Tiểu luận:Tôpô thương đối với một ánh xạ pot

CP
1
S
2
X Y
f : X −→ Y
f
= {U ⊂ Y, f
−1
(U) X}
f
Y f
• Ø = f
−1
(Ø) X Ø ∈
f
• f
−1
(Y ) = X − ⇒ Y ∈
f
• U
1
, U
2
∈
f
, f
−1

(U
1
∩ U
2
) = f
−1
(U
1
) ∩ f
−1
(U
2
) ⇒ U
1
∩ U
2
∈
f
• {U
i
, i ∈ I} ⊂
f
⇒ f
−1
(U
i
) ∀i ∈ I f
−1



i∈I
U
i

=

i∈I
f
−1
(U
i
)

i∈I
U
i
∈
f
f
Y
• RP
n
= { R
n+1
.}
p : R
n+1
\{0} −→ RP
n
, p(x) = [x] = αx, α ∈ R

RP
n
p
• RP
n
= { C
n+1
.}
q : C
n+1
\{0} −→ CP
n
, q(x) = [x] = {αx, α ∈ R}
CP
n
q
• X ∼ X Y =
X ∼ x ∈ X Y
p : X −→ Y
X, Z Y f : X −→ Y
Y f
g : Y −→ Z g ◦ f : X −→ Z
(⇒) Y f
U Y f
−1
(U) X f
g g ◦ f
g ◦ f V Z (g ◦ f)
−1
(V )

X (g ◦ f)
−1
(V ) = f
−1
(g
−1
(V )) Y g
−1
(V ) ∈
f
)
g
−1
(V ) Y g
σ X/ ∼
f  X/ ∼
f G  f
−1
(G) X
G ∈ σ  ≤ σ
X G X G
π : X −→ X/G, π(x) = Gx
∀U ⊂ X
π
−1
(π(U)) = {x ∈ X | π(x) ∈ π(U)}
= {x ∈ X | ∃y ∈ U : π(x) = π(y)}
= {x ∈ X | ∃g ∈ G, y ∈ Y : x = gy}
= {x
∪

g∈G
gU
θ
g
: X −→ X, θ(x) = gx U X gU
X
∪
g∈G
gU X π(U) π
X
i) G X G X/G
ii) A X X/A
S
n
= {∈ R
n+1
| x = 1} n
RP
1
∼
=
S
1
CP
1
∼
=
S
2
.

X/A, A ⊂ X
X A ⊂ X
X ∀x, y ∈ X, x ∼ y ⇔


x = y
x, y ∈ A
∼ X
∀x ∈ X, [x] =


{x}, x ∈ A
A, x ∈ A
X/ ∼= (X \ A) ∪ {A} = X/A
S
1
+
/A
∼
=
S
1
.
X = S
1
+
= {(x, y) ∈ R
2
, x
2

+ y
2
= 1, y ≥ 0}., A = {(−1, 0), (1, 0)}
f : S
1
−→ S
1
+
/A, f(x) = [x] f
• f
• ∀(x
1
, y
1
), (x
2
, y
2
) ∈ S
1
f(x
1
, y
1
) = f(x
2
, y
2
)
⇔ [(x

1
, y
1
)] = [(x
2
, y
2
)]
(x
1
, y
1
) = (x
2
, y
2
) (x
1
, y
1
), (x
2
, y
2
) ∈ A
(x
1
, y
1
) = (x

2
, y
2
) f
(x
1
, y
1
), (x
2
, y
2
) ∈ A [(x
1
, y
1
)] = [(x
2
, y
2
)] = A
(x
1
, y
1
) = (x
2
, y
2
) f

S
1
S
1
+
/A f
S
1
+
/A
∼
=
S
1
.
RP
1
∼
=
S
1
.
S
1
+
p

f
##
F

F
F
F
F
F
F
F
F
F
S
1
+
/A
g
//
RP
1
p f
f(x, y) = [x : y] = {α(x, y), α ∈ R} f
g g[(x, y)] = [x : y]
∀(x, y) ∈ R
2
, g(p(x, y)) = g[(x, y)] = α(x, y) = f(x, y) g ◦ p = f.
f g g
∀[(x
1
, y
1
)], [(x
2

