It is known that a quotient space of a Hausdorff space is not necessarily Hausdorff; however, in the book of Topology by Munkres, at page 140, it is given that

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But, we can always choose $Z = X$ and $g=i$ so that $g = i$ is a surjective continuous map. Hence, if $Z=X$ is Hausdorff, then by part $b$, $X^*$ must be Hausdorff, which is not true, as there are lots of counterexamples, by what is wrong with the above argument ?

  • $\begingroup$ In that case, $X^*$ is homemorphic to $X$ so it is Hausdorff $\endgroup$ – Dante Grevino Dec 11 '18 at 5:55
  • $\begingroup$ And by the existence of $f$, the topology in $X^*$ is finer than the initial topology with respecto to $f$, which is Hausdorff if $Z$ is Hausdorff. And we get $(b)$. $\endgroup$ – Dante Grevino Dec 11 '18 at 6:00

If $Z=X$ and $g=id$ then $$X^*=\{ g^{-1}(z) | z \in Z \}=X$$

I think that you are confusing $X^*$ with some quotient of $X$, which is not the case in this corollary.

  • $\begingroup$ Yeah, I assumed $X^*$ was some quotient space of $X$; didn't pay attention to the definition of $X^*$ there. Thanks for the answer. $\endgroup$ – onurcanbektas Dec 11 '18 at 6:08

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