5Seifert-van Kampen theorem
II Algebraic Topology
5.1 Seifert-van Kampen theorem
The Seifert-van Kampen theorem is the theorem that tells us what happens
when we glue spaces together.
Here we let X = A ∪ B, where A, B, A ∩ B are path-connected.
x
0
A B
We pick a basepoint
x
0
∈ A ∩ B
for convenience. Since we like diagrams, we can
write this as a commutative diagram:
A ∩ B B
A X
where all arrows are inclusion (i.e. injective) maps. We can consider what
happens when we take the fundamental groups of each space. Then we have the
induced homomorphisms
π
1
(A ∩ B, x
0
) π
1
(B, x
0
)
π
1
(A, x
0
) π
1
(X, x
0
)
We might guess that π
1
(X, x
0
) is just the free product with amalgamation
π
1
(X, x
0
) = π
1
(A, x
0
) ∗
π
1
(A∩B,x
0
)
π
1
(B, x
0
).
The Seifert-van Kampen theorem says that, under mild hypotheses, this guess is
correct.
Theorem (Seifert-van Kampen theorem). Let
A, B
be open subspaces of
X
such
that
X
=
A ∪ B
, and
A, B, A ∩ B
are path-connected. Then for any
x
0
∈ A ∩ B
,
we have
π
1
(X, x
0
) = π
1
(A, x
0
) ∗
π
1
(A∩B,x
0
)
π
1
(B, x
0
).
Note that by the universal property of the free product with amalgamation,
we by definition know that there is a unique map
π
1
(
A, x
0
)
∗
π
1
(A∩B,x
0
)
π
1
(
B, x
0
)
→
π
1
(X, x
0
). The theorem asserts that this map is an isomorphism.
Proof is omitted because time is short.
Example. Consider a higher-dimensional sphere
S
n
=
{
v
∈ R
n+1
:
|
v
|
= 1
}
for
n ≥ 2. We want to find π
1
(S
n
).
The idea is to write
S
n
as a union of two open sets. We let
n
= e
1
∈ S
n
⊆
R
n+1
be the North pole, and
s
=
−
e
1
be the South pole. We let
A
=
S
n
\ {n}
,
and
B
=
S
n
\ {s}
. By stereographic projection, we know that
A, B
∼
=
R
n
. The
hard part is to understand the intersection.
To do so, we can draw a cylinder
S
n−1
×
(
−
1
,
1), and project our
A ∩ B
onto the cylinder. We can similarly project the cylinder onto
A ∩ B
. So
A ∩ B
∼
=
S
n−1
× (−1, 1) ' S
n−1
, since (−1, 1) is contractible.
n
s
We can now apply the Seifert-van Kampen theorem. Note that this works only
if S
n−1
is path-connected, i.e. n ≥ 2. Then this tells us that
π
1
(S
n
)
∼
=
π
1
(R
n
) ∗
π
1
(S
n−1
)
π
1
(R
n
)
∼
=
1 ∗
π
1
(S
n−1
)
1
It is easy to see this is the trivial group. We can see this directly from the
universal property of the amalgamated free product, or note that it is the quotient
of 1 ∗ 1, which is 1.
So for n ≥ 2, π
1
(S
n
)
∼
=
1.
We have found yet another simply connected space. However, this is unlike
our previous examples. Our previous spaces were simply connected because they
were contractible. However, we will later see that
S
n
is not contractible. So this
is genuinely a new, interesting example.
Why did we go though all this work to prove that
π
1
(
S
n
)
∼
=
1? It feels like
we can just prove this directly — pick a point that is not in the curve as the
North pole, project stereographically to
R
n
, and contract it there. However,
the problem is that space-filling curves exist. We cannot guarantee that we can
pick a point not on the curve! It is indeed possible to prove directly that given
any curve on
S
n
(with
n ≥
1), we can deform it slightly so that it is no longer
surjective. Then the above proof strategy works. However, using Seifert-van
Kampen is much neater.
Example (
RP
n
). Recall that
RP
n
∼
=
S
n
/{± id}
, and the quotient map
S
n
→
RP
n
is a covering map. Now that we have proved that
S
n
is simply connected,
we know that S
n
is a universal cover of RP
n
.
