One of the reasons to introduce a topological space as a more general structure than a metric space is to be able to define continuous functions without a metric.

Let $X$,

𝑌

be sets, initially considered without a topological space structure. We write

𝑓 : 𝑋 \to 𝑌

to denote a function with domain

𝑋

and codomain

𝑌 .

The words function, map, mapping will mean the same thing. The following notation and terminology will be used.

Definition: image of an element, image of a set, preimage of a set

Let $𝑓 : 𝑋 \to 𝑌$ be a function. For $𝑥 \in 𝑋,$ the element $𝑓 (𝑥)$ of $𝑌$ is the image of $𝑥$ under $𝑓 .$ For a subset $𝐴 \subseteq 𝑋,$ the subset of $𝑌$ defined as

𝑓 (𝐴) = {𝑓 (𝑎) : 𝑎 \in 𝐴}

is the image of the set $𝐴$ under $𝑓 .$ For a subset $𝐵 \subseteq 𝑌,$ the subset of $𝑋$ defined as

𝑓^{- 1} (𝐵) = {𝑥 \in 𝑋 : 𝑓 (𝑥) \in 𝐵}

is called the preimage of the set $𝐵$ under $𝑓 .$

We are now going to define continuity, which does require a topology on both

𝑋

and

𝑌 .

Attention: we say “

𝑋

is a topological space” to mean that some topology

𝒯︁

𝑋

is chosen, and open sets are understood to be members of

𝒯︁ .

Definition: continuous function

Let $𝑋$ and $𝑌$ be topological spaces. A function $𝑓 : 𝑋 \to 𝑌$ is continuous if for every subset $𝑉$ of $𝑌,$ open in $𝑌,$ the preimage $𝑓^{- 1} (𝑉)$ is open in $𝑋 .$

Remark: in MATH21111 Metric Spaces, this was shown to be an equivalent definition of continuity.

This means that when

𝑋

and

𝑌

are metric spaces, considered with their metric topologies, we can use two equivalent definitions of a continuous function:

When

𝑋

𝑌

is not a metric space, we do not have the

𝜀

𝛿

definition, and can only use the topological definition.

It is not true for a general continuous function that images of open subsets of

𝑋

are open in

𝑌

Proposition: identity functions and constant functions are continuous

Let $𝑋, 𝑌$ be topological spaces. The identity map on $𝑋$ , ${i d}_{𝑋} : 𝑋 \to 𝑋$ defined by $𝑥 \mapsto 𝑥$ for all points $𝑥 \in 𝑋,$ is continuous.

Let $𝑦_{0}$ be a point of $𝑌 .$ The constant function ${c o n s t}_{𝑦_{0}} : 𝑋 \to 𝑌,$ $𝑥 \mapsto 𝑦_{0}$ for all $𝑥 \in 𝑋$ (i.e., the function that sends the whole of $𝑋$ to one point), is continuous.

Proof. A proof is given in the Tutorial. □

Proposition: composition of continuous functions is continuous

Suppose $𝑋 \overset{𝑓}{\to} 𝑌 \overset{𝑔}{\to} 𝑍$ where $𝑋, 𝑌, 𝑍$ are topological spaces, and the functions $𝑓, 𝑔$ are continuous. Then the composition $𝑋 \overset{𝑔 \circ 𝑓}{⟶} 𝑍$ is also continuous.

Proof. The key point of the proof is the formula for the preimage under composition, left as an exercise to the student:

\begin{equation*} \forall W \subseteq Z, \quad (g \circ f)^{-1}(W)= f^{- 1}(g^{- 1}(W)). \end{equation*}

Let

𝑊 \subseteq 𝑍

be any open subset of

𝑍 .

Since

𝑔

is continuous,

𝑔^{- 1} (𝑊)

is open in

𝑌 .

Since

𝑓

is continuous,

𝑓^{- 1} (𝑔^{- 1} (𝑊))

is open in

𝑋 .

We have proved that

(𝑔 \circ 𝑓)^{- 1} (𝑊)

is open in

𝑋,

and so we have verified the definition of “continuous” for

𝑔 \circ 𝑓 .

□

Closed sets

It is often useful to characterise continuous functions in terms of closed sets, which we will now define.

