Collapsing constant regions of path

Let $$f: [0,1] \to X$$ be some continuous map ($$X$$ some topological space). Let introduce the equivalence relation on $$[0,1]$$ $$y \sim_c z \iff$$ $$f$$ constant on $$[y,z]$$ (or on $$[z,y]$$ if $$z < y$$)
Is the space $$[0,1]/\sim_c$$ with quotient topology homeomorphic to some interval [a,b]?
And for which topological spaces $$X$$ is this true?

– user798113
Commented Dec 24, 2020 at 18:52
• then there are a finite number of such bad regions this is true by mathematical induction (we paste finite number of intervals). But when there are infinite many such regions I don't know what to do. Commented Dec 24, 2020 at 19:08

The statement is true iff $$X$$ is a $$T_1$$ space: See https://math.stackexchange.com/a/109333/58818 for a proof of the hard direction (that $$T_1$$ spaces have the desired property).
Let us show that if $$X$$ is not $$T_1$$, then there is a continuous function $$f\colon I\to X$$ such that the quotient $$X/\!\sim_c$$ is not homeomorphic to an interval. The idea is basically to find a sort of copy of the Sierpinski space within such $$X$$.
Since $$X$$ is not $$T_1$$, there are $$a\neq b$$ such that every neighbourhood of $$b$$ contains $$a$$.
Let $$f\colon[0,1]\to X$$ be given by $$f(t)=\begin{cases}b&\text{ if }t\leq 1/2\\a&\text{ if }t>1/2\end{cases}$$
Then $$f$$ is continuous, and there are precisely two $$\sim_c$$-equivalence classes: $$[0,1/2]$$ and $$(1/2,1]$$. This means that the quotient $$X/\!\sim_c$$ has precisely two points, and is immediatelly not homeomorphic to an interval.
(In fact, the open sets in this quotient are only $$X/\!\sim_c$$, $$\varnothing$$ and $$\left\{(1/2,1]\right\}$$ (it is another copy of the Sierpinski space), and it is not even Hausdorff.)