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Thus,
f
⊗
g
◦
g
=
T
f
◦
f
∩
T
g
◦
g
=
T
f
∩
f
∩
T
g
∩
g
f
◦
=
f
∩
f
∩
g
∩
g
=
f
∩
g
∩
(f
∩
g)
T
f
∩
g
∩
T(f
=
f
⊗
g
∩
f
=
∩
g)
⊗
g
f
⊗
g
◦
=
(f
⊗
g),
so that from Definition
23
,
f
◦
f
⊗
g
◦
g
=
f
⊗
g
◦
(f
⊗
g).
(8.1)
Let us show that this property of composition is valid for complex morphisms
f
:
C
,
f
:
D
and
g
:
A
→
C
→
E
,
g
:
B
→
B
→
F
as well. Then,
(f
⊗
g
)
◦
(f
⊗
g)
f
i
⊗
f
⊗
g
g
j
◦
f
⊗
g
f
i
⊗
g
j
∈
=
(f
i
⊗
g
j
)
|
,f
i
⊗
g
j
∈
,
g
j
dom
f
i
⊗
cod(f
i
⊗
g
j
)
=
f
i
⊗
f
⊗
g
g
j
◦
f
⊗
g
f
i
⊗
g
j
∈
=
(f
i
⊗
g
j
)
|
,f
i
⊗
g
j
∈
,
dom
g
j
dom
f
i
, cod(g
j
)
=
=
cod(f
i
)
f
i
⊗
f
◦
g
◦
g
j
◦
f
g
j
f
i
◦
,g
j
◦
=
(f
i
⊗
g
j
)
|
f
i
∈
g
j
∈
from the fact that all
f
i
,f
i
,g
j
,g
j
are simple arrows and (
8.1
)
f
i
◦
f
◦
g
◦
f
i
⊗
g
j
◦
g
j
|
f
g
j
f
i
◦
,g
j
◦
=
f
i
∈
g
j
∈
id
R
|
f
◦
g
◦
g
j
|
f
g
j
∈
f
i
◦
f
i
∩
g
j
◦
f
i
◦
,g
j
◦
=
R
f
i
∈
g
j
∈
(by Definition
22
)
(f
◦
(g
◦
f)
⊗
g)
=
.
Consequently,
(f
◦
(g
◦
(f
⊗
g
)
f)
⊗
g)
=
◦
(f
⊗
g)
, and hence the compositional
property for functor
⊗
is satisfied.