# $f:[0,2\pi]\rightarrow\mathbb{S}$, $f(t)=(\cos t, \sin t)$. Find a base for the finest Topology that makes $f$ continuous

I am solving the following question for a topology assignment.

Let $f:[0,2\pi]\rightarrow\mathbb{S}$, $f(t)=(\cos t, \sin t)$ where $[0,2\pi]$ is endowed with the induced topology by the euclidean topology on $\mathbb{R}$. Find a base for the finest topology on $\mathbb{S}$, the unit circle, that makes $f$ continuous.

It is clear that such a topology on $\mathbb{S}$ is given by $\tau=\{f^{-1}(V)\;\vert\; V\text{ open in } [0,2\pi]\}$. Since a base for the topology on $[0,2\pi]$ is given by sets of the type $(a,b)$ with $0\leq a<b\leq 2\pi$, $[0,a)$ and $(b,2\pi]$, I would say that a base for $\tau$ is given by sets of the type $f^{-1}((a,b)), f^{-1}([0,a))$ and $f^{-1}((b,2\pi])$. Is that correct?

Also, I am very tempted to say that $\tau$ is the quotient topology because $\mathbb{S}=[0,2\pi]/\sim$ where $x\sim y$ iff $x=y$ or $x=0, y=2\pi$ or $x=2\pi, y=0$ and the quotient map is exactly $f$. Also, the product topology is the finest that makes the quotient map continuous. However, $\tau$ and the quotient topology seem not to coincide because $\tau$ contains arches that are for example open on one end and closed at the other, but the product topology only contains open arches together with their intersections and unions. So, where I am going wrong?

Many thanks for the help!

• Do you mean $\tau=\{V|f^{-1}(V)\text{ open in }[0,2\pi]\}$? Commented Jun 12, 2017 at 18:58
• Yes, sorry for the miswriting Commented Jun 12, 2017 at 19:20

Hint: the map is equivalent to the map $f:\mathbb{R}/2\pi \mathbb{Z} \rightarrow \mathbb{S}^1$ given by $t \mapsto e^{it}$.

• I am not sure whether I understand your hint, can you give me some more details please? Commented Jun 12, 2017 at 19:30
• @Michela What is the status regarding the continuity of $t \mapsto e^{it}$ when we use the standard topology on any subspace of $\mathbb{R}$ and the circle inherits the standard topology on $\mathbb{C}$ as a subspace of the complexes?
– JMJ
Commented Jun 12, 2017 at 19:36
• $t\mapsto e^{it}$ with these two topologies is continuous, but is there a finer topology on the unit circle which makes such a function continuous? In my initial question I don't see why I have to exclude the subsets on the unit circle that are open on one side and close on the other... Commented Jun 12, 2017 at 20:00
• the only finer topology I can think of is the discrete topology. Continuity frequently fails for standard topology continuous functions in the clopen topology.
– JMJ
Commented Jun 12, 2017 at 20:34
• I understand now. Thanks a lot! Commented Jun 12, 2017 at 20:50

Using the corrected formula in the comments $$\tau = \{V \,\bigm|\, \text{f^{-1}(V) is open in [0,2\pi]}\}$$ your guess is correct, $\tau$ is indeed the quotient topology on $\mathbb{S}$ induced by the function $f$.

Certainly $\tau$ is a topology with the desired property that $f$ be continuous, certainly $\tau$ contains all topologies with this property, and certainly there cannot be any finer topology with these properties, because if $V \not\in\tau$ then $f^{-1}(V)$ is not open in $[0,2\pi]$ and so $f$ is not continuous.

You express some doubts about $\tau$ being the quotient topology induced by $f$. In some sense this is a matter of definitions; for instance, this exact formula for $\tau$ is used as the definition of "the quotient topology induced by $f$" as that definition is stated in Munkres book "Topology". But you also express a particular doubt:

• "$\tau$ and the quotient topology seem not to coincide because $\tau$ contains arches that are for example open on one end and closed at the other, but the product topology only contains open arches together with their intersections and unions."

I am not entirely sure that I understand your concern here, but I think perhaps this will allay your worries. Note that if $V \in \tau$ contains a half-open arc of the form $0 \le \theta < a$ for some small positive angle $a$ (here I am using polar coordinates), then it follows that $f^{-1}(V) \subset [0,2\pi]$ contains the point $0$ as well as the point $2\pi$. Since $f^{-1}(V)$ is open in $[0,2\pi]$, it must contain an interval of the form $[0,-\epsilon)$ as well as an interval of the form $(2\pi-\epsilon,2\pi]$. It follows that $V$ also contains a half-open arc of the form $b < \theta \le 0$ for some small negative angle $b$.

• @LeeMoscher Thanks a lot. My point was that I missed the fact that the preimage of (0,0) in the unit circle contains both 0 and $2\pi$. Commented Jun 13, 2017 at 19:45