7Simplicial homology
II Algebraic Topology
7.2 Some homological algebra
We will develop some formalism to help us compute homology groups
H
k
(
K
) in
lots of examples. Firstly, we axiomatize the setup we had above.
Definition (Chain complex and differentials). A chain complex
C
·
is a sequence
of abelian groups
C
0
, C
1
, C
2
, · · ·
equipped with maps
d
n
:
C
n
→ C
n−1
such that
d
n−1
◦ d
n
= 0 for all n. We call these maps the differentials of C
·
.
0 C
0
C
1
C
2
· · ·
d
0
d
1
d
2
d
3
Whenever we have some of these things, we can define homology groups in
exactly the same way as we defined them for simplicial complexes.
Definition (Cycles and boundaries). The space of n-cycles is
Z
n
(C) = ker d
n
.
The space of n-boundaries is
B
n
(C) = im d
n+1
.
Definition (Homology group). The
n
-th homology group of
C
·
is defined to be
H
n
(C) =
ker d
n
im d
n+1
=
Z
n
(C)
B
n
(C)
.
In mathematics, when we have objects, we don’t just talk about the objects
themselves, but also functions between them. Suppose we have two chain
complexes. For the sake of simplicity, we write the chain maps of both as d
n
.
In general, we want to have maps between these two sequences. This would
correspond to having a map
f
i
:
C
i
→ D
i
for each
i
, satisfying some commuta-
tivity relations.
Definition (Chain map). A chain map
f
·
:
C
·
→ D
·
is a sequence of homo-
morphisms f
n
: C
n
→ D
n
such that
f
n−1
◦ d
n
= d
n
◦ f
n
for all n. In other words, the following diagram commutes:
0 C
0
C
1
C
2
· · ·
0 D
0
D
1
D
2
· · ·
d
0
f
0
d
1
f
1
d
2
f
2
d
3
d
0
d
1
d
2
d
3
We want to have homotopies. So far, chain complexes seem to be rather rigid,
but homotopies themselves are rather floppy. How can we define homotopies for
chain complexes? It turns out we can have a completely algebraic definition for
chain homotopies.
Definition (Chain homotopy). A chain homotopy between chain maps
f
·
, g
·
:
C
·
→ D
·
is a sequence of homomorphisms h
n
: C
n
→ D
n+1
such that
g
n
− f
n
= d
n+1
◦ h
n
+ h
n−1
◦ d
n
.
We write f
·
' g
·
if there is a chain homotopy between f
·
and g
·
.
The arrows can be put in the following diagram, which is not commutative:
C
n−1
C
n
D
n
D
n+1
h
n−1
d
n
f
n
g
n
h
n
d
n+1
The intuition behind this definition is as follows: suppose
C
·
=
C
·
(
K
) and
D
·
=
C
·
(
L
) for
K, L
simplicial complexes, and
f
·
and
g
·
are “induced” by
simplicial maps f, g : K → L (if f maps an n-simplex σ to a lower-dimensional
simplex, then
f
•
σ
= 0). How can we detect if
f
and
g
are homotopic via the
homotopy groups?
Suppose
H
:
|K| × I → |L|
is a homotopy from
f
to
g
. We further suppose
that
H
actually comes from a simplicial map
K × I → L
(we’ll skim over the
technical issue of how we can make
K × I
a simplicial complex. Instead, you are
supposed to figure this out yourself in example sheet 3).
Let σ be an n-simplex of K, and here is a picture of H(σ × I):
Let
h
n
(σ) = H(σ × I).
We think of this as an (
n
+ 1)-chain. What is its boundary? We’ve got the
vertical sides plus the top and bottom. The bottom should just be
f
(
σ
), and
the top is just
g
(
σ
), since
H
is a homotopy from
f
to
g
. How about the sides?
They are what we get when we pull the boundary
∂σ
up with the homotopy,
i.e.
H
(
∂σ × I
) =
h
n−1
◦ d
n
(
σ
). Now note that
f
(
σ
) and
g
(
σ
) have opposite
orientations, so we get the result
d
n+1
◦ h
n
(σ) = h
n−1
◦ d
n
(σ) + g
n
(σ) − f
n
(σ).
Rearranging and dropping the σs, we get
d
n+1
◦ h
n
− h
n−1
◦ d
n
= g
n
− f
n
.
This looks remarkably like our definition for chain homotopies of maps, with the
signs a bit wrong. So in reality, we will have to fiddle with the sign of
h
n
a bit
to get it right, but you get the idea.
Lemma. A chain map f
·
: C
·
→ D
·
induces a homomorphism:
f
∗
: H
n
(C) → H
n
(D)
[c] 7→ [f(c)]
Furthermore, if f
·
and g
·
are chain homotopic, then f
∗
= g
∗
.
Proof.
Since the homology groups are defined as the cycles quotiented by the
boundaries, to show that
f
∗
defines a homomorphism, we need to show
f
sends
cycles to cycles and boundaries to boundaries. This is an easy check. If
d
n
(
σ
) = 0,
then
d
n
(f
n
(σ)) = f
n
(d
n
(σ)) = f
n
(0) = 0.
So f
n
(σ) ∈ Z
n
(D).
Similarly, if σ is a boundary, say σ = d
n
(τ), then
f
n
(σ) = f
n
(d
n
(τ)) = d
n
(f
n
(τ)).
So f
n
(σ) is a boundary. It thus follows that f
∗
is well-defined.
Now suppose
h
n
is a chain homotopy between
f
and
g
. For any
c ∈ Z
n
(
C
),
we have
g
n
(c) − f
n
(c) = d
n+1
◦ h
n
(c) + h
n−1
◦ d
n
(c).
Since c ∈ Z
n
(C), we know that d
n
(c) = 0. So
g
n
(c) − f
n
(c) = d
n+1
◦ h
n
(c) ∈ B
n
(D).
Hence
g
n
(
c
) and
f
n
(
c
) differ by a boundary. So [
g
n
(
c
)]
−
[
f
n
(
c
)] = 0 in
H
n
(
D
),
i.e. f
∗
(c) = g
∗
(c).
The following statements are easy to check:
Proposition.
(i) Being chain-homotopic is an equivalence relation of chain maps.
(ii) If a
·
: A
·
→ C
·
is a chain map and f
·
' g
·
, then f
·
◦ a
·
' g
·
◦ a
·
.
(iii) If f : C
·
→ D
·
and g : D
·
→ A
·
are chain maps, then
g
∗
◦ f
∗
= (g
·
◦ f
·
)
∗
.
(iv) (id
C
·
)
∗
= id
H
∗
(C)
.
The last two statements can be summarized in fancy language by saying that
H
n
is a functor.
Definition (Chain homotopy equivalence). Chain complexes
C
·
and
D
·
are
chain homotopy equivalent if there exist
f
·
:
C
·
→ D
·
and
g
·
:
D
·
→ C
·
such
that
f
·
◦ g
·
' id
D
·
, g
·
◦ f
·
' id
C
·
.
We should think of this in exactly the same way as we think of homotopy
equivalences of spaces. The chain complexes themselves are not necessarily the
same, but the induced homology groups will be.
Lemma. Let
f
·
:
C
·
→ D
·
be a chain homotopy equivalence, then
f
∗
:
H
n
(
C
)
→
H
n
(D) is an isomorphism for all n.
Proof.
Let
g
·
be the homotopy inverse. Since
f
·
◦ g
·
' id
D
·
, we know
f
∗
◦ g
∗
=
id
H
∗
(D)
. Similarly,
g
∗
◦ f
∗
=
id
H
∗
(C)
. So we get isomorphisms between
H
n
(
C
)
and H
n
(D).