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Mapping/Composition/Associativity/Fact
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Let
L
,
M
,
N
{\displaystyle {}L,M,N}
and
P
{\displaystyle {}P}
be sets, and let
F
:
L
⟶
M
,
x
⟼
F
(
x
)
,
{\displaystyle F\colon L\longrightarrow M,x\longmapsto F(x),}
G
:
M
⟶
N
,
y
⟼
G
(
y
)
,
{\displaystyle G\colon M\longrightarrow N,y\longmapsto G(y),}
and
H
:
N
⟶
P
,
z
⟼
H
(
z
)
,
{\displaystyle H\colon N\longrightarrow P,z\longmapsto H(z),}
be
mappings
.
Then
H
∘
(
G
∘
F
)
=
(
H
∘
G
)
∘
F
{\displaystyle {}H\circ (G\circ F)=(H\circ G)\circ F\,}
holds.
Proof
,
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