Mostre que a esfera
S
2
{\displaystyle S^{2}}
Demonstração
editar
Temos por definição que:
S
2
=
{
(
x
,
y
,
z
)
∈
R
3
/
x
2
+
y
2
+
z
2
=
1
}
{\displaystyle S^{2}=\{(x,y,z)\in \mathbb {R} ^{3}/x^{2}+y^{2}+z^{2}=1\}}
Assim considere
p
=
(
a
,
b
,
c
)
∈
S
2
{\displaystyle p=(a,b,c)\in S^{2}}
r
:
{
(
0
,
0
,
1
)
+
t
∗
(
a
,
b
,
c
−
1
)
/
t
∈
R
}
=
{
(
t
a
,
t
b
,
1
+
t
(
c
−
1
)
)
/
t
∈
R
}
{\displaystyle r:\{(0,0,1)+t*(a,b,c-1)/t\in \mathbb {R} \}=\{(ta,tb,1+t(c-1))/t\in \mathbb {R} \}}
e o plano:
π
:
{
(
x
,
y
,
z
)
∈
R
3
/
z
=
0
}
{\displaystyle \pi :\{(x,y,z)\in \mathbb {R} ^{3}/z=0\}}
Assim:
r
∩
π
:
{
(
x
,
y
,
z
)
∈
R
3
/
(
x
,
y
,
z
)
=
(
t
a
,
t
b
,
0
)
}
{\displaystyle r\cap \pi :\{(x,y,z)\in \mathbb {R} ^{3}/(x,y,z)=(ta,tb,0)\}}
1
+
t
(
c
−
1
)
=
0
t
(
c
−
1
)
=
−
1
t
=
−
1
c
−
1
t
=
1
1
−
c
{\displaystyle {\begin{aligned}1+t(c-1)&=0\\t(c-1)&=-1\\t&={\frac {-1}{c-1}}\\t&={\frac {1}{1-c}}\end{aligned}}}
Assim
φ
:
S
2
−
{
N
}
↦
R
2
(
a
,
b
,
c
)
↦
(
a
1
−
c
,
b
1
−
c
)
{\displaystyle {\begin{aligned}\varphi :S^{2}-\{N\}&\mapsto \mathbb {R} ^{2}\\(a,b,c)&\mapsto ({\frac {a}{1-c}},{\frac {b}{1-c}})\end{aligned}}}
Bem definida
Obervemos que
φ
{\displaystyle \varphi }
(
0
,
0
,
1
)
∈
S
2
{\displaystyle (0,0,1)\in S^{2}}
φ
.
{\displaystyle \varphi .}
Injetora
Sejam
p
1
=
(
a
1
,
b
1
,
c
1
)
{\displaystyle p_{1}=(a_{1},b_{1},c_{1})}
p
2
=
(
a
2
,
b
2
,
c
2
)
{\displaystyle p_{2}=(a_{2},b_{2},c_{2})}
p
1
≠
p
2
{\displaystyle p_{1}\neq p_{2}}
p
1
,
p
2
∈
S
2
∖
{
N
}
{\displaystyle p_{1},p_{2}\in S^{2}\setminus \{N\}}
φ
(
p
1
)
=
φ
(
(
a
1
,
b
1
,
c
1
)
)
=
(
a
1
1
−
c
1
,
b
1
1
−
c
1
)
.
{\displaystyle \varphi (p_{1})=\varphi ((a_{1},b_{1},c_{1}))=({\frac {a_{1}}{1-c_{1}}},{\frac {b_{1}}{1-c_{1}}}).}
φ
(
p
2
)
=
φ
(
(
a
2
,
b
2
,
c
2
)
)
=
(
a
2
1
−
c
2
,
b
2
1
−
c
2
)
.
{\displaystyle \varphi (p_{2})=\varphi ((a_{2},b_{2},c_{2}))=({\frac {a_{2}}{1-c_{2}}},{\frac {b_{2}}{1-c_{2}}}).}
Portanto,
φ
(
p
1
)
≠
φ
(
p
2
)
.
{\displaystyle \varphi (p_{1})\neq \varphi (p_{2}).}
φ
{\displaystyle \varphi }
Sobrejetora
Tome
(
x
,
y
,
0
)
∈
R
2
{\displaystyle (x,y,0)\in \mathbb {R} ^{2}}
N
=
(
0
,
0
,
1
)
∈
S
2
.
{\displaystyle N=(0,0,1)\in S^{2}.}
r
:
{
(
0
,
0
,
1
)
+
t
(
x
,
y
,
−
1
)
/
t
∈
R
}
=
{
(
t
x
,
t
y
,
1
−
t
)
/
t
∈
R
}
{\displaystyle r:\{(0,0,1)+t(x,y,-1)/t\in \mathbb {R} \}=\{(tx,ty,1-t)/t\in \mathbb {R} \}}
∃
!
