ Z ⊕ Z he0, e1, e2i e2 v0 e1 v1 C2 ' Z ⊕ Z hT1,T2i

∂1 = e0 → v1 − v0 e1 → v1 − v0 e2 → v0 − v0

∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H0 = Z0/B0 e0 T2 e0

e2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H0 = Z0/B0 e0 T2 e0 e Z0 ' C0 ' Z ⊕ Z hv0, v1i 2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H0 = Z0/B0 e0 T2 e0 e Z0 ' C0 ' Z ⊕ Z hv0, v1i 2 v0 e1 v1 B0 =im∂1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H0 = Z0/B0 e0 T2 e0 e Z0 ' C0 ' Z ⊕ Z hv0, v1i 2 v0 e1 v1 B0 =im∂1

∂1 = e0 → v1 − v0 e1 → v1 − v0 e2 → v0 − v0 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H0 = Z0/B0 e0 T2 e0 e Z0 ' C0 ' Z ⊕ Z hv0, v1i 2 v0 e1 v1 B0 =im∂1

∂1 = e0 → v1 − v0 e1 → v1 − v0 e2 → v0 − v0

∂1 = 0 ⇒ v1 = v0 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H0 = Z0/B0 e0 T2 e0 e Z0 ' C0 ' Z ⊕ Z hv0, v1i 2 v0 e1 v1 B0 =im∂1

∂1 = e0 → v1 − v0 e1 → v1 − v0 e2 → v0 − v0

∂1 = 0 ⇒ v1 = v0

B0 ' Z hv1 + v0i Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H0 = Z0/B0 e0 T2 e0 e Z0 ' C0 ' Z ⊕ Z hv0, v1i 2 v0 e1 v1 B0 =im∂1

∂1 = e0 → v1 − v0 e1 → v1 − v0 e2 → v0 − v0

∂1 = 0 ⇒ v1 = v0

B0 ' Z hv1 + v0i

H0 = Z0/B0 ' Z Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0

e2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0

Z1 =ker∂1 e2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0

Z1 =ker∂1 e2 v0 e1 v1 ker∂1 = n0e0 + n1e1 + n2e2 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0

Z1 =ker∂1 e2 v0 e1 v1 ker∂1 = n0e0 + n1e1 + n2e2

= n0(v1 − v0) + n1(v1 − v0) + n2(v0 − v0) Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0

Z1 =ker∂1 e2 v0 e1 v1 ker∂1 = n0e0 + n1e1 + n2e2

= n0(v1 − v0) + n1(v1 − v0) + n2(v0 − v0)

= v1(n0 + n1) + v0(n0 + n1 + n2 − n2) Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0

Z1 =ker∂1 e2 v0 e1 v1 ker∂1 = n0e0 + n1e1 + n2e2

= n0(v1 − v0) + n1(v1 − v0) + n2(v0 − v0)

= v1(n0 + n1) + v0(n0 + n1 + n2 − n2)

⇒ n0 = −n1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0

Z1 =ker∂1 e2 v0 e1 v1 ker∂1 = n0e0 + n1e1 + n2e2

= n0(v1 − v0) + n1(v1 − v0) + n2(v0 − v0)

= v1(n0 + n1) + v0(n0 + n1 + n2 − n2)

⇒ n0 = −n1

ker∂1 ' Z ⊕ Z he0 − e1, e2i Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 =ker∂1 ' Z ⊕ Z he0 − e1, e2i 2 v0 e1 v1 B1 =im∂2 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 =ker∂1 ' Z ⊕ Z he0 − e1, e2i 2 v0 e1 v1 B1 =im∂2

∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 =ker∂1 ' Z ⊕ Z he0 − e1, e2i 2 v0 e1 v1 B1 =im∂2

∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2

B1 ' Z ⊕ Z h−e0 + e1 + e2, e0 − e1 + e2i Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 =ker∂1 ' Z ⊕ Z he0 − e1, e2i 2 v0 e1 v1 B1 =im∂2

∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2

B1 ' Z ⊕ Z h−e0 + e1 + e2, e0 − e1 + e2i Terms are not similar.. What now? Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 =ker∂1 ' Z ⊕ Z he0 − e1, e2i 2 v0 e1 v1 B1 =im∂2

∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2

B1 ' Z ⊕ Z h−e0 + e1 + e2, e0 − e1 + e2i Terms are not similar.. What now?

he0 − e1, e2i ' he0 − e1 + e2, e2i Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 ' Z ⊕ Z he0 − e1 + e2, e2i 2 v0 e1 v1 B1 ' Z ⊕ Z h−e0 + e1 + e2, e0 − e1 + e2i Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 ' Z ⊕ Z he0 − e1 + e2, e2i 2 v0 e1 v1 B1 ' Z ⊕ Z h−e0 + e1 + e2, e0 − e1 + e2i

Goal: translate (−e0 + e1 + e2) into some term of (e2) Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 ' Z ⊕ Z he0 − e1 + e2, e2i 2 v0 e1 v1 B1 ' Z ⊕ Z h−e0 + e1 + e2, e0 − e1 + e2i

Goal: translate (−e0 + e1 + e2) into some term of (e2)

2e2 = (−e0 + e1 + e2) + (e0 − e1 + e2) Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 ' Z ⊕ Z he0 − e1 + e2, e2i 2 v0 e1 v1 B1 ' Z ⊕ Z h−e0 + e1 + e2, e0 − e1 + e2i

Goal: translate (−e0 + e1 + e2) into some term of (e2)

2e2 = (−e0 + e1 + e2) + (e0 − e1 + e2)

H1 = Z1/B1 ' he0 − e1 + e2, e2i/he0 − e1 + e2, 2e2i Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H1 = Z1/B1 e0 T2 e0 e Z1 ' Z ⊕ Z he0 − e1 + e2, e2i 2 v0 e1 v1 B1 ' Z ⊕ Z h−e0 + e1 + e2, e0 − e1 + e2i

Goal: translate (−e0 + e1 + e2) into some term of (e2)

2e2 = (−e0 + e1 + e2) + (e0 − e1 + e2)

H1 = Z1/B1 ' he0 − e1 + e2, e2i/he0 − e1 + e2, 2e2i

H1 ' Z2 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H2 = Z2/B2 e0 T2 e0

e2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H2 = Z2/B2 e0 T2 e0

Z2 =ker∂2 e2 v0 e1 v1 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H2 = Z2/B2 e0 T2 e0

Z2 =ker∂2 e2 v0 e1 v1 ∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H2 = Z2/B2 e0 T2 e0

Z2 =ker∂2 e2 v0 e1 v1 ∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2

n1(−e0 + e1 + e2) + n2(e0 − e1 + e2) = 0 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H2 = Z2/B2 e0 T2 e0

Z2 =ker∂2 e2 v0 e1 v1 ∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2

n1(−e0 + e1 + e2) + n2(e0 − e1 + e2) = 0

e0(−n1 + n2) + e1(n1 − n2) + e2(n1 + n2) = 0 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H2 = Z2/B2 e0 T2 e0

Z2 =ker∂2 e2 v0 e1 v1 ∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2

n1(−e0 + e1 + e2) + n2(e0 − e1 + e2) = 0

e0(−n1 + n2) + e1(n1 − n2) + e2(n1 + n2) = 0

n1 = n2 n1 + n2 = 0 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H2 = Z2/B2 e0 T2 e0

