Lecture Notes Topology (2301631) Phichet Chaoha Department of ...

28 Topology (2301631)
Example 1.126. Let L = R × {−1, 1}. Define an equivalence relation ∼ on L
by (x, −1) ∼ (x, 1) for all x ≠ 0. Then, the quotient space L/ ∼ is called the line
with double points. It is easy to see that L is Hausdorff, but L/ ∼ is not!
Definition 1.127. Let p : X → Y be a map and y ∈ Y . We will call the set
p −1 ({y}) the fiber of p over y.
Theorem 1.128. Let p : X → Y be a quotient map, Z a space, and g :
X → Z a continuous map. If g is constant on each fiber of p, then there exists a
unique continuous map f : Y → Z such that g = f ◦ p; i.e., we have the following
commutative diagram :
X
❅ ❅❘ p
g
✲ Z
♣ ♣ ♣ ♣
✒
∃!f
Y
Proof. Since g is constant on each fiber of p, we can simply define f : Y → Z
by f(y) = g(a) for each y ∈ Y and some point a in the fiber of p over y. Clearly,
g = f ◦ p. For each open subset V of Z, the set g −1 (V ) = p −1 (f −1 (V )) is open by
the continuity of g, and hence f −1 (V ) must be open in Y since p is a quotient map.
Therefore, f is continuous. Now suppose f ′ : Y → Z is a map such that g = f ′ ◦ p.
Then, for each y ∈ Y , we have f(y) = g(a) = f ′ ◦ p(a) = f ′ (y) where a is any point
in the fiber over y.
□
Example 1.129. Let p : I 2 → S 2 and q : I 2 → T 2 be the quotient maps
from the previous example. Then, by the previous theorem, there exists a unique
continuous map f : T 2 → S 2 such that p = f ◦ q. Morever, it is easy to see that f
is surjective.
Exercise 1.130. Prove that there is a surjective continuous map from the
Mobius strip onto the Klien bottle.