## 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:

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.

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).

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).

(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?