# A contradictory result from a corollary with the fact that the quotient space of a Hausdorff space is not necessary Hausdorff

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

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 ?

• In that case, $X^*$ is homemorphic to $X$ so it is Hausdorff – Dante Grevino Dec 11 '18 at 5:55
• 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 $(b)$. – 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.
• Yeah, I assumed $X^*$ was some quotient space of $X$; didn't pay attention to the definition of $X^*$ there. Thanks for the answer. – onurcanbektas Dec 11 '18 at 6:08