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 ${\displaystyle X.}$

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 ${\displaystyle {\mathcal {T}}_{1}.}$

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.

Next Lesson