While I'm here: on second thought I don't think $\omega$-groups are all that sensible, because of the swindle mentioned by both you and Theo. [The "logic" of an algebraic theory involving operations $p$ of infinite arity, like that of $\omega$-groups, means we must have $p(a_1, \ldots) = p(b_1, \ldots)$ whenever $a_i = b_i$. So the swindle is derivable in the theory.] In other words, while you can consider or construct the theory or monad of $\omega$-groups, it would unfortunately collapse to something trivial. Silly of me to have suggested otherwise. :-(

David, I've nothing intelligent yet to say about the $\mathbb{N}^+$ thing, but I don't think that forgetful functor has a right adjoint. It would have to preserve coproducts. The coproduct of two copies of the free $\omega$-monoid on the 1-element set, $F(1)$, is the free $\omega$-monoid on the 2-element set, $F(2)$. You'd need the canonical map $F(1) + F(1) \to F(2)$ on the underlying monoids, where $+$ is monoid coproduct, to be invertible. So, each countable word in two letters would need to be a finite concatenation of countable words, each in one or the other letter. Which is false. :-(

I don't know, Jenny. It might be worthwhile posting this as a separate question at MO (preferably with a link to a suitable document with relevant definitions). Or, perhaps you know an expert in domain theory you can ask directly? Good luck!

There are many weird examples where this can happen. A silly example is where you take a non-distributive lattice $L$ and define $T: L \to L$ to take the bottom element 0 to 0, and every other element to the top element 1. The Eilenberg-Moore category is then the two-element lattice which is cartesian closed. Slightly less silly: take $C = Top/X$, which is generally not cartesian closed, and $T$ to take a map $f: Y \to X$ to its image $im(f) \hookrightarrow X$. The EM category is then the lattice of subspaces of $X$ which is cartesian closed. But why do you ask?

It's hard to give reasonable general answers. I can say that if $C$ has coproducts, so does $Kl(T)$. Depending on $C$ and the monad $T$, it may be that $Kl(T)$ is complete and cocomplete; for example if $C$ is bicomplete and every $T$-algebra is free. This happens e.g. when $T$ is an idempotent monad; details at ncatlab.org/nlab/show/completion. But usually it comes down to individual cases and asking questions like: are products of free objects free? Are subobjects of free objects free? If you do have a specific example, maybe we can discuss that. Ans. to 2nd question next comment.

If you are willing to consider highly abstract approaches, you might consider certain toposes of smooth spaces, as developed in say the book by Moerdijk and Reyes. In such toposes one can interpret "the smooth space of smooth subspaces of $Y$", call it $P(Y)$, and then consider smooth maps $f: X \to P(Y)$. It seems possible that the specific types of $f$ you are after would be definable as smooth maps in a great many such toposes. (Unfortunately, I am not a specialist in synthetic differential geometry.)

And while I'm at it (this goes back to the motivation given at the end of the question), the category of cocommutative coalgebras is also cocomplete, cocomplete, extensive, and locally finitely presentable. So a lot of arguments that apply to $Set$ also apply there. One thing missing though is that quotients are not preserved by pulling back along morphisms.

It might be of interest to know that the category of cocommutative coalgebras is cartesian closed (cf. fact that finite-dimensional cocommutative coalgebras is the category opposite to f.d. commutative algebras).

In case anyone is interested, I wrote up an exposition of the discussion between Freyd and Pratt in the nLab: ncatlab.org/nlab/show/AT+category.

In most applications, where one speaks of sieves and covering sieves (with respect to a topology), the underlying category $C$ of a site is assumed to be small. Otherwise, the sieve $S$ won't actually be a set, but a class (as you observe). Which isn't necessarily problematic; it depends what you want to do.

The multiplication may as well be [[x, s], t] |-> [x, st], or even [[x, s], t] |-> [x, min(s, t)], and the unit x |-> [x, 1]. Under the min formulation, you can similarly form a monad C_I for any interval I, where C_I X is the pushout of the inclusion X -> X x I (at the endpoint "top" of I) along X -> 1. You can define a notion of I-contractibility as involving a retraction h: C_I X -> X of the unit (or even demand h to be an algebra structure). You can go on to define an I-analogue of the topological simplex category. So what is so canonical about the usual choice I = [0, 1]?

Well, your first comment on my answer gave me the impression that you were annoyed that I hadn't acknowledged your earlier mention (and if you were, then please accept my apologies). I think the point however is that any interval (toset with distinct top and bottom) induces a left exact geometric realization, so the question still remains: why choose this one? Is there some sort of abstract conceptual reason? The same question applies to the cone monad: for any interval there is an associated cone monad, so what's the reason for choosing [0, 1] as the interval?

No problem, José -- Peter's example is probably better in that it is immediately apprehended and "in nature". I do recommend the book by Freyd and Scedrov to your attention, if you are not already familiar with it. :-)

ncatlab.org/nlab/show/reflected+limit

I actually quite liked Peter's example. As for the objection, one might observe that the underlying functor from Peter's $Vect_n$ to $Set$ preserves any limits which exist in $Vect_n$ (since it is representable), and reflects them as well since it reflects isomorphisms (see the <a href="ncatlab.org/nlab/show/reflected+limit">nLab page</a>). So any product which exists in $Vect_n$ is the expected one.