6Simplicial complexes
II Algebraic Topology
6.2 Simplicial approximation
This is all very well, but we are really interested in continuous maps. So given a
continuous map
f
, we would like to find a related simplicial map
g
. In particular,
we want to find a simplicial map
g
that “approximates”
f
in some sense. The
definition we will write down is slightly awkward, but it turns out this is the
most useful definition.
Definition (Open star and link). Let
x ∈ |K|
. The open star of
x
is the union
of all the interiors of the simplices that contain x, i.e.
St
K
(x) =
[
x∈σ∈K
˚σ.
The link of
x
, written
Lk
K
(
x
), is the union of all those simplices that do not
contain x, but are faces of a simplex that does contain x.
Definition (Simplicial approximation). Let
f
:
|K| → |L|
be a continuous map
between the polyhedra. A function
g
:
V
K
→ V
L
is a simplicial approximation
to f if for all v ∈ V
K
,
f(St
K
(v)) ⊆ St
L
(g(v)). (∗)
The following lemma tells us why this is a good definition:
Lemma. If
f
:
|K| → |L|
is a map between polyhedra, and
g
:
V
K
→ V
L
is a simplicial approximation to
f
, then
g
is a simplicial map, and
|g| ' f
.
Furthermore, if
f
is already simplicial on some subcomplex
M ⊆ K
, then we get
g|
M
= f|
M
, and the homotopy can be made rel M.
Proof.
First we want to check
g
is really a simplicial map if it satisfies (
∗
). Let
σ
=
ha
0
, · · · , a
n
i
be a simplex in
K
. We want to show that
{g
(
a
0
)
, · · · , g
(
a
n
)
}
spans a simplex in L.
Pick an arbitrary
x ∈ ˚σ
. Since
σ
contains each
a
i
, we know that
x ∈ St
K
(
a
i
)
for all i. Hence we know that
f(x) ∈
n
\
i=0
f(St
K
(a
i
)) ⊆
n
\
i=0
St
L
(g(a
i
)).
Hence we know that there is one simplex, say,
τ
that contains all
g
(
a
i
) whose
interior contains
f
(
x
). Since each
g
(
a
i
) is a vertex in
L
, each
g
(
a
i
) must be a
vertex of τ. So they span a face of τ, as required.
We now want to prove that
|g| ' f
. We let
H
:
|K| × I → |L| ⊆ R
m
be
defined by
(x, t) 7→ t|g|(x) + (1 − t)f(x).
This is clearly continuous. So we need to check that
im H ⊆ |L|
. But we
know that both
|g|
(
x
) and
f
(
x
) live in
τ
and
τ
is convex. It thus follows that
H(x × I) ⊆ τ ⊆ |L|.
To prove the last part, it suffices to show that every simplicial approximation
to a simplicial map must be the map itself. Then the homotopy is
rel M
by
the construction above. This is easily seen to be true — if
g
is a simplicial
approximation to
f
, then
f
(
v
)
∈ f
(
St
K
(
v
))
⊆ St
L
(
g
(
v
)). Since
f
(
v
) is a vertex
and
g
(
v
) is the only vertex in
St
L
(
g
(
v
)), we must have
f
(
v
) =
g
(
v
). So done.
What’s the best thing we might hope for at this point? It would be great if
every map were homotopic to a simplicial map. Is this possible? Let’s take a
nice example. Let’s consider the following K:
How many homotopy classes of continuous maps are there
K → K
? Countably
many, one for each winding number. However, there can only be at most 3
3
= 27
simplicial maps. The problem is that we don’t have enough vertices to realize all
those interesting maps. The idea is to refine our simplicial complexes. Suppose
we have the following simplex:
What we do is to add a point in the center of each simplex, and join them up:
This is known as the barycentric subdivision. After we subdivide it once, we can
realize more homotopy classes of maps. We will show that for any map, as long
as we are willing to barycentrically subdivide the simplex many times, we can
find a simplicial approximation to it.
