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)
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