For any x
0
∈ RP
n
, we have a bijection
π
1
(RP
n
, x
0
) ↔ p
−1
(x
0
).
Since
p
−1
(
x
0
) has two elements by definition, we know that
|π
1
(
RP
n
, x
0
)
|
= 2.
So π
1
(RP
n
, x
0
) = Z/2.
You will prove a generalization of this in example sheet 2.
Example (Wedge of two circles). We are going to consider the operation of
wedging. Suppose we have two topological spaces, and we want to join them
together. The natural way to join them is to take the disjoint union. What if
we have based spaces? If we have (
X, x
0
) and (
Y, y
0
), we cannot just take the
disjoint union, since we will lose our base point. What we do is take the wedge
sum, where we take the disjoint union and then identify the base points:
X Y
x
0
y
0
X ∧ Y
x
0
∼ y
0
Suppose we take the wedge sum of two circles
S
1
∧ S
1
. We would like to pick
A, B
to be each of the circles, but we cannot, since
A
and
B
have to be open.
So we take slightly more, and get the following:
x
0
A B
Each of A and B now look like this:
We see that both
A
and
B
retract to the circle. So
π
1
(
A
)
∼
=
π
1
(
B
)
∼
=
Z
, while
A ∩ B is a cross, which retracts to a point. So π
1
(A ∩ B) = 1.
Hence by the Seifert-van Kampen theorem, we get
π
1
(S
1
∧ S
1
, x
0
) = π
1
(A, x
0
) ∗
π
1
(A∩B,x
0
)
π
1
(B, x
0
)
∼
=
Z ∗
1
Z
∼
=
Z ∗ Z
∼
=
F
2
,
where
F
2
is just
F
(
S
) for
|S|
= 2. We can see that
Z ∗ Z
∼
=
F
2
by showing that
they satisfy the same universal property.
Note that we had already figured this out when we studied the free group,
where we realized F
2
is the fundamental group of this thing.
More generally, as long as
x
0
, y
0
in
X
and
Y
are “reasonable”,
π
1
(
X ∧ Y
)
∼
=
π
1
(X) ∗ π
1
(Y ).
Next, we would exhibit some nice examples of the covering spaces of
S
1
∧ S
1
,
i.e. the “rose with 2 petals”.
Recall that π
1
(S
1
∧ S
1
, x
0
)
∼
=
F
2
∼
=
ha, bi.
Example. Consider the map
φ
:
F
2
→ Z/
3 which sends
a 7→
1
, b 7→
1. Note
that 1 is not the identity, since this is an abelian group, and the identity is 0.
This exists since the universal property tells us we just have to say where the
generators go, and the map exists (and is unique).
Now
ker φ
is a subgroup of
F
2
. So there is a based covering space of
S
1
∧ S
1
corresponding to it, say,
˜
X. Let’s work out what it is.
First, we want to know how many sheets it has, i.e. how many copies of
x
0
we have. There are three, since we know that the number of sheets is the index
of the subgroup, and the index of
ker φ
is
|Z/
3
|
= 3 by the first isomorphism
theorem.
˜x
0
Let’s try to lift the loop
a
at
˜x
0
. Since
a 6∈ ker φ
=
π
1
(
˜
X, ˜x
0
),
a
does not lift to
a loop. So it goes to another vertex.
˜x
0
a
Similarly, a
2
6∈ ker φ = π
1
(
˜
X, ˜x
0
). So we get the following
˜x
0
a
a
Since a
3
∈ ker φ, we get a loop labelled a.
˜x
0
a
a
a
Note that
ab
−1
∈ ker φ
. So
ab
−1
gives a loop. So
b
goes in the same direction as
a:
˜x
0
a
a
a
b
b
b
This is our covering space.
This is a fun game to play at home:
(i) Pick a group G (finite groups are recommended).
(ii) Then pick α, β ∈ G and let φ : F
2
→ G send a 7→ α, b 7→ β.
(iii) Compute the covering space corresponding to φ.