Z2 =ker∂2 e2 v0 e1 v1 ∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2

n1(−e0 + e1 + e2) + n2(e0 − e1 + e2) = 0

e0(−n1 + n2) + e1(n1 − n2) + e2(n1 + n2) = 0

n1 = n2 n1 + n2 = 0

n1 = n2 = 0 Ex. A homology group with a torsion coeﬃcient

∂2 ∂1 ∂0 v1 e1 v0 C2 −→ C1 −→ C0 −→ 0 T1 H2 = Z2/B2 e0 T2 e0

Z2 =ker∂2 e2 v0 e1 v1 ∂2 = T1 → −e0 + e1 + e2 T2 → e0 − e1 + e2

n1(−e0 + e1 + e2) + n2(e0 − e1 + e2) = 0

e0(−n1 + n2) + e1(n1 − n2) + e2(n1 + n2) = 0

n1 = n2 n1 + n2 = 0

n1 = n2 = 0

H2 = Z2/B2 = 0/B2 = 0 Ex. A homology group with a torsion coeﬃcient

• H0 ' Z hv0 + v1i v1 e1 v0

T1 • H1 ' Z2 he2i e0 T2 e0 • H2 = 0 e2 v0 e1 v1 Betti numbers

Bn = rank Hn Betti numbers

Bn = rank Hn

Hn ' Z ⊕ Z ⊕ .. ⊕ Z ⊕ Z2 ⊕ Zx Betti numbers

Bn = rank Hn

Hn ' Z ⊕ Z ⊕ .. ⊕ Z ⊕ Z2 ⊕ Zx • Inﬁnite pieces = Z ⊕ Z ⊕ ... ⊕ Z • Finite pieces = torsion co-eﬃcients = Z2 ⊕ Zx Betti numbers

Bn = rank Hn

Hn ' Z ⊕ Z ⊕ .. ⊕ Z ⊕ Z2 ⊕ Zx • Infinite pieces = Z ⊕ Z ⊕ ... ⊕ Z • Finite pieces = torsion co-efficients = Z2 ⊕ Zx Rank only counts the infinite numbers. Betti numbers

Bn = rank Hn

Hn ' Z ⊕ Z ⊕ .. ⊕ Z ⊕ Z2 ⊕ Zx • Infinite pieces = Z ⊕ Z ⊕ ... ⊕ Z • Finite pieces = torsion co-efficients = Z2 ⊕ Zx Rank only counts the infinite numbers.

Let’s look at some examples. Betti numbers

• H0 ' e2 Z v0 v2 • H1 ' Z • Hn = 0 where n ≥ 2 e0 e1

v1 Betti numbers

• H0 ' e2 Z v0 v2 • H1 ' Z • Hn = 0 where n ≥ 2 e0 e1

• B0 = 1 v1 • B1 = 1 • Bn = 0 where n ≥ 2 Betti numbers

• H0 ' e2 Z v0 v2 • H1 ' Z • Hn = 0 where n ≥ 2 e0 e1

• B0 = 1 v1 • B1 = 1 • Bn = 0 where n ≥ 2 v e v • H0 ' Z 1 1 0 • H1 ' Z2 T1 • Hn = 0 where n ≥ 2 e0 T2 e0

e2 v0 e1 v1 Betti numbers

• H0 ' e2 Z v0 v2 • H1 ' Z • Hn = 0 where n ≥ 2 e0 e1

• B0 = 1 v1 • B1 = 1 • Bn = 0 where n ≥ 2 v e v • H0 ' Z 1 1 0 • H1 ' Z2 T1 • Hn = 0 where n ≥ 2 e0 T2 e0

• B0 = 1 e2 • B1 = 0 v0 e1 v1 • Bn = 0 where n ≥ 2 Betti numbers Now what would be the betti numbers of the empty donut ? Betti numbers Now what would be the betti numbers of the empty donut ?

• All the vertices are connected, one connected group. Betti numbers Now what would be the betti numbers of the empty donut ?

• All the vertices are connected, one connected group.

• For the 1-dimensional holes, we have one which would go over ABC. We have another one which go from ICF . Betti numbers Now what would be the betti numbers of the empty donut ?

• All the vertices are connected, one connected group.

• For the 1-dimensional holes, we have one which would go over ABC. We have another one which go from ICF .

• For the 2-dimensional holes, we have the complete empty space in the middle. Betti numbers Now what would be the betti numbers of the empty donut ?

• All the vertices are connected, one connected group.

• For the 1-dimensional holes, we have one which would go over ABC. We have another one which go from ICF .

• For the 2-dimensional holes, we have the complete empty space in the middle.

So what are the betti numbers? Betti numbers Now what would be the betti numbers of the empty donut ?

• All the vertices are connected, one connected group.

• For the 1-dimensional holes, we have one which would go over ABC. We have another one which go from ICF .

• For the 2-dimensional holes, we have the complete empty space in the middle.

So what are the betti numbers?

• B0 = 1 • B1 = 2 • B2 = 1 • Bn = 0 for n > 2 Betti numbers And what about the question that was asked earlier? Can we deform a disk into a circle and the other way around? Betti numbers And what about the question that was asked earlier? Can we deform a disk into a circle and the other way around?

