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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?

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  • $\begingroup$ what do you think about this? $\endgroup$
    – user798113
    Commented Dec 24, 2020 at 18:52
  • $\begingroup$ 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. $\endgroup$
    – Nikolay
    Commented Dec 24, 2020 at 19:08

1 Answer 1

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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.)

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