# Topology/Continuity and Homeomorphisms
## Continuity

Continuity is the central concept of topology. Essentially, topological spaces have the minimum necessary structure to allow a definition of continuity. Continuity in almost any other context can be reduced to this definition by an appropriate choice of topology.

### Definition

Let $X,Y$ be topological spaces.
A function $f:X\to Y$ is continuous at $x\in X$ if and only if for all open neighborhoods $B$ of $f(x)$, there is a neighborhood $A$ of $x$ such that $A\subseteq f^{-1}(B)$. A function $f:X\to Y$ is continuous in a set $S$ if and only if it is continuous at all points in $S$.
The function $f:X\to Y$ is said to be continuous over $X$ if and only if it is continuous at all points in its domain.
$f:X\to Y$ is continuous if and only if for all open sets $B$ in $Y$, its inverse $f^{-1}(B)$ is also an open set. Proof: ($\Rightarrow$) The function $f:X\to Y$ is continuous. Let $B$ be an open set in $Y$. Because it is continuous, for all $x$ in $f^{-1}(B)$, there is a neighborhood $x\in A\subseteq f^{-1}(B)$, since B is an open neighborhood of f(x). That implies that $f^{-1}(B)$ is open. ($\Leftarrow$) The inverse image of any open set under a function $f$ in $Y$ is also open in $X$. Let $x$ be any element of $X$. Then the inverse image of any neighborhood $B$ of $f(x)$, $f^{-1}(B)$, would also be open. Thus, there is an open neighborhood $A$ of $x$ contained in $f^{-1}(B)$. Thus, the function is continuous.
If two functions are continuous, then their composite function is continuous. This is because if $f$ and $g$ have inverses which carry open sets to open sets, then the inverse $g^{-1}(f^{-1}(x))$ would also carry open sets to open sets.

### Examples

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

## Homeomorphism

When a homeomorphism exists between two topological spaces, then they are "essentially the same", topologically speaking.

### Definition

Let $X,Y$ be topological spaces A function $f:X\to Y$is said to be a homeomorphism if and only if
(i) $f$ is a bijection (ii) $f$ is continuous over $X$ (iii)$f^{-1}$ is continuous over $Y$
If a homeomorphism exists between two spaces, the spaces are said to be homeomorphic
If a property of a space $X$ applies to all homeomorphic spaces to $X$, it is called a topological property.

### Notes

* A map may be bijective and continuous, but not a homeomorphism. Consider the bijective map $f:[0,1)\rightarrow S^{1}$, where $f(x)=e^{2\pi ix}$ 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 $S^{1}$, like the set $\left[0,{\frac {1}{2}}\right)$.
* Homeomorphism is an equivalence relation

## Exercises

* Prove that the open interval $(a,b)$ is homeomorphic to $\mathbb {R}$.
* Establish the fact that a Homeomorphism is an equivalence relation over topological spaces.
* (i)Construct a bijection $f:[0,1]\to [0,1]^{2}$ (ii)Determine whether this $f$ is a homeomorphism.