, y
2
)] ∈ S
1
+
/A g[(x
1
, y
1
)] = g[(x
2
, y
2
)]
⇔ [(x
1
: y
1
] = [x
2
, y
2
]
⇔ (x
1
, y
1
) = α(x
2
, y

2
), α ∈ R
⇒



x
1
= αx
2
y
1
= αy
2
α = 1 [(x
1
, y
1
)] = [(x
2
, y
2
)]
α = 1 α
2
(x
2
2
+ y
2

2
) = x
2
1
+ y
2
1
= 1 ⇒ α = −1



x
1
= −x
2
y
1
= −y
2
y
2
≥ 0 y
2
= 0 ⇒ x
2
= 1 (x
1
, y
1
), (x

2
, y
2
) ∈ A
[(x
1
, y
1
)] = [(x
2
, y
2
)] g
f p g
RP
1
S
1
+
/A g
RP
1
∼
=
S
1
2. X = S
2
+
= {(x, y, z) ∈ R

3
, x
2
+ y
2
+ z
2
= 1, z ≥ 0},
S
1
+
= {(x, y, 0) ∈ R
3
, x
2
+ y
2
= 1}
CP
1
∼
=
S
2
S
2
∼
=
S
2

+
/S
1
+
f : S
2
−→ S
2
+
/S
1
+
f
f
∀(x
1
, y
1
, z
1
), (x
2
, y
2
, z
2
) ∈ S
2
, f(x
1

, y
1
, z
1
) = f(x
2
, y
2
, z
2
)
⇔ [(x
1
, y
1
, z
1
)] = [(x
2
, y
2
, z
2
)]
(x
1
, y
1
, z
1

) = (x
2
, y
2
, z
2
) (x
1
, y
1
, z
1
), (x
2
, y
2
, z
2
) ∈ S
1
+
(x
1
, y
1
, z
1
) = (x
2
, y

2
, z
2
) f
(x
1
, y
1
, z
1
), (x
2
, y
2
, z
2
) ∈ A [(x
1
, y
1
, z
1
)] = [(x
2
, y
2
, z
2
)] = A
(x

1
, y
1
, z
1
) = (x
2
, y
2
, z
2
) f
S
2
+
/S
1
+
∼
=
S
2
.
CP
1
∼
=
S
2
S

2
+
p

f
((
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
S
2
∼
=
S
2
+

/S
1
+
g
//
CP
1
[(x, y, z)]

//
(x + iy : z)
f
f : S
2
+
//
CP
1
(x, y, z)

//
(x + iy : z)
g : S
2
+
/S
1
+
//
CP

1
[(x, y, z)

//
[x + iy : z]
f = g ◦ p f g
g
∀[(x
1
, y
1
z
1
, )], [(x
2
, y
2
, z
2
)] ∈ S
2
+
/S
1
+
g[(x
1
, y
1
, z

1
)] = g[(x
2
, y
2
, z
2
)]
⇔ [x
1
+ iy
1
: z
1
] = [x
2
+ iy
2
: z
2
]
⇔ (x
1
+ iy
1
)z
2
= (x
2
+ iy

2
)z
1
⇔ (x
2
1
+ y
2
1
)z
2
2
= (x
2
2
+ iy
)
2
z
2
1
⇒ (1 − z
2
1
)z
2
2
= (1 − z
2
2

)z
2
1
⇒ z
2
= z
1
z
1
= z
2
= 0 (x
1
, y
1
z
1
, ), (x
2
, y
2
, z
2
) ∈ S
1
+
[(x
1
, y
1

z
1
, )] = [(x
2
, y
2
, z
2
)]
z
1
= z
2
> 0 x
1
= x
2
, y
1
= y
2
[(x
1
, y
1
z
1
)] = [(x
2
, y

2
, z
2
)]
g
f p g S
2
+
/S
1
+
CP
1
g
CP
1
∼
=
S
2
.