# Is there a monad on Set whose algebras are Tychonoff spaces?

Compact Hausdorff spaces are algebras of the ultrafilter monad on Set.

Is the category of Tychonoff spaces also monadic over Set?

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No. In fact any full subcategory of $\mathbf{Top}$ that contains all the discrete spaces cannot be monadic over $\mathbf{Set}$ unless it contains only discrete spaces. Indeed, for any such subcategory $\mathcal{C}$, the forgetful functor $\Gamma : \mathcal{C} \to \mathbf{Set}$ has a left adjoint $\Delta : \mathbf{Set} \to \mathcal{C}$ which sends a set to the corresponding discrete space. The monad induced by $\Delta \dashv \Gamma$ is the identity monad on $\mathbf{Set}$, so if $\Gamma$ were monadic, then $\mathcal{C}$ must be the full subcategory of discrete spaces.

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Of course, Zhen has answered the question correctly under the reasonable assumption that what is being asked is whether the usual forgetful functor $U: \mathrm{Tych} \to \mathrm{Set}$ is monadic (strictly speaking, it's functors that are or are not monadic, not categories!). One might wonder whether there is some other weird functor $G: \mathrm{Tych} \to \mathrm{Set}$ that is monadic (such a functor would of course have to be faithful and isomorphism-reflecting).

But the answer is that there isn't one. In general, if $G: C \to \mathrm{Set}$ is monadic, then we can infer lots about the category $C$, such as the fact $C$ is a Barr exact category. Suppose $\mathrm{Tych}$ were Barr exact; since it is also an extensive category, it would be a pretopos. But a pretopos is necessarily balanced, i.e., every monic epimorphism is an isomorphism (I believe this fact is proved for example in Paul Taylor's Practical Foundations of Mathematics). Since this last property clearly doesn't hold in $\mathrm{Tych}$ (for example, the identity function $\mathbb{R}_{\mathrm{disc}} \to \mathbb{R}$ from the discretely topologized $\mathbb{R}$ is a monic epi, but not an isomorphism), we conclude $\mathrm{Tych}$ is not Barr exact.

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Thank you both for the great answers; This helped me a lot! – Garlef Wegart Jun 28 '13 at 17:08

Even easier. Let $I$ be the unit interval and $E$ be the equivalence relation that identifies all the rationals. $I/E$ is certainly not Hausdorff. Is it a regular category? I'm not sure.

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