This definition means that every subset

𝐴

of a topological space

𝑋

falls into one of the four classes:

The collection of closed sets in

𝑋

has properties which mirror, but do not repeat, the properties of open sets – we need to exchange unions and intersections:

Proposition: properties of closed sets

If $𝑋$ is a topological space, then

$\varnothing$ and $𝑋$ are closed in $𝑋,$
arbitrary intersections of closed sets are closed,
finite unions of closed sets are closed.

Proof. Since closed sets are complements of open sets, these follow by applying the De Morgan laws to the properties of open sets. □

Proposition: closed set criterion of continuity

For a function $𝑓 : 𝑋 \to 𝑌$ between topological spaces $𝑋$ and $𝑌,$ the following are equivalent:

$𝑓$ is continuous;
the preimage of every closed subset of $𝑌$ is closed in $𝑋 .$

Proof. Implication 1. $\Rightarrow$ 2.: assume that $𝑓 : 𝑋 \to 𝑌$ is continuous. Let $𝐹 \subseteq 𝑌$ be any closed set in $𝑌 .$ Then $𝑉 = 𝑌 \ 𝐹$ is open in $𝑌 .$ Since preimage respects complements (see Tutorial), $𝑓^{- 1} (𝐹) = 𝑓^{- 1} (𝑌 \ 𝑉) = 𝑋 \ 𝑓^{- 1} (𝑉) .$ By continuity of $𝑓,$ $𝑓^{- 1} (𝑉)$ is open, so by definition of “closed”, $𝑋 \ 𝑓^{- 1} (𝑉)$ is closed. We proved that the preimage of every closed set is closed.

Reverse the logic to show 2. $\Rightarrow$ 1. □

Homeomorphisms and homeomorphic spaces

Definition: homeomorphism, homeomorphic spaces

Let $𝑋$ and $𝑌$ be topological spaces. A function $𝑓 : 𝑋 \to 𝑌$ is a homeomorphism if

$𝑓$ is a bijection,
$𝑓$ is continuous,
the inverse, $𝑓^{- 1} : 𝑌 \to 𝑋,$ of $𝑓,$ is continuous.

Two topological spaces $𝑋, 𝑌$ are homeomorphic if there exists a homeomorphism $𝑓 : 𝑋 \to 𝑌 .$

We will write

𝑋 \tilde{⟶} 𝑌

to indicate that there is a homeomorphism between

𝑋

and

𝑌,

𝑋 ⟶_{𝑓}^{\sim} 𝑌

to indicate that

𝑓 : 𝑋 \to 𝑌

is a homeomorphism.

In mathematics, the symbol

\sim

is typically used to denote an equivalence relation. This is also the case here:

Proof. We need to check the three axioms of equivalence relation.

Reflexive: we need to show that every space $𝑋$ is homeomorphic to itself. Indeed, ${id}_{𝑋} : 𝑋 \tilde{⟶} 𝑋$ is a homeomorphism because the identity map is a bijection, the identity map is continuous as shown earlier, and ${id}_{𝑋}^{- 1} = {id}_{𝑋}$ hence the inverse is also continuous.

Symmetric: We need to show that if $𝑋 \tilde{⟶} 𝑌,$ then $𝑌 \tilde{⟶} 𝑋 .$ Assume that $𝑓 : 𝑋 \tilde{⟶} 𝑌 .$ Since $𝑓$ is a homeomorphism, the inverse bijection $𝑓^{- 1}$ is, by definition, continuous. Furthermore, the inverse of $𝑓^{- 1}$ is $𝑓$ which is continuous. We conclude that $𝑓^{- 1} : 𝑌 \tilde{⟶} 𝑋$ is a homeomorphism.

Transitive: we need to show that if $𝑋 \tilde{⟶} 𝑌$ and $𝑌 \tilde{⟶} 𝑍,$ then $𝑋 \tilde{⟶} 𝑍 .$ Assume that $𝑋 ⟶_{𝑓}^{\sim} 𝑌$ and $𝑌 ⟶_{𝑔}^{\sim} 𝑍$ are homeomorphisms. Then $𝑋 ⟶_{𝑔 \circ 𝑓}^{\sim} 𝑍$ is a homeomorphism: it is a bijection as it has inverse $(𝑔 \circ 𝑓)^{- 1} = 𝑓^{- 1} \circ 𝑔^{- 1},$ and both $𝑔 \circ 𝑓$ and $𝑓^{- 1} \circ 𝑔^{- 1}$ are continuous as compositions of continuous maps. □

Let us now consider examples of homeomorphic spaces. First, introduce a topological space which appears in many examples.

