# Thread: Example of a non-Hausdorff surjective function

1. ## Example of a non-Hausdorff surjective function

Let X = [-1,1] and give X the usual topology.

Give an exmple of a surjective function g from X onto [0,1] such that the quotient topology on [0,1] induced by g is not Hausdorff.

Thanks for any help.

2. Originally Posted by Throughpoint
Let X = [-1,1] and give X the usual topology.

Give an exmple of a surjective function g from X onto [0,1] such that the quotient topology on [0,1] induced by g is not Hausdorff.

Thanks for any help.
What have you tried? Note that since $X$ is compact Hausdorff that saying the the topology induced on $[0,1]$ by some quotient map $q$ is equivalent to stating that $q$ is a closed map. So, find a quotient map $q:[-1,1]\to[0,1]$ which isn't closed and the resulting coinduced topology on $Y$ will be non-Hausdorff.

g(x) = x^2? Lol. Every element in [0,1] has a pre-image in [-1,-1].

I might be way off.

Thanks so much for the help so far btw. ^^

No wait, how about g(x) = 0 if x is a rational number, or 1 if x is not a rational number.

4. Originally Posted by Throughpoint

g(x) = x^2? Lol. Every element in [0,1] has a pre-image in [-1,-1].

I might be way off.

Thanks so much for the help so far btw. ^^

No wait, how about g(x) = 0 if x is a rational number, or 1 if x is not a rational number.
Ok, in your examples what closed set's image isn't closed?

5. Erm.. I guess g = x^2 does have a closed image.. as it maps onto [0,1]. That doesn't mean it's not closed IN [-1,1] does it?

My other example isn't closed in [-1,1]!