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hw8 Soln

The document contains solutions to homework problems from MATH 461, Fall 2024, covering topics such as quotient maps, product topology, and properties of compact spaces. It includes proofs demonstrating that a quotient map is a homeomorphism, that the projection map is a quotient map, and that closed subspaces of compact spaces are compact and vice versa in Hausdorff spaces. Each proof is logically structured and provides necessary definitions and reasoning to support the conclusions.

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16 views1 page

hw8 Soln

The document contains solutions to homework problems from MATH 461, Fall 2024, covering topics such as quotient maps, product topology, and properties of compact spaces. It includes proofs demonstrating that a quotient map is a homeomorphism, that the projection map is a quotient map, and that closed subspaces of compact spaces are compact and vice versa in Hausdorff spaces. Each proof is logically structured and provides necessary definitions and reasoning to support the conclusions.

Uploaded by

fzsjpxuwfm8279
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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MATH 461, Fall 2024

HW8 Solutions

1. If q : X → Y is a quotient map, let ∼ be the equivalence relation on X given by x ∼ y iff


q(x) = q(y). Prove that X/∼ with the quotient topology is homeomorphic to Y .
Let p : X → X/∼ be the quotient map from X to X modulo this relation. Then we define
q̄ : X/∼ → Y by sending each equivalence class [x] to q(x) ∈ Y . This is well-defined by
the definition of ∼: any two points in this equivalence class are equivalent, so x ∼ y, so
q(x) = q(y).
The function q̄ is surjective because q = q̄ ◦ p is surjective, and when a composite is surjective,
the second function in the composite is also surjective. It is injective because when q̄([x]) =
q̄([y]), q(x) = q(y), which implies x ∼ y, which implies [x] = [y]. Therefore q̄ is bijective.
Now consider a set U ⊆ Y and its preimages q̄ −1 (U ) ⊆ X/ ∼ and q −1 (U ) = p−1 q̄ −1 (U ) ⊆ X.
Since q is a quotient map, U is open iff q −1 (U ) = p−1 q̄ −1 (U ) is open. This is true iff q̄ −1 (U )
is open, since p is a quotient map. All together, U is open iff q̄ −1 (U ) is open, so q̄ is a
homeomorphism.
2. Let X and Y be nonempty topological spaces, and give X × Y the product topology. Prove
that the projection map π1 : X × Y → X is a quotient map.
Since Y is nonempty, each x ∈ X has a preimage (x, y) ∈ X × Y , so π1 is surjective.
If U ⊆ X is open, then its preimage U × Y is open, by the definition of the product topology.
S
On the other hand, if U × Y is open, then U × Y = α Uα × Vα for open sets Uα ⊆ X and
Vα ⊆ Y . Removing those terms where Vα is empty, each Uα is therefore a subset of U . On
the other hand, for each point x ∈ U there is some y such that (x, y) ∈ Uα × Vα , so that
S
x ∈ Uα . This shows that U = α Uα , so that U is open.
All together, we have shown that U is open iff its preimage U × Y is open, so π1 is a quotient
map.
3. Prove that a closed subspace of a compact space is compact.
Suppose X is compact and A ⊆ X is closed. Given an open cover {Uα } of A, with the Uα
open subsets of X, add in the open set X \ A to get an open cover of X. Since X is compact,
there is a finite subcover {U1 , . . . , Un , X \ A}. Then {U1 , . . . , Un } is still a cover of A, because
removing X \ A doesn’t affect whether every point of A is contained in the union. We have
arrived at a finite subcover of our original cover, so A is compact.
4. Prove that a compact subspace of a Hausdorff space is closed.
Suppose X is Hausforff and A ⊆ X is compact. Fix a point x ∈ Ac .
For each a ∈ A, we get open sets Ua ∋ a and Va ∋ x such that Ua ∩ Va = ∅. Since A is
compact, we can find finitely many a1 , . . . , an such that A ⊆ Ua1 ∪ · · · ∪ Uan . Notice that this
open set is disjoint from the intersection Va1 ∩ · · · ∩ Van . Therefore Va1 ∩ · · · ∩ Van is an open
set containing x, that has no points of A.
Repeating this for each point x ∈ Ac , we see that Ac is a union of open sets. Therefore Ac is
open, so A is closed.

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