Worksheet 9

Question 1

Consider a countable collection of topological spaces . Suppose for any ordered pair of +ve integers the following is true:

i) are open subsets in with .

ii) There exist homeomorphisms satisfying:

a)

b)

c) on

Consider the disjoint union of as , i.e., .

Let . Call if and .

Prove: is an equivalence relation.

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Solution

Claim: The relation on defined by iff and (where and ) is an equivalence relation.

Proof:

We verify the three properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Reflexivity

Let for some .

Since , we have .

By property (a), is the identity on , so .

Therefore, .

Symmetry

Suppose where and .

By definition of , we have and .

We must show that , i.e., and .

Step 1: Since is a homeomorphism from to , we have .

Step 2: Since is a homeomorphism, it is bijective and has an inverse .

Therefore:

Step 3: We claim that .

By property (c) with :

Since is the identity:

More directly, applying property (c) with different indices: by property (c) with and keeping :

Similarly:

This shows that and are inverses of each other.

Step 4: Therefore:

Thus .

Transitivity

Suppose and where , , and .

By definition:

• and
• and

We must show that , i.e., and .

Step 1: From , we have .

This means .

By property (b) (which is equivalent to , applied to the symmetric case):

Actually, let me reconsider. Property (b) states:

Since and (from the hypothesis), and maps onto :

We have , so (since bijectively maps this intersection to ).

Step 2: Now compute:

By property (c):

Since (from Step 1):

Step 3: By property (c), is a homeomorphism from to , so .

Therefore, .

Conclusion

Since is reflexive, symmetric, and transitive, it is an equivalence relation.

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Question 2

Let be equipped with the quotient topology. A set is said to be open if is open in for all .

Let denote the canonical map given by the composition as appropriate.

Prove: The image of is an open subset in .

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Solution

Claim: For each , the image is open in the quotient topology on .

Proof:

Recall that in the quotient topology on , a set is open if and only if its preimage is open in , where is the quotient map.

Step 1: Identify the preimage of

We need to find .

A point satisfies .

This means there exists such that .

But (by definition of as the composition), so , which means .

Therefore:

Step 2: Describe the equivalence class of points in

For , the equivalence class consists of all such that .

If , then (where ) iff:

• and and , or
• and and (by symmetry)

Therefore:

where we use the homeomorphisms to identify equivalent points across different spaces.

Step 3: Show this preimage is open in

We must verify that for each , the intersection is open in .

Case 1:

which is open in (it is the whole space).

Case 2:

By assumption, is a homeomorphism from to .

Since is an open subset of (by property (i)), and is a homeomorphism, the image is open in .

But the topology on may not be specified further. However, by the problem statement, we’re given that is an open subset in .

Actually, we need to reconsider. The problem says ” are open subsets.” This defines open sets within each , but we also need to understand the global topology on .

Since is a homeomorphism from (with its subspace topology from ) to (with its subspace topology from ), and is open in , the image is open in .

Therefore, is open in .

Step 4: Conclude

Since is open in for all , we have that is open in (by the definition of open sets in as stated in the problem).

By definition of the quotient topology, this means is open in .

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Question 3

Prove: The map is a homeomorphism.

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Solution

Claim: The restriction (where with codomain ) is a homeomorphism.

Proof:

We must verify that is continuous, bijective, and has a continuous inverse.

Step 1: Verify is continuous

The map is defined as the composition:

where is the inclusion of into the disjoint union , and is the quotient map.

Both (inclusion into a disjoint union) and (quotient map) are continuous.

Therefore, is continuous.

Step 2: Restrict codomain to

With codomain restricted to , the map remains continuous.

Step 3: Verify is bijective

Injectivity: Suppose for .

This means , so .

Since both and are in , and , we have (by definition of ):

• Either and (by reflexivity),
• Or they are related through other spaces.

But and is the identity, so the equivalence relation identifies a point in only with itself.

Wait, let me reconsider the definition of . The relation (where ) is defined only when and .

For , we have iff and .

Therefore, , so is injective.

Surjectivity: By definition, the codomain is , so every element in the codomain is of the form for some .

Therefore, is surjective.

Step 4: Show is an open map

To prove that is a homeomorphism, we can show that is an open map (i.e., the image of every open set in is open in ).

Alternatively, we can show that is continuous.

Method: Show is open

Let be an open set in .

We need to show that is open in (with the subspace topology from ).

In the subspace topology, is open in iff for some open set in .

By the quotient topology, is open in iff is open in .

Consider . An element satisfies iff .

Since , we have for some .

By the equivalence relation, for some .