Definition: circle, $𝑆^{1}$

The circle is the set

𝑆^{1} = {(𝑥, 𝑦) \in ℝ^{2} : 𝑥^{2} + 𝑦^{2} = 1}

viewed with the topology defined by the Euclidean metric in $ℝ^{2} .$

Example: some spaces homeomorphic to $ℝ$

Show that the Euclidean line $ℝ$ is homeomorphic to the following spaces, viewed with the Euclidean metric:

the open interval $(- 𝜋 / 2, 𝜋 / 2)$ (a subset of $ℝ$ );
an open half-line $(0, + \infty)$ ;
the right half-circle without the endpoints: $𝑆_{+}^{1} = {(𝑥, 𝑦) \in 𝑆^{1} : 𝑥 > 0} .$

Solution. We construct pairs of continuous functions which are mutual inverses between some pairs of the given spaces, see Figure 2

We do not need to construct homeomorphisms between each pair of the given spaces: since “homeomorphic” is an equivalence relation, we have constructed enough to show that all four spaces are homeomorphic to each other.

The homeomorphism between the right half-circle of the unit circle and

ℝ

can be described in purely geometric terms as in Figure 3: take a point of the half-circle and project it onto the vertical tangent line to the half-circle at the point

(1, 0)

in the plane.

The homeomorphisms shown above are not unique, and there are many other ways of showing that these four spaces are pairwise homeomorphic.

Remark. The four spaces shown above are quite different as metric spaces: for example,

Hence there is no way these metric spaces can be isometric (i.e., equivalent as metric spaces). We thus observe that being homeomorphic (i.e., equivalent as topological spaces) is a weaker requirement, for metric spaces, than being isometric.

Further examples and applications of homeomorphisms

Knot theory is a branch of mathematics which arose from an attempt to classify knots rigorously by Peter Guthrie Tait (1831–1901). In 1920s, following the work of J. Alexander, knot theory became part of Topology. Since 1980s, knot theory has been applied to the study of DNA and other molecules, and to the study of fluid vortices and has found further applications in Physics.

A knot is a smooth injective function

𝑆^{1} \to ℝ^{3}

where

𝑆^{1}

is the unit circle. The image of the function is a smooth closed curve in

ℝ^{3}

which has no self-intersections; such curves are themselves called knots. Two knots

𝐾_{1},

𝐾_{2}

are equivalent, or isotopic, if there is a continuous deformation of the space

ℝ^{3}

which transforms

𝐾_{1}

into

𝐾_{2} .

The main problem of knot theory is to classify all knots up to isotopy. For example, a basic result of knot theory is that a trefoil knot, the curve shown in Figure 4, is not isotopic to the unknot (a straightforward copy of the circle in

ℝ^{3}

Is the problem of deciding whether two knots are isotopic related to the notion of homeomorphism? After all, every knot is homeomorphic to a circle, isn’t it? Yes, but complements of knots are not homeomorphic. A highly non-obvious result is

Theorem: Gordon-Luecke theorem (1989)

Two knots $𝐾_{1},$ $𝐾_{2}$ in $ℝ^{3}$ are isotopic (up to taking a mirror image) if, and only if, there exists a homeomorphism $ℝ^{3} \ 𝐾_{1} \tilde{⟶} ℝ^{3} \ 𝐾_{2} .$

The theory of knots extends to links which are unions of disjoint knots in

ℝ^{3} .

An example of a non-trivial link is the famous Borromean rings: three interlinked circles in

ℝ^{3},

such that if any one circle is removed, the remaining two circles become unlinked, see Figure 5. Unlike knots, a link is not determined up to isotopy by a homeomorphism class of its complement.

Continuous functions and closed sets

Definition and basic properties of continuous functions

Closed sets

Homeomorphisms and homeomorphic spaces

Further examples and applications of homeomorphisms