Skip to main content

Topology: An Inquiry-Based Approach

Steven Schlicker

πŸ”—

Section Identifying Quotient Spaces

πŸ”—
Suppose X is a topological space and Y is a set, and let p:Xβ†’Y be a surjection. We can define a relation ∼p on X by x∼py if and only if p(x)=p(y). It is straightforward to show that ∼p is an equivalence relation, the details are left for Exercise 1. From this we can see that our two approaches to defining the quotient topology and quotient spaces are really the same.
πŸ”—
Oftentimes we have a topological space X and a relation ∼ on X, and we would like to have an effective way to be able to identify the quotient space X/∼ as homeomorphic to some familiar topological space Y. That is, we want to be able to show that there is a homeomorphism f from X/∼ to Y.
πŸ”—

Example 16.7.

πŸ”—
Consider the following situation. Let X=R with the standard topology and define the relation ∼ on R by x∼y if xβˆ’y∈Z. It is straightforward to show that ∼ is an equivalence relation. By this equivalence relation, we have xβˆ’1∼x for every real number x. This identifies R with the interval [0,1], where 0 and 1 are identified under the relation. So we might expect that R/∼ is homeomorphic to the circle S1={(x,y)∈R2∣x2+y2=1} as a subspace of R2 with the standard topology. Now the objective is to find a homeomorphism between S1 and R/∼. Since every point on the unit circle has the form (cos⁑(t),sin⁑(t)) for some real number t, we might try defining f:(R/∼)β†’S1 by f([t])=(cos⁑(t),sin⁑(t)). However, we have that 0∼1, which means that [0]=[1], but f([0])β‰ f([1]) and so f is not well-defined. Another option might be f([t])=(cos⁑(2Ο€t),sin⁑(2Ο€t)). In this case, if x∼y, then 2Ο€x and 2Ο€y differ by a multiple of 2Ο€ and so f([x])=f([y]). We could then show that f is a homeomorphism. We will continue this example shortly.
πŸ”—
The following theorem encapsulates the above example.
πŸ”—

Activity 16.6.

πŸ”—
Theorem 16.8 is a statement about sets and functions, and there is no topology involved. We prove the theorem in this activity. Use the conditions stated in Theorem 16.8.
πŸ”—

(a)

πŸ”—
Show that f― is well-defined. That is, show that whenever [x1]=[x2] in X/∼, then f―([x1])=f―([x2]).
πŸ”—

(c)

πŸ”—
Show that the uniqueness of f― comes from the equation f=fβ€•βˆ˜p.
πŸ”—
Now we present a final result that can be very helpful when working with quotient spaces.
πŸ”—

Proof.

πŸ”—
πŸ”—
The existence of f― as the unique function satisfying f=fβ€•βˆ˜p was established in Theorem 16.8. All that remains is to show that f― is continuous. Let O be an open set in Y. Since f is continuous, we know that fβˆ’1(O) is open in X. If x1∈fβˆ’1(O) and x1∼x2, then x2∈fβˆ’1(O) as well. Thus, we can write fβˆ’1(O) as
fβˆ’1(O)=⋃x∈fβˆ’1(O)[x].
πŸ”—
πŸ”—
That is, fβˆ’1(O) is a union of equivalence classes. Now f―([x])=f(x), so if x∈fβˆ’1(O), then [x]∈fβ€•βˆ’1(O). Thus,
fβˆ’1(O)=⋃x∈fβˆ’1(O)[x]=⋃[x]∈fβ€•βˆ’1(O)[x]=fβ€•βˆ’1(O).
πŸ”—
We conclude that fβ€•βˆ’1(O) is open in X and f― is continuous.
πŸ”—
Now we will see how to use Theorem 16.9 to establish a homeomorphism from a quotient space of a given topological space to another topological space
πŸ”—

Example 16.10.

πŸ”—
We return to the situation from Example 16.7 with X=R under the standard topology and equivalence relation ∼ defined by x∼y if xβˆ’y∈Z. We will use Theorem 16.9 to show that R/∼ is homeomorphic to the circle Y=S1.
Figure 16.11. A basis element for S1.
πŸ”—
Step 1
πŸ”—
Define a continuous surjection f:Xβ†’Y that respects the relation. That is, we need to ensure that f(x1)=f(x2) whenever x1∼x2 in X. We saw earlier that the function f defined by f(t)=(cos⁑(2Ο€t),sin⁑(2Ο€t)) respects the relation. Since every point on the unit circle is of the form (cos⁑(ΞΈ),sin⁑(ΞΈ)) for some real number ΞΈ, choosing t=ΞΈ2Ο€ makes f(t)=(cos⁑(ΞΈ),sin⁑(ΞΈ)) and f is a surjection. Now we need to demonstrate that f is continuous. A collection of basic open sets in S1 can be found by intersecting S1 with open balls in R2 as illustrated in Figure 16.11. We can see that the basic open sets are arcs of the form ab⏜ for a and b in S1. Suppose a=(cos⁑(2Ο€A),sin⁑(2Ο€A)) and b=(cos⁑(2Ο€B),sin⁑(2Ο€B)) for angles A and B. Then fβˆ’1(ab⏜) is the union of intervals (A+2Ο€k,B+2Ο€k) for k∈Z. As a union of open intervals, we have that fβˆ’1(ab⏜) is open in X. We have now found a continuous surjection from X to Y that respects the relation.
Step 2
πŸ”—
Find a continuous function from X/∼ to Y. Theorem 16.9 tells us that the function f―:(X/∼)β†’Y defined by f―([t])=f(t) is continuous. So f― is our candidate to be a homeomorphism.
Step 3
πŸ”—
Show that f― is a bijection. Let y∈Y. The fact that f is a surjection means that there is a t∈R such that f(t)=y. It follows that f―([t])=f(t)=y and f― is a surjection. To demonstrate that f― is an injection, suppose f―([s])=f―([t]) for some s,t∈R. Then (cos⁑(2Ο€s),sin⁑(2Ο€s))=f(s)=f(t)=(cos⁑(2Ο€t),sin⁑(2Ο€t)). It must be the case then that 2Ο€s and 2Ο€t differ by a multiple of 2Ο€. That is, 2Ο€sβˆ’2Ο€t=2Ο€k for some integer k. From this we have sβˆ’t=k∈Z, and so s∼t. This makes [s]=[t] and we conclude that f― is an injection.
Step 4
πŸ”—
Show that f― is a homeomorphism. At this point we already know that f― is a continuous bijection, so the only item that remains is to show that f―(O―) is open whenever O― is open in X/∼. Let p:Xβ†’(X/∼) be the standard map. Let O― be a nonempty open set in X/∼. Then O=pβˆ’1(O―) is open in X. Thus, O is a union of open intervals. Let (a,b) be an interval contained in O. From the definition of f we have that f(a,b) is the open arc f(a)f(b)⏜, which is open in Y. So f(O) is a union of open arcs in Y, which makes f(O) open in Y. Now f(O)=(fβ€•βˆ˜p)(O)=f―(p(O))=f―(O―), and f―(O―) is open in Y. We conclude that f― is a homeomorphism from X/∼ to S1, and so S1 is a quotient space of R.
<Prev^TopNext>