Therefore:

For each :

Since is open in :

• is open in
• is open in (intersection of open sets in , since is open in )
• is open in (image of open set under homeomorphism)

Therefore, is open in , so is open in , and hence open in .

Step 5: Conclusion

Since is continuous, bijective, and open, it is a homeomorphism.

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Question 4

Let where is some topological space. Prove: is continuous iff is continuous.

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Solution

Claim: A map is continuous if and only if for each , the restriction is continuous (after identifying with via the homeomorphism ).

Actually, let me interpret the problem more carefully. We have where is some subset of .

Interpretation: We want to show is continuous iff is continuous for all .

More specifically, the restriction of to composed with should be continuous.

Claim: A map is continuous if and only if, for each , the composition is continuous.

Proof:

Direction 1: If is continuous, then is continuous for all

Assume is continuous.

For each , the composition is the composition of continuous maps:

Therefore, is continuous.

Direction 2: If is continuous for all , then is continuous

Assume that for each , the map is continuous.

Let be an open set in . We must show that is open in .

By the quotient topology on , a set is open in iff its preimage under the quotient map is open in .

Therefore, we must show that is open in .

Since , we need to show this is open in .

Step 3: Show is open in

For each , we check whether is open in .

A point satisfies iff:

Now, (since ), so:

Therefore:

By assumption, is continuous, so is open in .

This holds for all .

Step 4: Conclude

Since is open in for all , we have that is open in (by the definition of open sets in ).

By the quotient topology, this means is open in .

Since was an arbitrary open set in , is continuous.

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Question 5

Suppose we replace countability with an arbitrary indexing set and frame the same for questions, how different are your answers?

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Solution

Analysis: Replacing the countable index set with an arbitrary index set changes some aspects of the analysis, but many key results remain valid.

Properties that remain unchanged:

1. Equivalence relation (Question 1)

The proof that is an equivalence relation depends only on:

• The definition of the relation using homeomorphisms and their properties
• Properties (a), (b), (c) of the homeomorphisms

None of these properties require countability. Therefore, the relation remains an equivalence relation for arbitrary index sets.

2. Homeomorphism of (Question 3)

The proof that is a homeomorphism depends only on:

• Continuity of inclusion and quotient maps
• Injectivity from the equivalence relation structure
• Surjectivity by definition
• Openness via the quotient topology

None of these require countability. Therefore, remains a homeomorphism for arbitrary index sets.

3. Continuity criterion (Question 4)

The proof that is continuous iff is continuous for all depends only on:

• The quotient topology definition
• The structure of the equivalence relation

None require countability. Therefore, the continuity criterion remains valid for arbitrary index sets.

Properties that may change:

1. Openness of (Question 2)

The proof uses the fact that we can intersect with each and verify openness.

With an arbitrary (possibly uncountable) index set :

This is still a countable union (one element per index in ) if is countable, but becomes an arbitrary union if is uncountable.

The definition of open sets in states: ” is open if is open in for all .”

If is uncountable, this requires verifying openness for uncountably many conditions. However, the result is:

For arbitrary index sets, this is a union of an uncountable number of sets (one for each ).

The openness still holds: Each is open in for all , so is open in , hence is open in .

The key difference is that we’re verifying uncountably many local openness conditions, but this is still valid.

Topological properties:

With countable index set:

• The quotient space might inherit certain properties from the countability of the structure
• Separability arguments might apply

With arbitrary index set:

• The results remain valid
• However, the quotient space may not be first-countable or separable unless additional conditions are imposed
• The topology can become more complicated

Summary of differences:

Aspect	Countable Index Set	Arbitrary Index Set
Equivalence relation	✓ Equivalence relation	✓ Equivalence relation
Homeomorphism	✓ Homeomorphism	✓ Homeomorphism
Openness of	✓ Open in	✓ Open in
Continuity criterion	✓ Valid	✓ Valid
Separability of	Depends on local spaces	Depends on local spaces
First-countability of	Depends on local spaces	Generally harder to ensure
Metrizable if local spaces are	✓ Often yes	✗ Likely no

Conclusion

Surprising result: Nearly all the topological statements remain valid for arbitrary index sets. The main proofs do not fundamentally depend on countability.

The key insight is that:

1. The equivalence relation structure is index-set independent
2. The quotient topology definition works for arbitrary partitions
3. The local-to-global continuity criterion (Question 4) is robust to arbitrary index sets

The main practical difference is that:

• With countable index sets, we can potentially apply countability-dependent arguments (separability, metrizability, etc.)
• With arbitrary index sets, the quotient space loses some nice properties but retains the fundamental topological structure

Therefore, the answers to questions 1-4 are essentially unchanged, but question 2’s answer may involve uncountable unions if the index set is uncountable.