Endsem

Part A

1.

The topology on a set in which every subset of is open is called the discrete topology.

2.

If is a basis for the topology on and , then a basis for the subspace topology on is

3.

A bijection between topological spaces is called a homeomorphism if both and are continuous.

4.

A topological space is said to be compact if every open cover of has a finite subcover.

5.

A topological space is said to be second countable if its topology can be generated by a basis with countable cardinality.

6.

Any nonempty connected subspace of is necessarily an interval.

7.

According to Urysohn’s lemma, in a normal topological space, continuous functions can separate disjoint closed sets.

8.

The map

for a fixed is a covering map.

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Part B

9.

Show that

is not an open set in with respect to the finite complement topology.

Solution

First compute the intersection:

In the finite complement topology on , the open sets are exactly the empty set and the complements of finite subsets of .

The set is neither empty nor the complement of a finite set. Therefore it is not open in the finite complement topology.

So

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10.

Let be a set and for be a countable collection of topologies on . Then

is also a topology on .

Solution

We verify the topology axioms.

1. Since each is a topology, both and lie in every . Hence

2. Let be a family of sets in . Then each lies in every . Because each is closed under arbitrary unions,

for every . Therefore

3. Let . Then for every , and since each is closed under finite intersections,

for every . Hence .

Thus is a topology on .

So

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11.

The identity map

is continuous, where and denote the trivial topology and the standard topology respectively.

Solution

This statement is false.

To be continuous, the identity map would have to satisfy that for every open set in , the preimage is open in .

But the trivial topology has only two open sets: and . Since is open in the standard topology but not open in the trivial topology, the identity map is not continuous.

So the correct conclusion is

If the direction were reversed, then the identity map would be continuous.

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12.

Any subspace of a connected topological space is also connected.

Solution

This statement is false.

Take , which is connected, and let

In the subspace topology, is discrete, because

are both open in .

Thus is separated into two nonempty disjoint open sets, so is disconnected.

Therefore

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13.

Any subspace of a locally compact topological space is locally compact in the subspace topology.

Solution

This statement is false in general.

A standard counterexample is

The space is locally compact, but is not locally compact in the subspace topology.

To see this, let and let be any neighborhood of in . Then there exists such that

If were locally compact, some neighborhood of would have compact closure in . But any such set contains a rational interval, and its closure in contains points of converging to an irrational limit. That closure is therefore not compact.

So is not locally compact.

Hence

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14.

Suppose for are covering maps. Then the map

is necessarily a covering map.

Solution

This statement is true.

Let . Since and are covering maps, there exist evenly covered neighborhoods of and of such that

where each restriction

is a homeomorphism.

Then

On each product piece , the map restricts to

and this restriction is the product of two homeomorphisms, hence itself a homeomorphism.

Therefore is evenly covered by , so is a covering map.

Thus

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Part C

15.

Prove that the map

is continuous. Hence show that every polynomial

with is a continuous function of the real variable .

Solution

We prove continuity of the product map first.

The multiplication map

is continuous. This is a standard fact from analysis: it follows, for example, because

and each term can be made small when and .

Now define recursively

Since a composition of continuous maps is continuous, each is continuous. In particular, is continuous.

So

Now let

Each map is continuous because it is the -fold product of with itself. Multiplication by a constant is continuous, and sums of continuous functions are continuous.

Therefore every polynomial function is continuous.

Hence

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16.

Let

and let

Prove that the restriction is a bijective continuous map between locally compact topological spaces, but is not a homeomorphism.

Solution

First note that is a subspace of consisting of isolated points.

Indeed, is isolated because

and each point is isolated because the numbers are separated from each other and tend to infinity, so each has a small interval in meeting in only that point.

Thus is discrete, hence locally compact.

Now compute the image:

So

The restriction is clearly injective because distinct points of map to distinct points of , and it is surjective by definition of .

Hence is a bijection. It is continuous because it is the restriction of the continuous map .

The space is locally compact as a subspace of the compact Hausdorff space ; in fact is compact because it contains the limit point of the sequence .

However, is not a homeomorphism because its inverse is not continuous. Indeed,

but the preimages are

in , and in the subspace they do not converge to (or to any point of ). In particular, the inverse map cannot send the convergent sequence to a convergent sequence in .

So is a continuous bijection between locally compact spaces, but it is not a homeomorphism.

Therefore

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17.

Define a space for . Give one example of a topological space that is for every .

Solution

There are several equivalent naming conventions in the literature; the standard hierarchy is as follows.