Definition (Barycenter). The barycenter of σ = ha
0
, · · · , a
n
i is
ˆσ =
n
X
i=0
1
n + 1
a
i
.
Definition (Barycentric subdivision). The (first) barycentric subdivision
K
0
of
K is the simplicial complex:
K
0
= {hˆσ
0
, · · · , ˆσ
n
i : σ
i
∈ K and σ
0
< σ
1
< · · · < σ
n
}.
If you stare at this long enough, you will realize this is exactly what we have
drawn above.
The
r
th barycentric subdivision
K
(r)
is defined inductively as the barycentric
subdivision of the r − 1th barycentric subdivision, i.e.
K
(r)
= (K
(r−1)
)
0
.
Proposition. |K| = |K
0
| and K
0
really is a simplicial complex.
Proof. Too boring to be included in lectures.
We now have a slight problem. Even though
|K
0
|
and
|K|
are equal, the
identity map from |K
0
| to |K| is not a simplicial map.
To solve this problem, we can choose any function
K → V
K
by
σ 7→ v
σ
with
v
σ
∈ σ
, i.e. a function that sends any simplex to any of its vertices. Then we can
define
g
:
K
0
→ K
by sending
ˆσ 7→ v
σ
. Then this is a simplicial map, and indeed
a simplicial approximation to the identity map
|K
0
| → |K|
. We will revisit this
idea later when we discuss homotopy invariance.
The key theorem is that as long as we are willing to perform barycentric
subdivisions, then we can always find a simplicial approximation.
Theorem (Simplicial approximation theorem). Let
K
and
L
be simplicial
complexes, and
f
:
|K| → |L|
a continuous map. Then there exists an
r
and
a simplicial map
g
:
K
(r)
→ L
such that
g
is a simplicial approximation of
f
.
Furthermore, if
f
is already simplicial on
M ⊆ K
, then we can choose
g
such
that |g||
M
= f|
M
.
The first thing we have to figure out is how far we are going to subdivide.
To do this, we want to quantify how “fine” our subdivisions are.
Definition (Mesh). Let K be a simplicial complex. The mesh of K is
µ(K) = max{kv
0
− v
1
k : hv
0
, v
1
i ∈ K}.
We have the following lemma that tells us how large our mesh is:
Lemma. Let dim K = n, then
µ(K
(r)
) ≤
n
n + 1
r
µ(K).
The key point is that as
r → ∞
, the mesh goes to zero. So indeed we can
make our barycentric subdivisions finer and finer. The proof is purely technical
and omitted.
Now we can prove the simplicial approximation theorem.
Proof of simplicial approximation theorem.
Suppose we are given the map
f
:
|K| → |L|
. We have a natural cover of
|L|
, namely the open stars of all vertices.
We can use f to pull these back to |K| to obtain a cover of |K|:
{f
−1
(St
L
(w)) : w ∈ V
L
}.
The idea is to barycentrically subdivide our
K
such that each open star of
K
is
contained in one of these things.
By the Lebesgue number lemma, there exists some
δ
, the Lebesgue number
of the cover, such that for each x ∈ |K|, B
δ
(x) is contained in some element of
the cover. By the previous lemma, there is an r such that µ(K
(r)
) < δ.
Now since the mesh
µ
(
K
(r)
) is the smallest distance between any two vertices,
the radius of every open star
St
K
(r)
(
x
) is at most
µ
(
K
(r)
). Hence it follows that
St
K
(r)
(
x
)
⊆ B
δ
(
x
) for all vertices
x ∈ V
K
(r)
. Therefore, for all
x ∈ V
K
(r)
, there
is some w ∈ V
L
such that
St
K
(r)
(x) ⊆ B
δ
(x) ⊆ f
−1
(St
L
(w)).
Therefore defining g(x) = w, we get
f(St
K
(r)
(x)) ⊆ St
L
(g(x)).
So g is a simplicial approximation of f.
The last part follows from the observation that if
f
is a simplicial map, then
it maps vertices to vertices. So we can pick g(v) = f(v).