# Topology/Lesson 2

So in the first lesson we learned what a topology is, what open sets, closed sets, and bases are. You should be comfortable with these concepts before beginning this second lesson.

## Continuous Function

We will define the notion of a continuous function below. (Note that in topological texts and papers it is common to use the word 'map' and even the word 'function' to mean a continuous function. To avoid ambiguity, in this course we will reserve the word 'function' to mean any function, but will use 'map' to mean a continuous function.)

### Definition

Let ${\displaystyle (X,{\mathcal {T}}_{X})}$ and ${\displaystyle (Y,{\mathcal {T}}_{Y})}$ be topological spaces. A function ${\displaystyle f:X\to Y}$ is called continuous if for every ${\displaystyle U\in {\mathcal {T}}_{Y},}$ we have ${\displaystyle f^{-1}(U)\in {\mathcal {T}}_{X}.}$ That is, f is continuous iff the f-preimage of every open set (in ${\displaystyle Y}$) is open (in ${\displaystyle X}$).

### Examples

It is immediate from the definition that the following two types of functions are always continuous. The proof of these two claims is left as an exercise.

1. If ${\displaystyle X}$ is a discrete space and ${\displaystyle Y}$ is any space, then any function ${\displaystyle f:X\to Y}$ is continuous.
2. If ${\displaystyle X}$ is any space and ${\displaystyle Y}$ has the indiscrete topology, then any function ${\displaystyle f:X\to Y}$ is continuous.

### Continuous at a point

It is also possible to talk about a function being continuous at a point of its domain. So, given a map ${\displaystyle f:X\to Y}$ and a point ${\displaystyle x_{0}\in X}$ we say that ${\displaystyle f}$ is continuous at ${\displaystyle x_{0}}$ if given any neighborhood ${\displaystyle V\subset Y}$ of ${\displaystyle f(x_{0})}$ there is a neighborhood ${\displaystyle U\subset X}$ of ${\displaystyle x_{0}}$ such that ${\displaystyle f(U)\subset V.}$

### Exercises

Let ${\displaystyle f:X\to Y}$ be a function. Show that the following are equivalent.

1. ${\displaystyle f}$ is continuous.
2. ${\displaystyle f}$ is continuous at ${\displaystyle x_{0}}$ for all ${\displaystyle x_{0}\in X.}$
3. For any closed set ${\displaystyle A\subset Y,}$ we have ${\displaystyle f^{-1}(A)\subset X}$ is closed in ${\displaystyle X.}$
4. If ${\displaystyle {\mathcal {B}}}$ is a basis for ${\displaystyle Y}$ then for any set ${\displaystyle B\in {\mathcal {B}}}$ we have ${\displaystyle f^{-1}(B)}$ is open in ${\displaystyle X.}$

## Open maps

The definition of a continuous map may seem awkward. Since it is a morphism in the category of topological spaces, one would expect it to preserve some property about open sets, but what one might first think is that open sets are preserved under the map of the function. But this gives a different concept.

### Definition (open map)

Let ${\displaystyle f:X\to Y}$ be a continuous function. Then we say that ${\displaystyle f}$ is an open map if for any open set ${\displaystyle U\subset X}$ we have ${\displaystyle f(U)}$ is open in ${\displaystyle Y.}$

Merely for the purposes of the discussion here, define an open function to be a function (not necessarily continuous) ${\displaystyle f:X\to Y}$ such that ${\displaystyle f(U)}$ is open in ${\displaystyle Y}$ whenever ${\displaystyle U}$ is open in ${\displaystyle X.}$

### Exercises

1. Construct finite-point spaces ${\displaystyle X}$ and ${\displaystyle Y}$ and a map ${\displaystyle f:X\to Y}$ that is continuous but not open.
2. Construct another function ${\displaystyle f:X\to Y}$ that is not continuous but is an 'open function'.
3. Show that the identity map ${\displaystyle id:X\to X}$ is always continuous and open.
4. Suppose that ${\displaystyle {\mathcal {T}}_{1}}$ and ${\displaystyle {\mathcal {T}}_{2}}$ are two distinct topologies on the set ${\displaystyle X.}$ Suppose that the identity map ${\displaystyle id:(X,{\mathcal {T}}_{1})\to (X,{\mathcal {T}}_{2})}$ is continuous. Show that ${\displaystyle {\mathcal {T}}_{2}\subset {\mathcal {T}}_{1}.}$ In this case, we say that ${\displaystyle {\mathcal {T}}_{1}}$ is finer than ${\displaystyle {\mathcal {T}}_{2}}$ or that ${\displaystyle {\mathcal {T}}_{2}}$ is coarser than ${\displaystyle {\mathcal {T}}_{1}.}$

## Homeomorphism

The "isomorphism" or "equivalence" of topological spaces is called "homeomorphism." This is analogous to bijection in the case of sets and group isomorphism in the case of groups. Topologically speaking, two spaces are indistinguishable if they are homeomorphic.

### Definition

Let ${\displaystyle f:X\to Y.}$ Then we say that ${\displaystyle f}$ is a homeomorphism if it is bijective and both ${\displaystyle f}$ and ${\displaystyle f^{-1}}$ are continuous. In this case we say that ${\displaystyle X}$ and ${\displaystyle Y}$ are homeomorphic and sometimes write ${\displaystyle X\cong Y}$ or ${\displaystyle X\simeq Y.}$

### Examples

1. If ${\displaystyle X}$ is any topological space then the identity map ${\displaystyle id_{X}:X\to X}$ is a homeomorphism.
2. If ${\displaystyle f:X\to Y}$ is injective and ${\displaystyle f}$ and ${\displaystyle f^{-1}}$ are both continuous then ${\displaystyle f}$ is called an embedding. In this case the map ${\displaystyle g:X\to f(X)}$ given by ${\displaystyle g(x)=f(x)}$ is a homeomorphism. That is, an embedding is a homeomorphism with its image in the target space.
3. The map ${\displaystyle f:(-1,1)\to \mathbb {R} }$ given by ${\displaystyle f(x)=\tan \left({\frac {\pi }{2}}x\right)}$ is a homeomorphism.

### Exercises

1. Show that the maps given in the examples are indeed homeomorphisms.
2. Show that if ${\displaystyle f:X\to Y}$ is a homeomorphism then ${\displaystyle f^{-1}:Y\to X}$ is also a homeomorphism.
3. Show that if ${\displaystyle X}$ is homeomorphic to ${\displaystyle Y}$ and ${\displaystyle Y}$ is homeomorphic to ${\displaystyle Z}$ then ${\displaystyle X}$ is homeomorphic to ${\displaystyle Z.}$ Note that these first 3 exercises show that homeomorphism is an equivalence relation.