Is Quotient Spaces of Hausdorff also Hausdorff where the Quotient Spaces is T1?

Question

Let X be Hausdorff space and X/~ be "~ Quotient Spaces",

Given that X/~ is T1,

Does it satisfy for X/~ to be Hausdorff?

Answer 1

Let $X$ be two copies of the real line, and let $X\to Y$ be the quotient map that identifies the two lines along the set $\{x\neq 0\}$.

$X$ is Hausdorff, but $Y$, the "line with doubled origin", is locally Hausdorff and therefore $T_1$, but not Hausdorff.

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In fact, every topological space is the quotient of a Hausdorff space. This appears to have been proved by M. Shimrat in 1956, though there may be a very short argument.

I think that something stronger should be true: every topological space $Y$ is a quotient of a Stonean space, which is a Hausdorff extremally disconnected space. I believe that it is possible to show this by embedding the complete lattice of opens in $Y$ into a complete Boolean algebra, though if $Y$ is not sober this may complicate the proof or weaken the conclusion.