Example of bijective continuous function that is not homeomorphism

Question

In the book Topology by Munkres, there is an example of a bijective continuous function that is not a homeomorphism.

I don't think that I completely understood the explanation. I understand that the function is a bijection and continuous.

---

Firstly, what is the topology specified on $[0, 1)$ so that $[0, \frac{1}{4})$ is open? And why does $f(0) = (1, 0)$ not lie in any open set $V$ such that $V \cap S^{1} \subset f(U)$? Is it similar to the reason that every open set containing $(6, 7]$ in standard topology on $\mathbb R$ contains some point greater than $7$ also?

---

Answer 1

Here is a simple example.
The identity map from the reals given the discrete topology, to the reals with the usual topology generated by open intervals.

Answer 2

The function $\varphi\colon\mathbb{R}\to S^1$ defined by $\varphi(t)=\bigl(\cos(2\pi t),\sin(2\pi t)\bigr)$ is continuous and surjective.

Therefore its restriction $f$ to the interval $[0,1)$ is continuous as well. It is also injective and surjective.

It cannot be a homeomorphism, because $C=S^1\setminus\{(-1,0)\}$ is connected, being equal to $\varphi\bigl((-1/2,1/2)\bigr)$, but $f^{-1}(C)=[0,1)\setminus\{1/2\}$ is not connected.

The proof in the book uses a different method. Here $U=[0,1/4)$ is open in $[0,1)$, because it is equal to $(-1/4,1/4)\cap[0,1/4)$.

However, $f\bigl([0,1/4)\bigr)$ is not open, because any open disk around $(1,0)$ contains points of $S^1$ that are not in $f\bigl([0,1/4)\bigr)$.

Answer 3

The topology on $[0,1)$ is a subspace topology on $\mathbb{R}$ whereas that on $S^1$ is a subspace topology on $ℝ^2$. Since $f$ is continuous, the inverse image of an open set in $S^1$ is an open set in $[0,1)$.

Again, $(− 1,1/4)$ is open in $ℝ$ and $[0,1/4) = [0,1) ∩ (− 1,1/4)$. So, $[0,1/4)$ is open in $[0,1)$. Any open disk $V$ of $\mathbb{R}^2$ around the point $f(0)$ contains points of $S^1$ that are not in $f([0,1/4))$. Thus, around the point $f(0)$ there is no open disk $V$ of $\mathbb{R}^2$ such that $V\cap f([0,1/4))\subset S^1$.

So, $f([0,1/4))$ is not open in $S^1$. That is, the image under $f$ of $[0,1/4)$ in the domain is not open in $S^1$.

Thus, $f^{-1}$ is not a continuous mapping.

Hence $f$ is not a homeomorphism.