{\displaystyle \exists !}
p
{\displaystyle p}
r
∩
S
2
=
{
p
}
{\displaystyle r\cap S^{2}=\{p\}}
t
2
x
2
+
t
2
y
2
+
(
1
−
t
)
2
=
1
t
2
x
2
+
t
2
y
2
+
t
2
−
2
t
=
0
t
2
∗
(
x
2
+
y
2
+
1
)
−
2
t
=
0
t
=
0
t
∗
(
x
2
+
y
2
+
1
)
−
2
=
0
t
=
2
x
2
+
y
2
+
1
{\displaystyle {\begin{aligned}t^{2}x^{2}+t^{2}y^{2}+(1-t)^{2}&=1\\t^{2}x^{2}+t^{2}y^{2}+t^{2}-2t&=0\\t^{2}*(x^{2}+y^{2}+1)-2t&=0\\t&=0\\t*(x^{2}+y^{2}+1)-2&=0\\t&={\frac {2}{x^{2}+y^{2}+1}}\end{aligned}}}
Logo o ponto
p
{\displaystyle p}
p
=
(
2
x
x
2
+
y
2
+
1
,
2
y
x
2
+
y
2
+
1
,
x
2
+
y
2
−
1
x
2
+
y
2
+
1
)
{\displaystyle p=({\frac {2x}{x^{2}+y^{2}+1}},{\frac {2y}{x^{2}+y^{2}+1}},{\frac {x^{2}+y^{2}-1}{x^{2}+y^{2}+1}})}
É facilmente verificável que
p
∈
S
2
,
{\displaystyle p\in S^{2},}
|
p
|
=
1.
{\displaystyle |p|=1.}
Assim concluímos que
φ
{\displaystyle \varphi }
φ
−
1
:
R
2
↦
S
2
∖
{
N
}
(
x
,
y
)
↦
(
2
x
x
2
+
y
2
+
1
,
2
y
x
2
+
y
2
+
1
,
x
2
+
y
2
−
1
x
2
+
y
2
+
1
)
{\displaystyle {\begin{aligned}\varphi ^{-1}:\mathbb {R} ^{2}&\mapsto S^{2}\setminus \{N\}\\(x,y)&\mapsto ({\frac {2x}{x^{2}+y^{2}+1}},{\frac {2y}{x^{2}+y^{2}+1}},{\frac {x^{2}+y^{2}-1}{x^{2}+y^{2}+1}})\end{aligned}}}
Observe que
φ
−
1
(
x
,
y
)
≠
(
0
,
0
,
1
)
;
∀
x
,
y
∈
R
2
.
{\displaystyle \varphi ^{-1}(x,y)\neq (0,0,1);\forall x,y\in \mathbb {R} ^{2}.}
Mostrar que,
(
φ
∘
φ
−
1
)
(
p
)
=
(
φ
−
1
∘
φ
)
(
p
)
=
p
.
{\displaystyle (\varphi \circ \varphi ^{-1})(p)=(\varphi ^{-1}\circ \varphi )(p)=p.}
φ
−
1
(
p
)
=
(
2
x
x
2
+
y
2
+
1
,
2
y
x
2
+
y
2
+
1
,
x
2
+
y
2
−
1
x
2
+
y
2
+
1
)
φ
(
φ
−
1
(
p
)
)
=
(
2
x
x
2
+
y
2
+
1
∗
x
2
+
y
2
+
1
2
,
2
y
x
2
+
y
2
+
1
∗
x
2
+
y
2
+
1
2
)
=
(
x
,
y
)
{\displaystyle {\begin{aligned}\varphi ^{-1}(p)&=({\frac {2x}{x^{2}+y^{2}+1}},{\frac {2y}{x^{2}+y^{2}+1}},{\frac {x^{2}+y^{2}-1}{x^{2}+y^{2}+1}})\\\varphi (\varphi ^{-1}(p))&=({\frac {2x}{x^{2}+y^{2}+1}}*{\frac {x^{2}+y^{2}+1}{2}},{\frac {2y}{x^{2}+y^{2}+1}}*{\frac {x^{2}+y^{2}+1}{2}})\\&=(x,y)\end{aligned}}}
Analogamente, mostra-se que
(
φ
−
1
∘
φ
)
(
p
)
=
p
.
{\displaystyle (\varphi ^{-1}\circ \varphi )(p)=p.}
Portanto
φ
−
1
{\displaystyle \varphi ^{-1}}
φ
.
{\displaystyle \varphi .}
Continuidade
Como
φ
{\displaystyle \varphi }
φ
−
1
{\displaystyle \varphi ^{-1}}
φ
{\displaystyle \varphi }
