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Let $X$ be a topological space. Is there a Hausdorff space $HX$ and a continuous function $i:X\rightarrow HX$ such that for any Hausdorff space $A$ and a continuous function $j:X\rightarrow A$, there exists a unique continuous function $f:HX\rightarrow A$ satisfying $fi=j$.

$$\begin{array}{ccccccccc} X & \xrightarrow{i} & HX & \\\ & \searrow{j} & \downarrow{f} \\&&A \end{array}$$

Note: I heard before that free objects and "left adjoints" are really the same thing (not sure). Since I don't know the definition of left adjoints and did not study category theory yet, I chose to phrase my question in the way I am most comfortable with.

Thank you

Amr
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    If you do not stipulate uniqueness of $f$ (which is usually done for "free objects"), I think you can take $HX$ to be $X$ endowed with the discrete topology; then $j:HX \to A$ is automatically continuous. – Lord_Farin Aug 19 '13 at 14:00
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    @Lord_Farin Yes I forgot this. Thanks – Amr Aug 19 '13 at 14:01
  • See also: http://math.stackexchange.com/questions/120636/an-inductive-construct-of-the-hausdorff-reflection – Martin Sleziak Dec 27 '14 at 14:58

4 Answers4

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Let $HX$ be the quotient space $X/\sim$, where $x\sim y$ iff $f(x)=f(y)$ for each $f$ from $X$ to a Hausdorff space. Let $q:X\to HX$ be the quotient map.

Let us show that $HX$ is Hausdorff: Take $[x]\ne[y]\in HX$, i.e. for each $x,y$ representing these classes there is a map $f:X\to Y$ into a Hausdorff space $Y$ such that $f(x)\ne f(y)$. There are disjoint open sets $f(x)\in U,f(y)\in V$. Then $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open neighborhoods of $x$ and $y$. Assume that $z\in f^{-1}(U)$ and $z\sim v$. Then by definition of '$\sim$' we have $f(z)=f(v)$, so we conclude that $f^{-1}(U)$ and $f^{-1}(V)$ are $\sim$-saturated disjoint open sets. It follows that $q(f^{-1}(U))$ and $q(f^{-1}(V))$ are disjoint open neighborhoods of $[x]$ and $[y]$.

Now, assume that $f:X\to Y$ is a continuous map into a Hausdorff space. Whenever $x\sim y$ we also have $f(x)=f(y)$, hence there is a unique induced map $\tilde f:HX\to Y$ such that $\tilde f\circ q=f$.

This shows that the category $\mathbf{Top}_2$ of Hausdorff spaces is a full reflective subcategory of $\mathbf{Top}$.

Stefan Hamcke
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  • "Let $HX$ be the quotient space $X/\sim$, where $x\sim y$ iff $f(x)=f(y)$ for each $f$ from $X$ to a Hausdorff space. Let $q:X\to HX$ be the quotient map." I like this a lot. As it stresses the idea that continuous functions from $X$ to hausdorff spaces do not differentiate between points in $X$ which can't be separated by open sets. – Amr Aug 19 '13 at 15:32
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    @Amr: Note that "$x$ and $y$ can't be separated by open sets" is not a transitive relation, but is a subset of the relation "same image under each $f$ to Hausdorff space". Maybe the later is the transitive closure of the former. – Stefan Hamcke Aug 19 '13 at 16:59
  • Yes I already noted that – Amr Aug 19 '13 at 20:07
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What you are saying is that the full subcategory of Hausdorff spaces is a reflective subcategory of the category of topological spaces. That is to say, the inclusion functor $i: Haus \to Top$ has a left adjoint $H: Top \to Haus$, which is sometimes called Hausdorffification.

One way to see there exists such a thing is by the general adjoint functor theorem, whose hypothesis are easily checked in this case.

Bruno Stonek
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  • +1 Thanks for your answer. Does the theorem tell me that the space $HX$ exists only or gives me a way to construct $HX$ as well ? – Amr Aug 19 '13 at 14:31
  • @Amr: you're welcome! The theorem guarantees the existence of such a functor. Since it is an adjoint, it is unique up to natural isomorphism, so the Hausdorffification is unique up to homeomorphism (as you can in fact easily check just using the universal property you have very well described). – Bruno Stonek Aug 19 '13 at 14:35
  • @Amr: an explicit construction $HX$ is by a quotient of $X$ by an equivalence relation, the largest one that makes it Hausdorff. Say that two points are equivalent if they can't be separated by open sets, etc. – Bruno Stonek Aug 19 '13 at 14:36
  • Hmm No this is enough :) – Amr Aug 19 '13 at 14:37
  • @Bruno: No, this quotient is not Hausdorff in general. See also http://mathoverflow.net/questions/11191 – Martin Brandenburg Aug 20 '13 at 07:21
  • @MartinBrandenburg: interesting! Thanks for the link. – Bruno Stonek Aug 20 '13 at 08:41
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There are various constructions for the maximal Hausdorff quotient, see MO/78175 and MO/11191. I quite like the transfinite construction. Actually this is a special case of Kelly's paper on transfinite constructions, see also the corresponding nlab article.

Community
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Martin Brandenburg
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A construction suggested on Wikipedia in fact produces a left adjoint for the forgetful functor $\mathcal{H}\to Top$ for any full subcategory $\mathcal{H}$ of $Top$ consisting of Hausdorff spaces that is closed under taking products and taking closed subspaces - for example $\mathcal{H}=Haus$ itself, or $\mathcal{H}=\mathcal{T_{3\frac{1}{2}}}$ (completely regular Hausdorff spaces), or $\mathcal{H}=CompHaus$ (compact Hausdorff spaces, the case to which Wikipedia applies the construction to find the Stone-Čech compactification).

Indeed, if $f:X\to Y$ is a continuous map with $Y$ Hausdorff, and $f(X)$ is dense in $Y$, there is an injection $y\mapsto\{f^{-1}(U)\mid y\in U \,\&\,U \text{ open in }Y\}$ from $Y$ into the double power set $\mathcal{P}^{2}(X)=\mathcal{P}(\mathcal{P}(X))$ of $X$.

So for $X\in Top$, let $\mathcal{F}(X)$ denote the set of triples $(Y,\tau,f)$ with $Y\in\mathcal{P}^{3}(X)$, $\tau$ is a topology on $Y$ such that $(Y,\tau)\in\mathcal{H}$, and $f:X\to (Y,\tau)$ is continuous with dense image. And let $g:X\to\prod_{(Y,\tau,f)\in\mathcal{F}(X)}(Y,\tau)$ be the natural map. Then the closure $\mathcal{G}(X):=\overline{g(X)}$ of the image of $g$ is in $\mathcal{H}$, and the corestriction $g:X\to\mathcal{G}(X)$ of $g$ to $\mathcal{G}(X)$ is universal among continuous maps from $X$ to objects of $\mathcal{H}$.

In the case $\mathcal{H}=Haus$, by universality $g:X\to\mathcal{G}(X)$ is just $nat:X\to HX$, which is surjective. It follows that here the image of $g$ is closed in $\prod_{(Y,\tau,f)\in\mathcal{F}(X)}(Y,\tau)$, a fact not immediately clear from the definitions.

Matthé van der Lee
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