Graphics Reference
In-Depth Information
by
[
()
=
[
()
]
()
j
z
j
z
,
z
Œ
Z
K
.
*
q
q
q
7.2.2.3. Lemma.
j
*q
is a well-defined homomorphism.
Proof.
First of all, by Lemma 7.2.2.2(1), the definition makes sense, since j
q
(z) Œ
Z
q
(L). To show that j
*q
is well defined, let a Œ H
q
(K) and assume that a = [z] = [z¢], z,
z¢ŒZ
q
(K). Then z - z¢ belongs to B
q
(K). Therefore,
()
-
()
=
(
)
Œ
()
j
z
j
z
¢
j
z
-
z
¢
BL
q
q
q
q
by Lemma 7.2.2.2(2), that is, [j
q
(z)] = [j
q
(z¢)]. This proves that j
*q
is well defined.
Next, let [z
i
] = z
i
+ B
q
(K) be elements of H
q
(K). Then
[]
+
[]
(
)
=
(
(
)
+
()
)
j
zz
j
j
j
zzBK
zz
+
*
q
1
2
*
q
1
2
q
(
)
+
()
=
+
B
L
q
12
q
(
()
+
()
)
+
()
=
z
j
z
B
L
q
1
q
2
q
(
()
+
()
)
+
(
()
+
()
)
=
j
zBL
j
z BL
q
1
q
q
2
q
[]
[[]
(
)
+
(
)
.
=
j
z
j
z
*
q
1
*
q
2
Thus, j
*q
is a homomorphism and Lemma 7.2.2.3 is proved.
Definition.
The maps j
*q
are called the
homomorphisms on homology induced by the
chain map
j. In particular, if f : K Æ L is a simplicial map, we shall let
()
Æ
()
f
:
HK HL
*
q
q
q
denote the map on the homology group induced by the chain map f
#
.
Consider the simplicial complex K =∂ ·
v
0
v
1
v
2
Ò. The next two examples compute
f
*q
for two simplicial maps f : K Æ K.
7.2.2.4. Example.
To compute f
*q
when f is the constant map defined by f(
v
i
) =
v
0
.
Solution.
The given f induces the constant map ΩfΩ:ΩKΩÆΩKΩ, ΩfΩ(x) =
v
0
. We know
from Example 7.2.1.5 that
()
ª
()
ª
Z
HK HK
0
1
and
q
()
=
H
K
0
for
q
>
1,
so that we only have to worry about what happens in dimensions 0 and 1. The map
f
#1
:C
1
(K) Æ C
1
(K) is obviously the zero map by definition, and so f
*1
:H
1
(K) Æ H
1
(K)