7Simplicial homology

II Algebraic Topology



7.5 Continuous maps and homotopy invariance
This is the most technical part of the homology section of the course. The goal is
to see that the homology groups
H
(
K
) depend only on the polyhedron, and not
the simplicial structure on it. Moreover, we will show that they are homotopy
invariants of the space, and that homotopic maps
f ' g
:
|K| |L|
induce
equal maps
H
(
K
)
H
(
L
). Note that this is a lot to prove. At this point, we
don’t even know arbitrary continuous maps induce any map on the homology.
We only know simplicial maps do.
We start with a funny definition.
Definition (Contiguous maps). Simplicial maps
f, g
:
K L
are contiguous if
for each
σ K
, the simplices
f
(
σ
) and
g
(
σ
) (i.e. the simplices spanned by the
image of the vertices of σ) are faces of some some simplex τ L.
σ
τ
g
f
The significance of this definition comes in two parts: simplicial approxima-
tions of the same map are contiguous, and contiguous maps induce the same
maps on homology.
Lemma. If
f, g
:
K L
are simplicial approximations to the same map
F
,
then f and g are contiguous.
Proof.
Let
σ K
, and pick some
s ˚σ
. Then
F
(
s
)
˚τ
for some
τ L
.
Then the definition of simplicial approximation implies that for any simplicial
approximation f to F , f(σ) spans a face of τ.
Lemma. If f, g : K L are continguous simplicial maps, then
f
= g
: H
n
(K) H
n
(L)
for all n.
Geometrically, the homotopy is obvious. If
f
(
σ
) and
g
(
σ
) are both faces of
τ
,
then we just pick the homotopy as
τ
. The algebraic definition is less inspiring,
and it takes some staring to see it is indeed what we think it should be.
Proof. We will write down a chain homotopy between f and g:
h
n
((a
0
, · · · , a
n
)) =
n
X
i=0
(1)
i
[f(a
0
), · · · , f(a
i
), g(a
i
), · · · , g(a
n
)],
where the square brackets means the corresponding oriented simplex if there are
no repeats, 0 otherwise.
We can now check by direct computation that this is indeed a chain homotopy.
Now we know that if
f, g
:
K L
are both simplicial approximations to the
same
F
, then they induce equal maps on homology. However, we do not yet know
that continuous maps induce well-defined maps on homologies, since to produce
simplicial approximations, we needed to perform barycentric subdivision, and
we need to make sure this does not change the homology.
K
K
0
Given a barycentric subdivision, we need to choose a simplicial approximation
to the identity map
a
:
K
0
K
. It turns out this is easy to do, and we can do
it almost arbitrarily.
Lemma. Each vertex
ˆσ K
0
is a barycenter of some
σ K
. Then we choose
a
(
ˆσ
) to be an arbitrary vertex of
σ
. This defines a function
a
:
V
K
0
V
K
.
This a is a simplicial approximation to the identity. Moreover, every simplicial
approximation to the identity is of this form.
Proof. Omitted.
Next, we need to show this gives an isomorphism on the homologies.
Proposition. Let
K
0
be the barycentric subdivision of
K
, and
a
:
K
0
K
a simplicial approximation to the identity map. Then the induced map
a
:
H
n
(K
0
) H
n
(K) is an isomorphism for all n.
Proof.
We first deal with
K
being a simplex
σ
and its faces. Now that
K
is just
a cone (with any vertex as the cone vertex), and
K
0
is also a cone (with the
barycenter as the cone vertex). Therefore
H
n
(K)
=
H
n
(K
0
)
=
(
Z n = 0
0 n > 0
So only
n
= 0 is (a little) interesting, but it is easy to check that
a
is an
isomorphism in this case, since it just maps a vertex to a vertex, and all vertices
in each simplex are in the same homology class.
To finish the proof, note that
K
is built up by gluing up simplices, and
K
0
is built by gluing up subdivided simplices. So to understand their homology
groups, we use the Mayer-Vietoris sequence.
Given a complicated simplicial complex
K
, let
σ
be a maximal dimensional
simplex of
K
. We let
L
=
K \ {σ}
(note that
L
includes the boundary of
σ
).
We let S = {σ and all its faces} K and T = L S.
We can do similarly for
K
0
, and let
L
0
, S
0
, T
0
be the corresponding barycentric
subdivisions. We have
K
=
L S
and
K
0
=
L
0
S
0
(and
L
0
S
0
=
T
0
). By the
previous lemma, we see our construction of
a
gives
a
(
L
0
)
L
,
a
(
S
0
)
S
and
a(T
0
) T . So these induce maps of the corresponding homology groups
H
n
(T
0
) H
n
(S
0
) H
n
(L
0
) H
n
(K
0
) H
n1
(T
0
) H
n1
(S
0
) H
n1
(L
0
)
H
n
(T ) H
n
(S) H
n
(L) H
n
(K) H
n1
(T ) H
n1
(S) H
n1
(L)
a
a
a
a
a
a
a
By induction, we can assume all but the middle maps are isomorphisms. By
the five lemma, this implies the middle map is an isomorphism, where the five
lemma is as follows:
Lemma (Five lemma). Consider the following commutative diagram:
A
1
B
1
C
1
D
1
E
1
A
2
B
2
C
2
D
2
E
2
a
b
c
d
e
If the top and bottom rows are exact, and
a, b, d, e
are isomorphisms, then
c
is
also an isomorphism.
Proof. Exercise in example sheet.
