Background
----------


Recall that a functor $G\colon A\to X$ is called
[monadic](http://ncatlab.org/nlab/show/monadic+functor) 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](http://books.google.com/books?id=SGwwDerbEowC&lpg=PP1&dq=moerdijk%20sheaves&pg=PA178#v=onepage&q&f=false) 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?



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


