5Seifert-van Kampen theorem

II Algebraic Topology



5.3 A refinement of the Seifert-van Kampen theorem
We are going to make a refinement of the theorem so that we don’t have to
worry about that openness problem. We first start with a definition.
Definition (Neighbourhood deformation retract). A subset
A X
is a neigh-
bourhood deformation retract if there is an open set
A U X
such that
A
is
a strong deformation retract of
U
, i.e. there exists a retraction
r
:
U A
and
r ' id
U
rel A.
This is something that is true most of the time, in sufficiently sane spaces.
Example. If
Y
is a subcomplex of a cell complex, then
Y
is a neighbourhood
deformation retract.
Theorem. Let
X
be a space,
A, B X
closed subspaces. Suppose that
A
,
B
and
A B
are path connected, and
A B
is a neighbourhood deformation
retract of A and B. Then for any x
0
A B.
π
1
(X, x
0
) = π
1
(A, x
0
)
π
1
(AB,x
0
)
π
1
(B, x
0
).
x
0
A B
This is just like Seifert-van Kampen theorem, but usually easier to apply, since
we no longer have to “fatten up” our A and B to make them open.
Proof.
Pick open neighbourhoods
A B U A
and
A B V B
that strongly deformation retract to
A B
. Let
U
be such that
U
retracts
to
A B
. Since
U
retracts to
A
, it follows that
U
is path connected since
path-connectedness is preserved by homotopies.
Let
A
0
=
AV
and
B
0
=
BU
. Since
A
0
= (
X \B
)
V
, and
B
0
= (
X \A
)
U
,
it follows that A
0
and B
0
are open.
Since
U
and
V
retract to
A B
, we know
A
0
' A
and
B
0
' B
. Also,
A
0
B
0
= (
A V
)
(
B U
) =
U V ' A B
. In particular, it is path connected.
So by Seifert van-Kampen, we get
π
1
(A B) = π
1
(A
0
, x
0
)
π
1
(A
0
B
0
,x
0
)
π
1
(B
0
, x
0
) = π
1
(A, x
0
)
π
1
(AB,x
0
)
π
1
(B, x
0
).
This is basically what we’ve done all the time when we enlarge our
A
and
B
to become open.
Example. Let
X
=
S
1
S
1
, the rose with two petals. Let
A, B
=
S
1
be the
circles.
x
0
A B
Then since
{x
0
}
=
A B
is a neighbourhood deformation retract of
A
and
B
,
we know that
π
1
X
=
π
1
S
1
π
1
S
1
.