MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The inclusion of the full subcategory of Hausdorff topological spaces into the category of topological spaces has a left adjoint, which can be proven easily by the Adjoint Functor Theorem (see for example, S. MacLane, Categories for Working Mathematicians). To every topological space this left adjoint associates a Hausdorff space called the largest Hausdorff quotient.

Do you know a reference in which this left adjoint is constructed explicitly?

share|cite|improve this question
up vote 14 down vote accepted

Consider the equivalence relation $\sim$ on your space $X$ such that $x\sim y$ iff $x$ and $y$ have the same image under all surjective continuous maps $f:X\to Y$ with codomain $Y$ a Hausdorff space. Put on the set $X/\sim$ the least topology which makes all those maps continuous, and you have the space you want. I doubt there is any actual reference for this.

share|cite|improve this answer
(One can try to mod out $X$ by the relation "$x$ and $y$ cannot be separeted by disjoint open sets" —or its transtive closure, really— but the result is not Hausdorff; you can iterate this transfinitely, though, and you do get the space largest quotient. This is much more complicated/annoying/long to carry out) – Mariano Suárez-Alvarez Oct 15 '11 at 1:47
There are some set-theoretic issues here, though they can be dealt with. (You are currently quantifying over all compact Hausdorff spaces, which do not form a set.) – Daniel Litt Oct 15 '11 at 2:59
@Daniel, that is why I restricted to surjective maps. (In any case, the relation is well-defined even if one considers all maps and all $T_2$ sets, and one can easily show that there is a least topology satisfying the condition. One can quantify over all spaces!) – Mariano Suárez-Alvarez Oct 15 '11 at 3:03
Thanks, Marino! I was thinking in the way you describe in your second post and got stuck once I saw that the result is not Hausdorff. :) – mbasic Oct 15 '11 at 14:32

Mariano already answered the question, but let me make two additional remarks:

1) Actually the proof of GAFT is constructive and can be (at least sometimes) used to get explicitly a left adjoint functor. In this case, you can see directly the "big coequalizer" whose set-theoretic existence issue is dealt with the solution set condition: just consider all maps from your space $X$ into Hausdorff spaces which are surjective; this bounds the cardinality of the Hausdorff spaces, and thus up to isomorphism there is only a set of them.

2) Another construction of the left adjoint to $\mathsf{Haus} \to \mathsf{Top}$ works as follows: Let $X$ be a topological space, and consider the equivalence relation $\sim$ generated by: If $x,y$ cannot be separated by disjoint open sets, then $x \sim y$. Then $H(X):=X / \sim$ has the property that every map from $X$ into a Hausdorff space uniquely factors through $X \to H(X)$. If $H(X)$ was Hausdorff, we would be done. But this is not always the case. Instead, we have to repeat this construction: $X \to H(X) \to H(H(X)) \to H(H(H(X))) \to \dotsc$, then take the colimit $H^{\omega}(X)$, and make again $H^{\omega}(X) \to H(H^{\omega}(X)) \to \dotsc ...$. You can continue this for every ordinal number. Since $X$ is a set and all these maps are quotient maps, at some stage we get an isomorphism, which is the desired Hausdorff quotient.

It is interesting when we arrive at this stage, see my question about the nonhausdorff dimension.

share|cite|improve this answer

To augment very slightly Mariano's nice answer, Hausdorff quotients (as opposed to surjections) suffice.

To obtain the finest Hausdorff quotient of an arbitrary space $X$, take the quotient of $X$ by the intersection of all partitions of $X$ determined by quotient maps from $X$ with $T_{2}$ image.

share|cite|improve this answer

A good reference is §4.3 of : D. Huybrechts, A Global Torelli theorem for hyperkähler manifolds (after Verbitsky). Seminaire Bourbaki Exp. No. 1040 Juin 2011 Astérisque No. 348 (2012), 375-403; arxiv:1106.5573.

share|cite|improve this answer
That doesn't look lke a good reference. It deals with a rather special case; the construction of $\approx$ certainly does not apply to a general topological space, and in fact the situation is such that the naive relation $\sim$ of non-separation is an equivalence relation, which seldom happens, and the quotient is Hausdorff, which happens even less :-) – Mariano Suárez-Alvarez Nov 18 '13 at 14:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.