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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.
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