# Example of bijective continuous function that is not homeomorphism

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?

• The topology on $[0,1)$ is the induced topology from $\mathbb{R}$. A subset $U$ of $[0,1)$ is open if and only if there is an open subset $\mathcal{O}$ of $\mathbb{R}$ such that $U=\mathcal{O}\cap[0,1)$. Yes: any open subset of $S^1$ that contains $(1,0)$ will necessarily include points corresponding to small negative angles. (Alternatively: you can disconnect $[0,1)$ by removing a single point, but if you remove a single point from $[0,1)$, the result is never disconnected, so the two cannot be homeomorphic). Feb 4 '19 at 19:21
• @ArturoMagidin Thanks for the explanation and the alternative approach. Now I get it Feb 4 '19 at 19:23
• (that should be "if you remove a single point from $\mathbf{S}^1$, the result is never disconnected"...) Feb 4 '19 at 19:30
• @ArturoMagidin oh yeah, I got the gist of it Feb 4 '19 at 19:31

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 $$f\bigl([0,1/4)\bigr)$$.

• +1 @egreg sir.. Aug 15 '19 at 13:12

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.

The topology on [0,1) is a subspace topology on ℝ whereas that on S¹ is a subspace topology on ℝ². Since f is continuous, the inverse image of an open set in S¹ is an open set in [0,1). Again, (− 1,¼) is open in ℝ and [0,¼) = [0,1) ∩ (− 1,¼). So, [0,¼) is open in [0,1). Any open disk around the point (1,0) contains points of S¹ that are not in f([0,1/4)). So, f([0,1/4)) is not open in S¹. Thus, f⁻¹ doesn't preserve the openness so that it is not a continuous mapping. Hence f is not a homeomorphism.