Counter-examples of homeomorphism

Question

Briefly speaking, we know that a map $f$ between $2$ topological spaces is homeomorphic if $f$ is a bijection and the inverse of $f$ and itself are both continuous.

So, can anyone give me $2$ counter examples(preferably simple ones) of non-homeomorphic maps $f$ between 2 topological spaces that satisfy the properties I give? (Only one of them is satisfied and 3 examples for each property.)

1. $f$ is bijective and continuous, but its inverse is not continuous.
2. $f$ is bijective and the inverse is continuous, but $f$ itself is not continuous.

In addition, can we think about some examples of topologies that are path-connected?

I will understand the concept of homeomorphism much better if I know some simple counterexamples. I hope you can help me out. Thanks!

Answer 1

$1.$ Let $X$ be the set of real numbers, with the discrete topology, and let $Y$ be the reals, with the ordinary topology. Let $f(x)=x$. Then $f$ is continuous, since every subset of $X$ is open. But $f^{-1}$ is not continuous.

$2.$ In doing $1$, we have basically done $2$.

Answer 2

Consider $f:[0,2\pi)\to S^1$ given by $t\mapsto\langle\cos t,\sin t\rangle,$ where $S^1$ is the unit circle in the plane, and $[0,2\pi)$ is the real interval, both considered under their appropriate subspace topology as subsets of Euclidean spaces. Then $f$ is bijective and continuous, but its inverse is not continuous, providing an example for 1. Thus, its inverse is an example for 2.

Be careful with 3, as you need to specify what you mean by "inverse" in cases where $f$ isn't bijective. An equivalent way to define homeomorphism is as a bijective, continuous, open map (maps open sets to open sets). This avoids the need to worry about inverse functions--indeed, open maps need not be injective. For an example of a surjective, continuous, open map that is not a homeomorphism (since it is not injective), consider $p:\Bbb R^2\to\Bbb R$ given by $\langle x,y\rangle\mapsto x$, where $\Bbb R^2$ and $\Bbb R$ are in their usual topologies. For an example of an injective, continuous, open map that is not a homeomorphism (since it is not surjective), consider the inclusion $(0,1)\hookrightarrow[0,1].$

Answer 3

1 and 2 are really the same thing. Also, what do you mean by the inverse of a function if it is not a bijection (your question 3)? The inverse of a function is not defined for something that is not bijective.

For an example of 1 take $X = \mathbb{Q}$ with the discrete topology and $Y = \mathbb{Q}$ with the usual topology inherited from $\mathbb{R}$. Let $Id: X \rightarrow Y$ be the identity map on the level of sets. This map is clearly continuous and a bijection. But, it is not open hence the inverse is not continuous.

Answer 4

$f:[0,2\pi)\to S^1$ given by $f(t)=(\cos(t),\sin(t))$ is continuous and bijective but its inverse is not continuous.

The inverse of $f$ is an example of a bijection $f^{-1}:S^1\to [0,2\pi)$ that is not continuous, but its inverse is continuous.

It is very illustrative to think geometrically about this example. It just involves gluing the interval $[0,2\pi)$ to get the circle.

If $f$ has an inverse then $f$ is a bijection.

Answer 5

Let $(X,\mathcal{O_1})$ and $(X,\mathcal{O_2})$ be topological spaces such that $\mathcal{O_1}\supset\mathcal{O_2}$ and $\mathcal{O_1}\neq\mathcal{O_2}$.
Let $f:(X,\mathcal{O_1})\to (X,\mathcal{O_2})$ be a mapping such that $f(x)=x$ for all $x\in X$.
Then, $f$ is bijection.
Then, for any $O\in\mathcal{O_2}$, $f^{-1}(O)=O\in\mathcal{O_1}$ because $\mathcal{O_1}\supset\mathcal{O_2}$.
So, $f$ is continuous.
Since $\mathcal{O_1}\supset\mathcal{O_2}$ and $\mathcal{O_1}\neq\mathcal{O_2}$, there exists a set $O_3\in\mathcal{O_1}$ and $O_3\notin\mathcal{O_2}$.
Then, $O_3\in\mathcal{O_1}$ but $(f^{-1})^{-1}(O_3)=O_3\notin\mathcal{O_2}$.
So, $f^{-1}$ is not continuous.

Example:
Let $X=\{1,2\}$.
Let $\mathcal{O_1}=\{\emptyset, X, \{1\}\}$.
Let $\mathcal{O_2}=\{\emptyset, X\}$.
Then, $(X,\mathcal{O_1})$ and $(X,\mathcal{O_2})$ are topological spaces.
And, $\mathcal{O_1}\supset\mathcal{O_2}$ and $\mathcal{O_1}\neq\mathcal{O_2}$.