I am currently working through John M. Lee's book on 'Introduction to Topological Manifolds' and am looking at the following question:
Exercise 2.28. Let $X$ be the half-open interval $[0,1) \subseteq \mathbb{R}$, and let $\mathbb{S}^{1}$ be the unit circle in $\mathbb{C}$ (both with their Euclidean metric topologies, as usual). Define a map $a: \boldsymbol{X} \rightarrow \mathbb{S}^{1}$ by $a(s)=e^{2 \pi i s}=\cos 2 \pi s+i \sin 2 \pi s$. Show that $a$ is continuous and bijective but not a homeomorphism.
I understand that $a$ is continuous and bijective, with an inverse $a^{-1}:\mathbb{S}^{1}\rightarrow X$ given by $a^{-1}(k)=\text{arg}(k)/2\pi$ where $\text{arg}$ is the complex argument.
It seems to me that the last part of the question wants me to say something like this:
Let $a,b \in (0,0.5)$. Define $I_{a,b}$ be the set:
$$I_{a,b}=\{k\in\mathbb{S}^{1}: k = e^{2 \pi i s} \text{ for some } s \in [-a,b]\}$$
The function $a^{-1}$ maps $I_{a,b}$ to $J_{a,b}=[0,b) \cup (1-a,1)$. $J_{a,b}$ is not open in $X$. Therefore, as $I_{a,b}$ is open but it's image under $a^{-1}$ is not open, it follows that $a^{-1}$ cannot be continuous. Consequently, $a$ cannot be a homeomorphism, by the definition of homeomorphism.
Firstly, is the above answer correct? What makes me doubt it is that I am not convinced that $J_{a,b}$ is not open in $X$. Earlier in the book (in the prerequisite Appendix $B$), the following definitions of open were given for open sets in the Euclidean Metric topology. For any set $M$:
- For any $x \in M$ and $r>0$, the (open) ball of radius $r$ around $x$ is the set
$$B_{r}(x)=\{y \in M: d(y, x)<r\}.$$- A subset $A \subseteq M$ is said to be an open subset of $M$ if it contains an open ball around each of its points.
From this, it follows that (noting $M=X$ in this instance);
$J_{a,b}=B_b\{0\}\cup B_{\frac{a}{2}}(1-\frac{a}{2}).$
As $J_{a,b}$ is a union of open sets in $X$, it therefore follows that $J_{a,b}$ is an open set in $X$.
I am very confused; Is my answer above wrong, or have I misunderstood the definition of 'open' in this context?
Best Answer
I think your having the right intuition, but some of your defintions are not correct.
There are a couple of things going on here. Firstly, your definition of an open subset is not correct. It should be $A\subseteq M$ is open if and only if for all $x\in A$ there exists an open ball $B_r (x)$ centered around $x$ such that the ball $B_r (x)$ is contained in $A$.
Secondly, your criterion for checking continuity is wrong. By definition a map $f:X\to Y$ between topological spaces $X$ and $Y$ is called continuous if for every open subset $U\subseteq Y$ the set $f^{-1}(U)$ is open in $X$. Here this set is defined as: $$f^{-1}(U) = \{x\in X \colon f(x)\in U \}$$ Maps for which the opposite holds, i.e. for $V\subseteq X$ open we have also $f(V)\subseteq Y$ open, are called open. But they need not be continuous.
Now to begin answering your question, it is important to realise that $X$ has the subspace topology with respect to the usual topology in $\mathbb{R}$. I.e. $A\subseteq X$ is open if and only if there is an open subset $U\subseteq \mathbb{R}$ such that $A=X\cap U$. Now note that subsets of the form $(-x,x)$ are open in $\mathbb{R}$. This means that subsets of the form $[0,x)$ are open in $X$, since $[0,x)=(-x,x)\cap X$, for $0<x<1$.
Consider now what happens if we look at the pre-image of such a set under your function $a^{-1}$. Since $a$ is a bijection we have: $$(a^{-1})^{-1}([0,x))=a([0,x))=\{e^{2\pi i s}\in \mathbb{S}^1 \colon s\in [0,x) \}$$ You can (and should) check that this set is not open. This means that $a^{-1}$ is not continuous, and thus that $a$ is not a homeomorphism.
For completion, if we define:$$I_{a,b}=\{e^{2\pi i s}\in \mathbb{S}^1 \colon s\in (-a,b) \}$$ then this is an open set in $\mathbb{S}^1$ (Note the round brackets as oposed to your square brackets!) Now your $J_{a,b}$ is an open set in $X$ and we have: $$(a^{-1})^{-1}(J_{a,b})=I_{a,b}$$ So sets of this form are not helpfull here. It is the closed part of the boundary of $X$ that leads to trouble and to use that fact we should use open sets that also have this closed part as their boundary.