Quotient map on S^1 such that that the quotient is an uncountable space with the indiscrete toplogy. Based on Leinster Basic Category Theory page 135, problem 5.2.22 (b)
I am trying to find a quotient map on the circle $S^1$ which results in the quotient having the indiscrete topology.
Using the following parameterisation of $S^1 = \{e^{i2\pi x}\;, \; x \in [0,1)\}$, take $x$ and define an equivalence relation $x \sim y$ if $y - x \in \mathbb{Q}$, leading to the quotient map $q\colon S^1 \rightarrow S^1/\mathbb{Q}$. Since $x \in [0,1)$, and we know more generally the quotient topology of $\mathbb{R}/\mathbb{Q}$ is indiscrete see this Answer - for example, then we have our required indiscrete quotient.
Is the informal answer I have outlined above sound? If not what other approaches could you suggest for me to solve this problem?
Thanks
 A: It does seem like your answer depends on the statement that $S^1/\mathbb Q$ has the subspace topology of $\mathbb R/\mathbb Q$, which is not clear a priori.
It seems like you are using a statement along the line of: assume $X$ is a space with a subspace $A$ and $\sim$ is an equivalence relation on $X$. Then $\sim$ restricts to an equivalence relation on $A$, and the induced map $A/\sim \to X/\sim$ is a topological embedding.
I actually don’t know if the above statement which you are implicitly using is true. Colimits and limits do not behave well with each other in general (for example products of quotient maps are not always quotient maps).
But you can check directly that the space $S^1/\mathbb Q$ has indiscreet topology. Take some open set $U$ in the space and without loss of generality assume it is nonempty. Then $q^{-1}U$ is open in the circle. Given any point $x$ on the circle, there is some $y$ in $q^{-1}U $ such that $x$ and $y$ differ by a rational angle. Hence $qx=qy$ is in $U$. This shows that any point of the form $qx$ is in $U$, and since $q$ is surjective these are all of the points.