The betti numbers of the two are diﬀerent, as the circle has B1 = 1 and the disk has B1 = 0. Betti numbers And what about the question that was asked earlier? Can we deform a disk into a circle and the other way around?

The betti numbers of the two are diﬀerent, as the circle has B1 = 1 and the disk has B1 = 0. No, we can not deform the one into the other. Euler characteristics P n χ(K) = n(−1) rank Hn(K) Euler characteristics P n χ(K) = n(−1) rank Hn(K)

χ(K) = b0 − b1 + b2 − b3 + b4 − b5... Euler characteristics P n χ(K) = n(−1) rank Hn(K)

χ(K) = b0 − b1 + b2 − b3 + b4 − b5... So let’s take a sphere, which is isomorphic to a tetrahedron. Euler characteristics P n χ(K) = n(−1) rank Hn(K)

χ(K) = b0 − b1 + b2 − b3 + b4 − b5... So let’s take a sphere, which is isomorphic to a tetrahedron.

• H0 ' Z • H1 = 0 • H2 ' Z • Hn = 0 for n > 2 Euler characteristics P n χ(K) = n(−1) rank Hn(K)

χ(K) = b0 − b1 + b2 − b3 + b4 − b5... So let’s take a sphere, which is isomorphic to a tetrahedron.

• H0 ' Z • H1 = 0 • H2 ' Z • Hn = 0 for n > 2 χ(K) = 1 − 0 + 1 = 2 Euler characteristics P n χ(K) = n(−1) rank Hn(K)

χ(K) = b0 − b1 + b2 − b3 + b4 − b5...

What about this one? Euler characteristics P n χ(K) = n(−1) rank Hn(K)

χ(K) = b0 − b1 + b2 − b3 + b4 − b5...

What about this one? v1 e1 v0

• H0 ' Z T1 • H1 ' Z2 e0 T2 e0 • Hn = 0 where n ≥ 2 e2 v0 e1 v1 Euler characteristics P n χ(K) = n(−1) rank Hn(K)

χ(K) = b0 − b1 + b2 − b3 + b4 − b5...

What about this one? v1 e1 v0

• H0 ' Z T1 • H1 ' Z2 e0 T2 e0 • Hn = 0 where n ≥ 2 e2 χ(X) = 1 − 0 + 0 = 1 v0 e1 v1 Homotopy equivalance Remember the replacement of the circle with the triangle? Homotopy equivalance Remember the replacement of the circle with the triangle?

Possible because it is homeomorphic to each other. Homotopy equivalance Remember the replacement of the circle with the triangle?

Possible because it is homeomorphic to each other.

Recall: A homeomorphism is a bijective continuous function between topological spaces that has a continuous inverse function. Homotopy equivalance Remember the replacement of the circle with the triangle?

Possible because it is homeomorphic to each other.

Homotopy equivalence • Given two spaces X and Y . • Given two continuous maps f : X → Y and g : Y → X • Then X = g(f(X)) and Y = f(g(Y )) holds. Homotopy equivalance Remember the replacement of the circle with the triangle?

Possible because it is homeomorphic to each other.

Homotopy equivalence • Given two spaces X and Y . • Given two continuous maps f : X → Y and g : Y → X • Then X = g(f(X)) and Y = f(g(Y )) holds. Homeomorphism ⇒ Homotopy equivalance. Homotopy equivalance Remember the replacement of the circle with the triangle?

Possible because it is homeomorphic to each other.

Homotopy equivalence • Given two spaces X and Y . • Given two continuous maps f : X → Y and g : Y → X • Then X = g(f(X)) and Y = f(g(Y )) holds. Homeomorphism ⇒ Homotopy equivalance. Note: Homotopy equivalance ; Homeomorphism. A disk is not homeomorphic to a single point, although the disk and the point are homotopy equivalant. Homotopy invariance Any homotopy equivalance are subject to the homotopy invariance. Which states that if X and Y are homotopy equivalant spaces: Homotopy invariance Any homotopy equivalance are subject to the homotopy invariance. Which states that if X and Y are homotopy equivalant spaces:

• If X is path-connected, so is Y. Homotopy invariance Any homotopy equivalance are subject to the homotopy invariance. Which states that if X and Y are homotopy equivalant spaces: