Since
τ
0
must be closed, we have
d
(
ξ
∧
τ
0
) = (
dξ
)
∧
τ
0
+ (

1)
k

1
·
ξ
∧
(
dτ
0
) = (
dξ
)
∧
τ
0
by
part (c) of the problem. Similarly, since
ω
is closed, we have
d
(
ω
∧
η
) = (

1)
k
·
ω
∧
(
dη
).
Therefore
ω
0
∧
τ
0
=
ω
∧
τ
+
d
(
ξ
∧
τ
0
+ (

1)
k
·
ω
∧
η
)
. It remains to observe that
ξ
∧
τ
0
and
ω
∧
η
both have compact support because
τ
0
and
η
have compact support by assumption.
So we see that [
ω
0
∧
τ
0
]
c
= [
ω
∧
τ
]
c
.
This implies that the formula [
ω
]
∧
[
τ
]
c
:= [
ω
∧
τ
]
c
yields a well defined map (6.8).
The last step is checking bilinearity.
However, this is trivial, because of the way that
the quotient vector space operations are defined, and because of the fact that the wedge
product at the level of differential forms is already bilinear.
6.3.
Since
∂M
=
∅
, we have
R
∂M
τ
= 0. So Stokes’ theorem implies that
Z
M
ω
=
Z
M
dτ
=
Z
∂M
τ
= 0
.
Now since
M
is
n
dimensional, every differential (
n
+ 1)form on
M
must vanish, and
hence every
n
form on
M
is closed. Therefore
H
n
c
(
M
) equals the quotient of Ω
n
c
(
M
) by
the subspace
W
⊂
Ω
n
c
(
M
) consisting of all forms that can be written as
dτ
for some
τ
∈
Ω
n

1
c
(
M
). We just saw that if
ω
∈
W
, then
R
M
ω
= 0. Finally, integration is a linear
map, so by the standard property of quotient vector spaces, we see that there is a (unique
and well defined) linear map
R
:
H
n
c
(
M
)
→
R
such that the composition
Ω
n
c
(
M
)
→
H
n
c
(
M
)
R
→
R
equals the map
ω
7→
R
M
ω
(the first map above is the quotient map).
6.4.
(a)
By assumption,
{
U, V
}
is an open cover of
M
. Let
{
ρ
U
, ρ
V
}
be a partition of
unity corresponding to this open cover (we know that such a partition of unity always
exists). The general definition of a “partition of unity corresponding to a given open cover”
was mentioned in class a couple of times, but in our situation, since the cover is finite, the
formulation simplifies a little bit. The defining properties are that
ρ
U
, ρ
V
:
M
→
[0
,
1]
7
are smooth functions such that
ρ
U
(
p
) +
ρ
V
(
p
) = 1 for all
p
∈
M
, and
1
supp(
ρ
U
)
⊂
U
and
supp(
ρ
V
)
⊂
V
. Write
S
U
= supp(
ρ
U
) and
S
V
= supp(
ρ
V
) for simplicity. Now for every
p
∈
U
, define (
ω
1
)
p
as follows. If
p
∈
U
∩
V
, then set (
ω
1
)
p
=
ρ
V
(
p
)
·
ω
p
, and if
p
6∈
U
∩
V
,
set (
ω
1
)
p
= 0. The key claim is that
ω
1
is smooth everywhere on
U
, i.e., (
ω
1
)
p
depends
smoothly on
p
∈
U
.
To check this, use the standard idea: the restriction
ω
1
U
∩
V
is smooth on
U
∩
V
because
both
ω
and
ρ
V
U
∩
V
are smooth.
On the other hand, the restriction
ω
1
U
\
S
V
is smooth
because it is identically zero (recall that
ρ
V
is identically zero on the complement of
S
V
).
But
U
\
S
V
is open because
S
V
is closed in
M
, and
U
= (
U
∩
V
)
∪
(
U
\
S
V
) because
S
V
⊂
V
by assumption. Since being smooth is a local property, it follows that
ω
1
is smooth on
U
.
Likewise, for every
p
∈
V
, define (
ω
2
)
p
as follows. If
p
∈
U
∩
V
, set (
ω
2
)
p
=

ρ
U
(
p
)
·
ω
p
,
and if
p
6∈
U
∩
V
, set (
ω
2
)
p
= 0. A similar argument shows that
ω
2
is smooth everywhere
on
V
. Thus we obtain
ω
1
∈
Ω
k
(
U
) and
ω
2
∈
Ω
k
(
V
).
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 Winter '11
 Mitya
 Math, Topology, Open set, Closed set, Ωk