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At this link a counterexample has been provided. I would like to ask for a clarification of why the space $X/A$ is contractible. Is it related to the fact the topology on the quotient is too coarse? How do I see it is contractible? Thank for your explanations!

Porcupine
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  • Do you believe in the description of the open subsets given in the link? If so, this space is homeomorphic to https://en.wikipedia.org/wiki/Sierpi%C5%84ski_space . It is not difficult to see that this is contractible, and indeed it comes from the fact that the topology is rather coarse (it is not the discrete space with two points). – Captain Lama Mar 23 '23 at 08:28

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You can see that it is contractible by finding a homotopy between the identity and a constant map. Abstractly, the space can be written as $\{a,b\}$ with $\{a\}$ open but $\{b\}$ not open, a.k.a. the Sierpiński space. In such situation consider

$$H:X\times [0,1]\to X$$ $$H(x,t)=\begin{cases} x&\text{if } t=1\\ a&\text{otherwise} \end{cases}$$

It is continuous because $H^{-1}(\{b\})=\{(b,1)\}$ is closed, and $\{b\}$ is the only interesting closed subset of $X$.

freakish
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