We now have everything we need to move from simplical maps to continuous
maps. Putting everything we have proved so far together, we obtain the following
result:
Proposition. To each continuous map
f
:
|K| |L|
, there is an associated
map f
: H
n
(K) H
n
(L) (for all n) given by
f
= s
ν
1
K,r
,
where
s
:
K
(r)
L
is a simplicial approximation to
f
, and
ν
K,r
:
H
n
(
K
(r)
)
H
n
(
K
) is the isomorphism given by composing maps
H
n
(
K
(i)
)
H
n
(
K
(i1)
)
induced by simplical approximations to the identity.
Furthermore:
(i) f
does not depend on the choice of r or s.
(ii) If g : |M| |K| is another continuous map, then
(f g)
= f
g
.
Proof. Omitted.
Corollary. If
f
:
|K| |L|
is a homeomorphism, then
f
:
H
n
(
K
)
H
n
(
L
) is
an isomorphism for all n.
Proof. Immediate from (ii) of previous proposition.
This is good. We know now that homology groups is a property of the space
itself, not simplicial complexes. For example, we have computed the homology
groups of a particular triangulation of the
n
-sphere, and we know it is true for
all triangulations of the n-sphere.
We’re not exactly there yet. We have just seen that homology groups are
invariant under homeomorphisms. What we would like to know is that they are
invariant under homotopy. In other words, we want to know homotopy equivalent
maps induce equal maps on the homology groups.
The strategy is:
(i) Show that “small” homotopies don’t change the maps on H
n
(ii) Note that all homotopies can be decomposed into “small” homotopies.
Lemma. Let
L
be a simplicial complex (with
|L| R
n
). Then there is an
ε
=
ε
(
L
)
>
0 such that if
f, g
:
|K| |L|
satisfy
kf
(
x
)
g
(
x
)
k < ε
, then
f
= g
: H
n
(K) H
n
(L) for all n.
The idea of the proof is that if
kf
(
x
)
g
(
x
)
k
is small enough, we can
barycentrically subdivide
K
such that we get a simplicial approximation to both
f and g.
Proof.
By the Lebesgue number lemma, there is an
ε >
0 such that each ball of
radius 2ε in |L| lies in some star St
L
(w).
Now apply the Lebesgue number lemma again to
{f
1
(
B
ε
(
y
))
}
y∈|L|
, an open
cover of |K|, and get δ > 0 such that for all x |K|, we have
f(B
δ
(x)) B
ε
(y) B
2ε
(y) St
L
(w)
for some y |L| and St
L
(w). Now since g and f differ by at most ε, we know
g(B
δ
(x)) B
2ε
(y) St
L
(w).
Now subdivide r times so that µ(K
(r)
) <
1
2
δ. So for all v V
K
(r)
, we know
St
K
(r)
(v) B
δ
(v).
This gets mapped by both
f
and
g
to
St
L
(
w
) for the same
w V
L
. We define
s : V
K
(r)
V
L
sending v 7→ w.
Theorem. Let f ' g : |K| |L|. Then f
= g
.
Proof.
Let
H
:
|K| × I |L|
. Since
|K| × I
is compact, we know
H
is uniformly
continuous. Pick
ε
=
ε
(
L
) as in the previous lemma. Then there is some
δ
such
that |s t| < δ implies |H(x, s) H(x, t)| < ε for all x |K|.
Now choose 0 =
t
0
< t
1
< · · · < t
n
= 1 such that
t
i
t
i1
< δ
for all
i
.
Define
f
i
:
|K| |L|
by
f
i
(
x
) =
H
(
x, t
i
). Then we know
kf
i
f
i1
k < ε
for all
i. Hence (f
i
)
= (f
i1
)
. Therefore (f
0
)
= (f
n
)
, i.e. f
= g
.
This is good, since we know we can not only deal with spaces that are
homeomorphic to complexes, but spaces that are homotopic to complexes. This
is important, since all complexes are compact, but we would like to talk about
non-compact spaces such as open balls as well.
Definition (
h
-triangulation and homology groups). An
h
-triangulation of a
space
X
is a simplicial complex
K
and a homotopy equivalence
h
:
|K| X
.
We define H
n
(X) = H
n
(K) for all n.
Lemma. H
n
(X) is well-defined, i.e. it does not depend on the choice of K.
Proof. Clear from previous theorem.
We have spent a lot of time and effort developing all the technical results
and machinery of homology groups. We will now use them to do stuff.
Digression historical motivation
At first, we motivated the study of algebraic topology by saying we wanted to
show that
R
n
and
R
m
are not homeomorphic. However, historically, this is not
why people studied algebraic topology, since there are other ways to prove this
is true.
Initially, people were interested in studying knots. These can be thought of
as embeddings
S S
3
. We really should just think of
S
3
as
R
3
{∞}
, where
the point at infinity is just there for convenience.
The most interesting knot we know is the unknot U:
A less interesting but still interesting knot is the trefoil T .
One question people at the time asked was whether the trefoil knot is just the
unknot in disguise. It obviously isn’t, but can we prove it? In general, given two
knots, is there any way we can distinguish if they are the same?
The idea is to study the fundamental groups of the knots. It would obviously
be silly to compute the fundamental groups of
U
and
T
themselves, since they
are both isomorphic to
S
1
and the groups are just
Z
. Instead, we look at the
fundamental groups of the complements. It is not difficult to see that
π
1
(S
3
\ U)
=
Z,
and with some suitable machinery, we find
π
1
(S
3
\ T )
=
ha, b | a
3
b
2
i.
Staring at it hard enough, we can construct a surjection
π
1
(
S
3
\ T
)
S
3
. This
tells us
π
1
(
S
3
\ T
) is non-abelian, and is certainly not
Z
. So we know
U
and
T
are genuinely different knots.