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I'm aware that the image of a connected set is connected and the preimage of a disconnected set is disconnected. However, I'm struggling to find an example of a disconnected set such that the image of the disconnected set is also disconnected. Can the preimage of a connected set be disconnected?

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HINT: I assume that you want $f$ to be continuous. Consider the function $f:\Bbb R\to\Bbb R:x\mapsto x^2$. What is the image of $\{-1,1\}$? What is the pre-image of $[1,2]$?

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  • $\begingroup$ Image of $\{-1,1\}$ is connected, should we take $\{0,1\}$ instead? $\endgroup$ – Akash Gaur Jun 24 '20 at 16:07
  • $\begingroup$ @AkashGaur: Although the question shows no signs of having been edited, I’m reasonably sure that I gave that as an example of a disconnected set whose image is connected. Whether that was in response to a comment that has since been deleted or in response to an original version of the question that was changed too quickly to count as an edit I can no longer say after four years. It’s even possible that I misread the question at the time. $\endgroup$ – Brian M. Scott Jun 24 '20 at 16:43
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If $f:X \to Y$ is continuous, then the contrapositive of $A$ connected $\implies f(A)$ connected is $f(A)$ disconnected $\implies A$ disconnected. Pre-image of a disconnected set is disconnected is incorrect, eg. let $f:\mathbb R \to \mathbb R$ be the constant zero function, then the pre-image of $\{0,1\}$ is connected.

For the other examples, take the image of $\{0,1\}$ under the identity function on $\mathbb R$ and the pre-image of $\{0\}$ under the constant zero function $f:\{0,1\} \to \mathbb R$ under the subspace topology of $\mathbb R$.

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