If G is monadic and the comparison functor is an equivalence that is not an isomorphism, does G create limits?

Question

Background

Recall that a functor $G\colon A\to X$ is called monadic if it has a left adjoint $F$ for which the Eilenberg--Moore comparison functor $K\colon A\to X^{\mathbb{T}}$ is an equivalence of categories, where $\mathbb{T}$ is the monad in $X$ defined by the adjunction $\langle F,G,\ldots\rangle\colon X\rightharpoonup A$, and $X^{\mathbb{T}}$ is the category of $\mathbb{T}$-algebras in $X$.

This means that a monadic functor is the forgetful functor $G^{\mathbb{T}}\colon X^{\mathbb{T}}\to X$ up to composition with an equivalence of categories (the comparison functor $K$). Now, it can be verified that $G^{\mathbb{T}}$ creates
limits (Ex. 6.2.2 of Mac Lane). If the comparison functor is an isomorphism, then it is straightforward to verify that $G$ creates limits. In fact, I think that even if it is only assumed that $K$ is an equivalence for which the object function is surjective, then $G$ creates limits.

However, in Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, it is stated that any monadic functor creates limits. The proof starts with the following words (with minor omissions):

Let $G$ be monadic. Then by definition, $G$ is the forgetful functor $G^\mathbb{T}$ up to an equivalence of categories. It thus suffices to show that such a forgetful functor $G^\mathbb{T}$ creates limits.

I simply do not understand this statement: In general, the composition of an equivalence and a functor that creates limits need not create limits. For example, the identity $\mathbf{Set}\to\mathbf{Set}$ creates limits, and for any skeleton $X$ of $\mathbf{Set}$ the inclusion $X\subseteq \mathbf{Set}$ is an equivalence. Let $X$ be some skeleton of $\mathbf{Set}$ (for which I am happy to assume any necessary axiom of choice), and take a one-element set $1$ that is not in $X$. Then $1$ is a limit of the functor obtained by composing the unique functor from the empty category to $X$ with $X\subset\mathbf{Set}\stackrel{\operatorname{Id}}{\to}\mathbf{Set}$, but $1$ has no lifting in $X$.

So it seems that there are 4 possibilities:

1. The above Proposition 4.4.1, as stated, is wrong. There is a counter example where a monadic functor (for which the comparison functor is not an isomorphism) does not create limits.

2. The proof in ML-M covers just some of the cases, and for the other cases it is not known if the assertion is true (namely, for a monadic functor $G$ for which the comparison functor is not an isomorphism, it is not known whether in general $G$ creates limits).

3. The proposition is correct because the comparison functor has some additional special property (e.g., its object function must be surjective whenever it is an equivalence).

4. (Most likely) I am wrong, and the quoted argument from Mac Lane--Moerdijk is correct.

I would like to note that in Theorem 3.4.2 on p. 105 of Barr-Wells, it is only claimed that monadic functors reflect limits.

Question

Which one of the above 4 possibilities is true? In essence, my question is: If $G$ is monadic and the comparison functor is an equivalence that is not an isomorphism, does $G$ create limits?

Answer 1

Mac Lane-Moerdijk slipped up; they really should have said "reflects limits". Now it's true that the forgetful functor from the literal category of algebras to the underlying category creates limits (according to the definition of "creates" in CWM), but this notion doesn't transfer across equivalences, as you have observed.

An example where the distinction is important is the statement that for a topos E, the power object functor P: E^{op} --> E is monadic. It wouldn't make much sense to say that this creates limits in Mac Lane's sense of the term.

Answer 2

According to the remarks on the nLab's created limit page, Categories for the Working Mathematician has a nonstandard definition of creating limits; the standard definition only requires lifts up to isomorphism.