5Seifert-van Kampen theorem

II Algebraic Topology



5.2 The effect on π
1
of attaching cells
We have started the course by talking about cell complexes, but largely ignored
them afterwards. Finally, we are getting back to them.
The process of attaching cells is as follows: we start with a space
X
, get a
function
S
n1
X
, and attach
D
n
to
X
by gluing along the image of
f
to get
X
f
D
n
:
0
y
0
x
0
Since we are attaching stuff, we can use the Seifert-van Kampen theorem to
analyse this.
Theorem. If
n
3, then
π
1
(
X
f
D
n
)
=
π
1
(
X
). More precisely, the map
π
1
(
X, x
0
)
π
1
(
X
f
D
n
, x
0
) induced by inclusion is an isomorphism, where
x
0
is a point on the image of f .
This tells us that attaching a high-dimensional disk does not do anything to
the fundamental group.
Proof.
Again, the difficulty of applying Seifert-van Kampen theorem is that we
need to work with open sets.
Let 0
D
n
be any point in the interior of
D
n
. We let
A
=
X
f
(
D
n
\ {
0
}
).
Note that
D
n
\{
0
}
deformation retracts to the boundary
S
n1
. So
A
deformation
retracts to X. Let B =
˚
D, the interior of D
n
. Then
A B =
˚
D
n
\ 0
=
S
n1
× (1, 1)
We cannot use
y
0
as our basepoint, since this point is not in
A B
. Instead,
pick an arbitrary
y
1
A B
. Since
D
n
is path connected, we have a path
γ
:
y
1
y
0
, and we can use this to recover the fundamental groups based at
y
0
.
Now Seifert-van Kampen theorem says
π
1
(X
f
D
n
, y
1
)
=
π
1
(A, y
1
)
π
1
(AB,y
1
)
π
1
(B, y
1
).
Since
B
is just a disk, and
A B
is simply connected (
n
3 implies
S
n1
is
simply connected), their fundamental groups are trivial. So we get
π
1
(X
f
D
n
, y
1
)
=
π
1
(A, y
1
).
We can now use γ to change base points from y
1
to y
0
. So
π
1
(X
f
D
n
, y
0
)
=
π
1
(A, y
0
)
=
π
1
(X, y
0
).
The more interesting case is when we have smaller dimensions.
Theorem. If
n
= 2, then the natural map
π
1
(
X, x
0
)
π
1
(
X
f
D
n
, x
0
) is
surjective, and the kernel is
hh
[
f
]
ii
. Note that this statement makes sense, since
S
n1
is a circle, and f : S
n1
X is a loop in X.
This is what we would expect, since if we attach a disk onto the loop given
by f, this loop just dies.
Proof. As before, we get
π
1
(X
f
D
n
, y
1
)
=
π
1
(A, y
1
)
π
1
(AB,y
1
)
π
1
(B, y
1
).
Again,
B
is contractible, and
π
1
(
B, y
1
)
=
1. However,
π
1
(
A B, y
1
)
=
Z
. Since
π
1
(A B, y
1
) is just (homotopic to) the loop induced by f, it follows that
π
1
(A, y
1
)
π
1
(AB,y
1
)
1 = (π
1
(A, y
1
) 1)/hhπ
1
(A B, y
1
)ii
=
π
1
(X, x
0
)/hhfii.
In summary, we have
π
1
(X
f
D
n
) =
(
π
1
(X) n 3
π
1
(X)/hhfii n = 2
This is a useful result, since this is how we build up cell complexes. If we want to
compute the fundamental groups, we can just work up to the two-cells, and know
that the higher-dimensional cells do not have any effect. Moreover, whenever
X
is a cell complex, we should be able to write down the presentation of π
1
(X).
Example. Let
X
be the 2-torus. Possibly, our favorite picture of the torus is
(not a doughnut):
This is already a description of the torus as a cell complex!
We start with our zero complex X
(0)
:
We then add our 1-cells to get X
(1)
:
a
b
We now glue our square to the cell complex to get X = X
(2)
:
matching up the colors and directions of arrows.
So we have our torus as a cell complex. What is its fundamental group?
There are many ways we can do this computation, but this time we want to do
it as a cell complex.
We start with
X
(0)
. This is a single point. So its fundamental group is
π
1
(X
(0)
) = 1.
When we add our two 1-cells, we get π
1
(X
(1)
) = F
2
=
ha, bi.
Finally, to get
π
1
(
X
), we have to quotient out by the boundary of our square,
which is just aba
1
b
1
. So we have
π
1
(X
(2)
) = F
2
/hhaba
1
b
1
ii = ha, b | aba
1
b
1
i
=
Z
2
.
We have the last congruence since we have two generators, and then we make
them commute by quotienting the commutator out.
This procedure can be reversed given a presentation of a group, we can just
add the right edges and squares to produce a cell complex with that presentation.
Corollary. For any (finite) group presentation
hS | Ri
, there exists a (finite)
cell complex (of dimension 2) X such that π
1
(X)
=
hS | Ri.
There really isn’t anything that requires that finiteness in this proof, but
finiteness makes us feel more comfortable.
Proof.
Let
S
=
{a
1
, · · · , a
m
}
and
R
=
{r
1
, · · · , r
n
}
. We start with a single point,
and get our
X
(1)
by adding a loop about the point for each
a
i
S
. We then get
our 2-cells
e
2
j
for
j
= 1
, · · · , n
, and attaching them to
X
(1)
by
f
i
:
S
1
X
(1)
given by a based loop representing r
i
F (S).
Since all groups have presentations, this tells us that all groups are funda-
mental groups of some spaces (at least those with finite presentations).