mathamatical structure where a bijection is not an isomorphism

Question

During a lecture in module theory, my lecturer mentioned that not in all mathematical structures have the property where a function is an isomorphism if and only if it is a bijection, although this is true in set and module theory.

No example was given but it was mentioned that functions between topological spaces require an isomorphism to be a bijection and continuous and being merely a bijection is not enough.

I am looking for an example where this is true, that is; a bijection function that is not an isomorphism. And do these bijection not isomorphisic functions have a name?

Answer 1

The statement is that an isomorphism of two topological spaces $X$ and $Y$ (a homeomorphism) is a bijective continuous map $f:X\to Y$ such that $f^{-1}$ is also continuous. Now, when you're dealing with algebraic homomorphisms, their inverses are automatically homomorphisms, there is no reason why this should hold for general maps with structure. Hence, in a sense, the algebraic isomorphisms are the weird ones.

To see that inverses of continuous maps are not necessarily continuous, let $X=[0,2\pi)$ and $Y=S^1$, the circle. Then, $f(\theta):= \exp(i\theta)$ is a bijective continuous map from $X$ to $Y$, but its inverse is clearly not continuous since $f^{-1}(1)=0,$ but $\lim_{\theta\to 2\pi^-} f^{-1}(\exp(i\theta))=2\pi$ in $\mathbb{R}$ and hence, doesn't even exist in $X$.

The other thing that's typically done is take a set $X$ and let $\tau$ and $\tau'$ be two topologies on $X$ such that $\tau'$ is strictly stronger that $\tau$. Then, the identity map $id: (X,\tau')\to(X,\tau)$ is continuous and, by definition, bijective, but the inverse map is not continuous as can be seen by considering $U\in \tau'\setminus \tau$.

Answer 2

This is a very easy example:

Consider the real numbers $\mathbb{R}$ with the discrete topology $\tau_1$ (every set is open) and the topology $\tau_2=\{\emptyset,\mathbb{R}\}$. Then the identity mapping is a bijection, but not continuous. Therefore it is not an isomorphism between topological spaces.

Answer 3

A simple example of a bijection of $\mathbb R$ to $\mathbb R$ which is not an isomorphism of topological spaces (a "homeomorphism") because it is not continuous:

$$f(x)=\begin{cases}x+\frac{1}{2}&\{x\}<\frac{1}{2}\\x-\frac{1}{2}&\{x\}\ge\frac{1}{2}\end{cases}$$

where $\{x\}$ is the fractional part of $x$. The idea is that it maps $[n,n+\frac{1}{2})$ to $[n+\frac{1}{2},n)$ by adding $\frac{1}{2}$ and maps $[n+\frac{1}{2},n+1)$ to $[n,n+\frac{1}{2})$ by subtracting $\frac{1}{2}$, for all $n\in\mathbb Z$.

Answer 4

Examples are everywhere:

Translations (e.g. $x\mapsto x+1$) are bijections in fields which are not isomorphisms of fields

The map $f:\Bbb R\rightarrow\Bbb R$ defined by $$ f(x)=\left\{ \begin{array}{ll} x & \text{if $x<0$ or $x>1$}\\ 1-x & \text{if $0\leq x\leq1$} \end{array}\right. $$ is a bijection which is not a homeomorphism for the natural topology.