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I am trying to learn topology but I dont know how to proof this problem.

Let $X$ be a Hausdorff space, ~ an equivalence relation and $\pi:X \to X/{\sim}$ the canonical map. $X/{\sim}$ is also Hausdorff, if there exists a continuous function $s:X/{\sim} \to X$, such that $\pi \circ s=\textrm{Id}_{X/{\sim}}$

I've read that the diagonal of a Hausdorff space is closed regarding the product topology but I don't know how to proceed or where to start

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I would go about it this way (letting $Y$ represent the quotient space $X/\sim$):

  • First, use the fact that $\pi\circ s$ is the identity on $Y$ (so is one-to-one) to prove that $s$ is one-to-one.
  • Next, letting $Z$ be the image of $s,$ note that $Z$ is a subspace of $X,$ so is Hausdorff.
  • Finally, show that $\pi\restriction Z$ is the inverse of $s.$ Since both $s$ and $\pi\restriction Z$ are continuous, that means that $s$ is a homeomorphism from $Y$ to $Z,$ so since $Z$ is Hausdorff, so is $Y.$
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  • $\begingroup$ Thank you, your answer really helped me a lot $\endgroup$ – Kugen Apr 27 '19 at 13:02
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Notice that $s:X/_\sim\to \operatorname{im}s$ and $\left.\pi\right\rvert_{\operatorname{im} s}:\operatorname{im} s\to X/_\sim$ are continuous maps, and they are one the inverse of the other (this is true set-theoretically speaking). Therefore, $X/_\sim$ is homeomorphic to $\operatorname{im}s\subseteq X$, which inherits the Hausdorff property.

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