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

1. A map may be bijective and continuous, but not a homeomorphism. Consider the bijective map $f:[0,1)\to S^1$, 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 $[0,\frac{1}{2})$.
   
2. Homeomorphism is an equivalence relation

Exercises

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