[<< wikibooks] Differentiable Manifolds/Orientation
Proof: The given family of functions defines an atlas on

∂
M

{\displaystyle \partial M}
, so that the proposition will be proven once it is demonstrated that the requirement regarding the positivity of the determinant is satisfied.
Indeed, let

α
,
β
∈
A

{\displaystyle \alpha ,\beta \in A}
. Then

π

2
,
…
,
n

∘

φ

α

↾
∂
M
∘
(

π

2
,
…
,
n

∘

φ

β

↾
∂
M

)

−
1

=

π

2
,
…
,
n

∘
(

φ

α

∘

φ

β

−
1

)
∘
(

π

2
,
…
,
n

↾
{
(

x

1

,
…
,

x

n

)
∈

R

n

|

x

1

=
0
}

)

−
1

{\displaystyle \pi _{2,\ldots ,n}\circ \varphi _{\alpha }\upharpoonright \partial M\circ (\pi _{2,\ldots ,n}\circ \varphi _{\beta }\upharpoonright \partial M)^{-1}=\pi _{2,\ldots ,n}\circ (\varphi _{\alpha }\circ \varphi _{\beta }^{-1})\circ (\pi _{2,\ldots ,n}\upharpoonright \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\})^{-1}}
.Since

φ

α

∘

φ

β

−
1

{\displaystyle \varphi _{\alpha }\circ \varphi _{\beta }^{-1}}
maps the set

{
(

x

1

,
…
,

x

n

)
∈

R

n

|

x

1

=
0
}

{\displaystyle \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\}}
to itself, the first row of

D
(

φ

α

∘

φ

β

−
1

)

{\displaystyle D(\varphi _{\alpha }\circ \varphi _{\beta }^{-1})}
is zero so long as

x
∈
{
(

x

1

,
…
,

x

n

)
∈

R

n

|

x

1

=
0
}

{\displaystyle x\in \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\}}
, except the very first entry. Yet the very last entry must be non-negative so long as

x
∈
{
(

x

1

,
…
,

x

n

)
∈

R

n

|

x

1

=
0
}

{\displaystyle x\in \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\}}
, since otherwise the fundamental theorem of calculus and the continuity of the derivative would imply that for a

y
:=
(

y

1

,
…
,

y

n

)
∈

R

n

{\displaystyle y:=(y_{1},\ldots ,y_{n})\in \mathbb {R} ^{n}}
such that

y

n

>
0

{\displaystyle y_{n}>0}
was sufficiently small,

φ

α

∘

φ

β

−
1

(
y
)

{\displaystyle \varphi _{\alpha }\circ \varphi _{\beta }^{-1}(y)}
would be contained within

{
(

x

1

,
…
,

x

n

)
∈

R

n

|

x

n

<
0
}

{\displaystyle \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{n}<0\}}
,contrary to the definition of charts that contain a piece of the boundary. Hence, upon carrying out a Leibniz expansion of

D
(

φ

α

∘

φ

β

−
1

)

{\displaystyle D(\varphi _{\alpha }\circ \varphi _{\beta }^{-1})}
along the first row, we obtain that the matrix obtained from

D
(

φ

α

∘

φ

β

−
1

)

{\displaystyle D(\varphi _{\alpha }\circ \varphi _{\beta }^{-1})}
by removing the first row and the first column has positive determinant. Yet the definitions of the respective partial derivatives as limits show that this matrix is exactly the matrix

D
(

π

2
,
…
,
n

∘
(

φ

α

∘

φ

β

−
1

)
∘
(

π

2
,
…
,
n

↾
{
(

x

1

,
…
,

x

n

)
∈

R

n

|

x

1

=
0
}

)

−
1

)

{\displaystyle D(\pi _{2,\ldots ,n}\circ (\varphi _{\alpha }\circ \varphi _{\beta }^{-1})\circ (\pi _{2,\ldots ,n}\upharpoonright \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\})^{-1})}
.

◻

{\displaystyle \Box }