Graphics Reference
In-Depth Information
q
Ê
Á
ˆ
˜
(
)
Â
t
()
(
[
]
)
=
()
()
[
ˆ
]
f
o
∂
vv
◊◊◊
v
f
1
v
◊◊◊
v
◊◊◊
v
#
q
01
q
#
0
t
q
q
-
1
q
-
1
t
=
0
q
Â
t
[
]
=
=
()()
ˆ
1
f
vvv
◊◊◊
◊◊◊
0
.
#
0
t
q
q
-
1
t
=
0
The lemma is proved.
Now the maps f
#q
are no more interesting by themselves than were the chain groups
C
q
(K). What will be important are the maps that they induce on the homology groups,
and Lemma 7.2.2.1 is essential for that. We generalize the construction somewhat.
Definition.
A
chain map
()
Æ
()
j :
CK
CL
#
#
is a “vector” j=(...,j
-1
,j
0
,j
1
,...) of homomorphisms j
q
:C
q
(K) Æ C
q
(L) satisfying
∂
q
°
j
q
=j
q-1
°
∂
q
.
Having a chain map (. . . ,j
-1
,j
0
,j
1
, . . .) is equivalent to having a commutative
diagram
∂
∂
∂
q
+
2
()
æÆ
q
+
1
()
ææ
q
()
Æ ◊◊◊
◊◊◊ æ
ææ
Æ
CK
ææ
CK
CK
q
+
1
q
q
-
1
Ø
j
Ø
j
Ø
j
q
+
1
q
q
-
1
()
æÆ
()
ææ
()
Æ ◊◊◊
◊◊◊ æ
ææ
Æ
CL
ææ
CL
CL
q
+
1
q
q
-
1
∂
∂
∂
q
+
2
q
+
1
q
Note that f
#
= (..., f
#-1
,f
#0
,f
#1
, . . .) is a chain map by Lemma 7.2.2.1 called the
chain
map induced by the simplicial map f
.
7.2.2.2. Lemma.
Let j :C
#
(K) Æ C
#
(L) be an arbitrary chain map. Then
(1) j(Z
q
(K)) Õ Z
q
(L), for all q.
(2) j(B
q
(K)) Õ B
q
(L), for all q.
Proof.
This follows easily from the definition of a chain map. If z ΠZ
q
(K), then
(
()
)
=
(
()
)
=
()
=
∂j
z
j
∂
z
j
1
0 ,
qq
q
-
1
q
q
-
which proves (1). To prove (2), note that
(
()
)
=
(
()
)
j∂
c
∂
j
c
.
qp
+
1
p
+
1
q
+
1
Definition.
Let j :C
#
(K) Æ C
#
(L) be an arbitrary chain map. Define maps
()
Æ
()
j
*
:
HK
HL
q
q
q
