Definition

Let ${\displaystyle X,Y}$ be topological spaces.

A function ${\displaystyle f:X\to Y}$ is continuous at ${\displaystyle x\in X}$ if and only if for all open neighborhoods ${\displaystyle B}$ of ${\displaystyle f(x)}$, there is a neighborhood ${\displaystyle A}$ of ${\displaystyle x}$ such that ${\displaystyle A\subseteq f^{-1}(B)}$.
A function ${\displaystyle f:X\to Y}$ is continuous in a set ${\displaystyle S}$ if and only if it is continuous at all points in ${\displaystyle S}$.

The function ${\displaystyle f:X\to Y}$ is said to be continuous over ${\displaystyle X}$ if and only if it is continuous at all points in its domain.

${\displaystyle f:X\to Y}$ is continuous if and only if for all open sets ${\displaystyle B}$ in ${\displaystyle Y}$, its inverse ${\displaystyle f^{-1}(B)}$ is also an open set.
Proof:
(${\displaystyle \Rightarrow }$)
The function ${\displaystyle f:X\to Y}$ is continuous. Let ${\displaystyle B}$ be an open set in ${\displaystyle Y}$. Because it is continuous, for all ${\displaystyle x}$ in ${\displaystyle f^{-1}(B)}$, there is a neighborhood ${\displaystyle x\in A\subseteq f^{-1}(B)}$, since B is an open neighborhood of f(x). That implies that ${\displaystyle f^{-1}(B)}$ is open.
(${\displaystyle \Leftarrow }$)
The inverse image of any open set under a function ${\displaystyle f}$ in ${\displaystyle Y}$ is also open in ${\displaystyle X}$. Let ${\displaystyle x}$ be any element of ${\displaystyle X}$. Then the inverse image of any neighborhood ${\displaystyle B}$ of ${\displaystyle f(x)}$, ${\displaystyle f^{-1}(B)}$, would also be open. Thus, there is an open neighborhood ${\displaystyle A}$ of ${\displaystyle x}$ contained in ${\displaystyle f^{-1}(B)}$. Thus, the function is continuous.

If two functions are continuous, then their composite function is continuous. This is because if ${\displaystyle f}$ and ${\displaystyle g}$ have inverses which carry open sets to open sets, then the inverse ${\displaystyle g^{-1}(f^{-1}(x))}$ would also carry open sets to open sets.

Examples

• Let ${\displaystyle X}$ have the discrete topology. Then the map ${\displaystyle f:X\rightarrow Y}$ is continuous for any topology on ${\displaystyle Y}$.
• Let ${\displaystyle X}$ have the trivial topology. Then a constant map ${\displaystyle g:X\rightarrow Y}$ is continuous for any topology on ${\displaystyle Y}$.

Definition

Let ${\displaystyle X,Y}$ be topological spaces
A function ${\displaystyle f:X\to Y}$is said to be a homeomorphism if and only if

(i) ${\displaystyle f}$ is a bijection
(ii) ${\displaystyle f}$ is continuous over ${\displaystyle X}$
(iii)${\displaystyle f^{-1}}$ is continuous over ${\displaystyle Y}$

If a homeomorphism exists between two spaces, the spaces are said to be homeomorphic

If a property of a space ${\displaystyle X}$ applies to all homeomorphic spaces to ${\displaystyle X}$, it is called a topological property.

Notes

1. A map may be bijective and continuous, but not a homeomorphism. Consider the bijective map ${\displaystyle f:[0,1)\rightarrow S^{1}}$, where ${\displaystyle f(x)=e^{2\pi ix}}$ mapping the points in the domain onto the unit circle in the plane. This is not a homeomorphism, because there exist open sets in the domain that are not open in ${\displaystyle S^{1}}$, like the set ${\displaystyle \left[0,{\frac {1}{2}}\right)}$.
   
2. Homeomorphism is an equivalence relation