4Reduced homology

III Algebraic Topology



4 Reduced homology
Definition
(Reduced homology)
.
Let
X
be a space, and
x
0
X
a basepoint.
We define the reduced homology to be
˜
H
(X) = H
(X, {x
0
}).
Note that by the long exact sequence of relative homology, we know that
˜
H
n
(
X
)
=
H
n
(
X
) for
n
1. So what is the point of defining a new homology
theory that only differs when n = 0, which we often don’t care about?
It turns out there is an isomorphism between
H
(
X, A
) and
˜
H
(
X/A
) for
suitably “good” pairs (X, A).
Definition
(Good pair)
.
We say a pair (
X, A
) is good if there is an open set
U
containing
¯
A
such that the inclusion
A U
is a deformation retract, i.e. there
exists a homotopy H : [0, 1] × U U such that
H(0, x) = x
H(1, x) A
H(t, a) = a for all a A, t [0, 1].
Theorem. If (X, A) is good, then the natural map
H
(X, A) H
(X/A, A/A) =
˜
H
(X/A)
is an isomorphism.
Proof. As i : A U is in particular a homotopy equivalence, the map
H
(A) H
(U)
is an isomorphism. So by the five lemma, the map on relative homology
H
(X, A) H
(X, U)
is an isomorphism as well.
As i : A U is a deformation retraction with homotopy H, the inclusion
{∗} = A/A U/A
is also a deformation retraction. So again by the five lemma, the map
H
(X/A, A/A) H
(X/A, U/A)
is also an isomorphism. Now we have
H
n
(X, A) H
n
(X, U) H
n
(X \ A, U \ A)
H
n
(X/A, A/A) H
n
(X/A, U/A) H
n
X
A
\
A
A
,
U
A
\
A
A
excise A
excise A/A
We now notice that
X \ A
=
X
A
\
A
A
and
U \ A
=
U
A
\
A
A
. So the right-hand
vertical map is actually an isomorphism. So the result follows.