I know the contravariant functor $\mathrm{Top}\to\mathrm{Set}$ sending a topological space to its set of open sets is representable, with representing object being the two point space with precisely one singleton being the only nontrivial open set.

Why do things not work in the category of Hausdorff spaces?

Towards a contradiction, I suppose the functor $F$ is representable, so $F(-)\cong\operatorname{Hom}_{\mathrm{Haus}}(-,A)$ for some representing space $A$. In particular, $F(A)\cong\operatorname{Hom}(A,A)$. Let $U\in F(A)$ be the universal element corresponding to $\mathrm{id}_A$.

Then for any space $B$, and any $V\in F(B)$, there is a unique morphism $f\colon A\to B$ such that $F(f)(V)=U$, i.e., $f^{-1}(V)=U$. This unique morphism is $\eta_B^{-1}(\eta_A(U))$ where $\eta_X$ is the component morphism $F(X)\to\mathrm{Hom}(X,A)$.

This claim certainly seems unlikely, so that $F$ is not representable. However, I'm having trouble coming up with a concrete example to show this cannot be.


Let $O : \mathbf{Haus}^\mathrm{op} \to \mathbf{Set}$ be the functor that sends a Hausdorff space $X$ to the set of open subsets of $X$. Then $O (1)$ has two elements, so any representation of $O$ must be a two-point space. But the only two-point Hausdorff space is discrete – which clearly won't work.

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  • $\begingroup$ Thanks. Why is it clear that such a representative won't work? I know any finite Hausdorff space is discrete, so $O(X)\simeq\mathrm{Hom}(X,A)$ for all finite Hausdorff $X$, with $A$ the discrete two point space, as they both have cardinality $2^{|X|}$. So is there some infinite Hausdorff space where this fails? Comparing the number of open sets and continuous maps into $A$ for an infinite Hausdorff space is not easy for me. $\endgroup$ – Hana Bailey Mar 3 '14 at 12:34
  • $\begingroup$ Every open subset of a discrete space is also closed. So if $O$ were represented by a discrete space then every open subset is also closed, which is nonsense. $\endgroup$ – Zhen Lin Mar 3 '14 at 12:46
  • $\begingroup$ Sorry, obviously that's nonsense, but how does that follow just from being represented by a discrete space? If you were take $(-\infty,b)\in\mathbb{R}$ for instance, this corresponds to some morphism $\mathrm{Hom}(\mathbb{R},A)$. How does this imply that $(b,\infty)$ is closed? $\endgroup$ – Hana Bailey Mar 3 '14 at 20:16
  • $\begingroup$ You are forgetting how $O$ is defined on morphisms. Take the universal element: by definition, every open subset of every Hausdorff space occurs as the inverse image of the universal one. But if $O$ is represented by a discrete space... $\endgroup$ – Zhen Lin Mar 3 '14 at 20:22
  • $\begingroup$ Or is it something like, every open set in $B$ occurs as the inverse image of some universal element $U$ in $A$, but $U$ is both open and closed, so its inverse image in $B$ is both open and closed? $\endgroup$ – Hana Bailey Mar 3 '14 at 20:54

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