(a) Let
G
be the subset of (
A
×
B
)
×
(
C
×
D
) given by
G
=
{
(
s,t
)

s
= (
x,y
) and
t
= (
α
(
x
)
,β
(
y
)) for some
x
∈
A,y
∈
B
}
,
and
γ
= (
A
×
B,C
×
D,G
).
We verify that
γ
is a bijective function:
•
Pick any
s
∈
A
×
B
. By definition
s
= (
x,y
) for some
x
∈
A,y
∈
B
. We have
α
(
x
)
∈
C
,
β
(
y
)
∈
D
,
(
α
(
x
)
,β
(
y
))
∈
C
×
D
. Write
t
= (
α
(
x
)
,β
(
y
)). We have (
s,t
) = ((
x,y
)
,
(
α
(
x
)
,β
(
y
)))
∈
G
.
•
Pick any
s
∈
A
×
B
and
t,t
′
∈
C
×
D
. Suppose (
s,t
)
,
(
s,t
′
)
∈
G
.
Since (
s,t
)
∈
G
, we have
s
= (
x,y
)
,t
= (
α
(
x
)
,β
(
y
)) for some
x
∈
A,y
∈
B
. Since (
s
′
,t
′
)
∈
G
, we
have
s
= (
x
′
,y
′
)
,t
′
= (
α
(
x
′
)
,β
(
y
′
)) for some
x
′
∈
A,y
′
∈
B
.
We have (
x,y
) =
s
= (
x
′
,y
′
) (with
x,x
′
∈
A
and
y,y
′
∈
B
). Then
x
=
x
′
and
y
=
y
′
. Therefore
α
(
x
) =
α
(
x
′
),
β
(
y
) =
β
(
y
′
), and
t
= (
α
(
x
)
,β
(
y
)) = (
α
(
x
′
)
,β
(
y
′
)) =
t
′
.
•
Pick any
t
∈
C
×
D
.
Then
t
= (
u,v
) for some
u
∈
C,v
∈
D
.
Since
α
is a surjective function,
u
=
α
(
x
) for some
x
∈
A
. Since
β
is a surjective function,
v
=
β
(
y
) for some
y
∈
B
. We have
(
x,y
)
∈
A
×
B
.
Write
s
= (
x,y
). We have (
s,t
) = ((
x,y
)
,
(
α
(
x
)
,β
(
y
)))
∈
G
.
•
Pick any
t
∈
C
×
D
and any
s,s
′
∈
A
×
B
. Suppose (
s,t
)
,
(
s
′
,t
)
∈
G
.
Since (
s,t
)
∈
G
, we have
s
= (
x,y
)
,t
= (
α
(
x
)
,β
(
y
)) for some
x
∈
A,y
∈
B
. Since (
s
′
,t
)
∈
G
, we
have
s
′
= (
x
′
,y
′
)
,t
= (
α
(
x
′
)
,β
(
y
′
)) for some
x
′
∈
A,y
′
∈
B
.
We have (
α
(
x
)
,β
(
y
)) =
t
= (
α
(
x
′
)
,β
(
y
′
)) (with
α
(
x
)
,α
(
x
′
)
∈
C
,
β
(
y
)
,β
(
y
′
)
∈
D
). Then
α
(
x
) =
α
(
x
′
) and
β
(
y
) =
β
(
y
′
). Since
α
is an injective function, we have
x
=
x
′
. Since
β
is an injective
function, we have
y
=
y
′
. Therefore
s
= (
x,y
) = (
x
′
,y
′
) =
s
′
.
We have verified that
γ
= (
A
×
B,C
×
D,G
) is a bijective function. (The function
γ
is given ‘explicitly’
by
γ
(
x,y
) = (
α
(
x
)
,β
(
y
)) for any
x
∈
A,y
∈
B
.)
(b) We make some observations first:
•
Since
α
is a bijective function from
A
to
C
,
α
−
1
is a bijective function from
C
to
A
.
•
Since
β
is a bijective function from
B
to
D
,
β
−
1
is a bijective function from
D
to
B
.
•
If
ϕ
is a function from
A
to
B
then
β
◦
ϕ
◦
α
−
1
is a function from
C
to
D
.
•
If
ψ
is a function from
C
to
D
then
β
−
1
◦
ψ
◦
α
is a function from
A
to
B
.
Let
H
be the subset of
Map
(
A,B
)
×
Map
(
C,D
) given by
H
=
{
(
ϕ,β
◦
ϕ
◦
α
−
1
)

ϕ
∈
Map
(
A,B
)
}
,
and Γ = (
Map
(
A,B
)
,
Map
(
C,D
)
,H
). By definition, Γ is a function. (The function Γ is given ‘explicitly’
by Γ(
ϕ
) =
β
◦
ϕ
◦
α
−
1
for any
ϕ
∈
Map
(
A,B
).)
We verify that Γ is a bijective function:
•
Pick any
ψ
∈
Map
(
C,D
). We have
β
−
1
◦
ψ
◦
α
∈
Map
(
A,B
). Moreover,
Γ(
β
−
1
◦
ψ
◦
α
) =
β
◦
(
β
−
1
◦
ψ
◦
α
)
◦
α
−
1
= (
β
◦
β
−
1
)
◦
ψ
◦
(
α
◦
α
−
1
) =
id
D
◦
ψ
◦
id
C
=
ψ.
•
Pick any
ϕ,ϕ
′
∈
Map
(
C,D
). Suppose Γ(
ϕ
) = Γ(
ϕ
′
). Then we have
ϕ
=
β
−
1
◦
β
◦
ϕ
◦
α
−
1
◦
α
=
β
−
1
◦
Γ(
ϕ
)
◦
α
=
β
−
1
◦
Γ(
ϕ
′
)
◦
α
=
β
−
1
◦
β
◦
ϕ
′
◦
α
−
1
◦
α
=
ϕ
′
.