The separation axioms

1. A space is if for any two distinct points, at least one of them has an open neighborhood not containing the other.
2. A space is if for any two distinct points, each has a neighborhood not containing the other. Equivalently, every singleton is closed.
3. A space is if it is Hausdorff: any two distinct points have disjoint neighborhoods.
4. A space is if it is regular and : given a point and a disjoint closed set, there are disjoint open neighborhoods separating them.
5. A space is if it is normal and : any two disjoint closed sets have disjoint open neighborhoods.
6. A space is if it is completely normal and .
7. A space is if it is perfectly normal.

Example

Any discrete topological space is for every .

Indeed, in a discrete space every subset is open and closed, so all separation properties hold automatically.

Thus one example is

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18.

State and prove the Tietze extension theorem.

Solution

Statement

Let be a normal topological space and let be closed. Then every continuous function

extends to a continuous function

such that .

Equivalently, if is continuous and bounded, then extends to a continuous function on .

Proof idea

The proof is constructive and uses Urysohn’s lemma repeatedly.

First reduce to the case by scaling and translating a bounded real-valued map.

The strategy is to build continuous functions on that approximate on more and more accurately, with errors .

Step 1: Separate upper and lower level sets

For a continuous , consider the closed subsets of

These are closed in , hence of the form and for closed sets in . Because is normal, Urysohn’s lemma gives a continuous function that is on and on .

Subtracting this first approximation from leaves a new continuous function on whose range is smaller. Repeating the procedure with finer and finer bounds yields a uniformly convergent series of continuous functions on .

Step 2: Uniform convergence

One constructs continuous functions such that

and the partial sums

converge uniformly on to .

Since the are continuous and the series converges uniformly, the sum

is continuous on .

By construction, and .

Thus every continuous function on a closed subset of a normal space extends to the whole space.

So the Tietze extension theorem holds:

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19.

Consider the topological space and the action of on given by

Define a relation on by if and only if there exists such that .

(a) Prove that is an equivalence relation.

(b) Prove that is compact and path connected in the quotient topology.

Solution

(a) is an equivalence relation

We verify the three properties.

Reflexive: for every , take . Then , so .

Symmetric: if , then for some . Since , we also have , so .

Transitive: if and , then and for some nonzero reals . Hence

so .

Therefore is an equivalence relation.

(b) Compactness and path connectedness of the quotient

Each equivalence class is a line through the origin with the origin removed. Thus the quotient space is naturally the real projective space

To see compactness, observe that every equivalence class meets the unit sphere

Indeed, if , then is equivalent to . The quotient map

is continuous and surjective. Since is compact, its continuous image is compact. Therefore is compact.

For path connectedness, consider two equivalence classes represented by .

If , then has only one equivalence class under scaling, so the quotient is a single point and hence path connected.

If , the sphere is path connected. Choose a path in joining to either or ; its projection to the quotient gives a path from to .

Hence is path connected.

Therefore

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20.

Let . Let be a covering map where and are topological spaces. Let

be a continuous map. Suppose and . Prove that there exists a unique map

such that and .

Solution

This is the standard lifting theorem for coverings applied to the square .

The key facts are:

• is path connected.
• is simply connected.

Because is a covering map, every path in starting at has a unique lift starting at .

Existence

For each point , choose any path in from to . Then is a path in starting at .

Lift this path uniquely to a path in starting at . Define

Because is simply connected, any two paths from to are homotopic rel endpoints. The homotopy lifting property for coverings implies that their lifted endpoints agree. Hence is well defined.

Continuity

Continuity follows from local triviality of the covering map: around each point of , there is an evenly covered neighborhood, and the lift varies continuously over such neighborhoods. This is the standard local lifting argument.

Uniqueness

If and are two lifts with , then for each point the paths and are lifts of the same path starting at the same point. By uniqueness of path lifting, they coincide. Hence .

Therefore there exists a unique lift such that

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Part D

21.

Let be a locally compact Hausdorff space, a normal space and be a covering map such that is a finite set for every . Prove that every pair determines a continuous function such that , where is closed and is continuous. What is the significance of the hypothesis that is finite for every ?

Solution

The intended conclusion is a Tietze-extension-type statement for .

The key point is that a covering map with finite fibers is a finite-sheeted covering. For locally compact Hausdorff spaces, finite-sheeted coverings are proper and hence closed maps. This is the role of the finiteness assumption: it forces the covering to behave like a proper map rather than an infinite-sheeted one.

Once the map is closed and the target space is normal, the standard perfect-map theorem applies: the source space inherits normality from .

After that, the Tietze extension theorem applies to : if is closed and is continuous, then there exists a continuous extension

with

So the conclusion is

The significance of the finiteness hypothesis is that it ensures the covering has finitely many sheets over each point, which is what makes the covering map closed/proper and allows normality to be transferred to .

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Endsem Question Paper