====== A Counterexample to the Barr–Beck Monadicity Theorem Without Preservation of Split Colimits ======

===== 1. The Barr–Beck theorem =====

Suppose $F:\mathcal C\to\mathcal D$ admits a left adjoint $F^L:\mathcal D\to\mathcal C$. Then

$$
T=FF^L:\mathcal D\to\mathcal D
$$

is a monad, and we have a canonical comparison functor

$$
\mathcal C\longrightarrow \operatorname{Alg}_T(\mathcal D).
$$

The **Barr–Beck monadicity theorem** gives sufficient conditions for this comparison to be an equivalence:

  - $F$ is conservative (i.e. detects isomorphisms).
  - $F$ preserves geometric realizations of $F$-split simplicial objects, assuming these realizations exist.

For ordinary categories, a common formulation of the second condition is that $\mathcal C$ admits coequalizers of $F$-split pairs, and $F$ preserves these coequalizers.

We give an explicit example where $F$ is conservative and admits a left adjoint, but fails the second condition.

===== 2. The category of torsion-free abelian groups =====

Let

$$
\mathcal C=\mathrm{Ab}_{\mathrm{tf}},
\qquad
\mathcal D=\mathrm{Set},
$$

where $\mathrm{Ab}_{\mathrm{tf}}$ denotes the category of torsion-free abelian groups.

Consider the forgetful functor

$$
F:\mathrm{Ab}_{\mathrm{tf}}\longrightarrow\mathrm{Set}.
$$

**Claim 1.** The functor $F$ is conservative and admits a left adjoint.

Indeed:

  - A homomorphism of torsion-free abelian groups is an isomorphism if its underlying map of sets is bijective. Thus $F$ is conservative.
  - The free abelian group functor $F^L(S)=\mathbb Z^{(S)}$ is left adjoint to $F$, since a free abelian group is torsion-free.

The associated monad

$$
T=FF^L
$$

is the usual free abelian group monad on $\mathrm{Set}$. Consequently,

$$
\operatorname{Alg}_T(\mathrm{Set})\simeq\mathrm{Ab}.
$$

In particular,

$$
\mathrm{Ab}_{\mathrm{tf}}
\not\simeq
\operatorname{Alg}_T(\mathrm{Set})
$$

via the canonical comparison functor: its essential image consists only of torsion-free groups.

Thus $F$ is not monadic. Let us see explicitly how the second Barr–Beck condition fails.

===== 3. A coequalizer that is not preserved =====

Consider the following pair of morphisms in $\mathrm{Ab}_{\mathrm{tf}}$:

$$
\mathbb Z^2
\underset{g}{\overset{f}{\rightrightarrows}}
\mathbb Z,
$$

where

$$
f(a,b)=a+2b,
\qquad
g(a,b)=a.
$$

**Coequalizer in $\mathrm{Set}$.**

The coequalizer identifies all integers differing by an even number:

$$
a+2b\sim a.
$$

Therefore,

$$
\operatorname{coeq}_{\mathrm{Set}}(f,g)
=
\mathbb Z/2\mathbb Z
$$

as a set with two elements.

**Coequalizer in $\mathrm{Ab}_{\mathrm{tf}}$.**

In the category of all abelian groups, the coequalizer would be

$$
\operatorname{coeq}_{\mathrm{Ab}}(f,g)
=
\mathbb Z/2\mathbb Z.
$$

But $\mathbb Z/2\mathbb Z$ is not torsion-free.

In fact, the coequalizer in $\mathrm{Ab}_{\mathrm{tf}}$ is the zero group:

$$
\operatorname{coeq}_{\mathrm{Ab}_{\mathrm{tf}}}(f,g)=0.
$$

To verify this, suppose $A$ is torsion-free and $h:\mathbb Z\to A$ satisfies $hf=hg$. Then

$$
2h(1)=0.
$$

Since $A$ is torsion-free, this forces $h(1)=0$, hence $h=0$.

Thus the zero group has the required universal property.

We obtain

$$
F\left(\operatorname{coeq}_{\mathcal C}(f,g)\right)
=
\{*\},
$$

whereas

$$
\operatorname{coeq}_{\mathcal D}(Ff,Fg)
=
\{0,1\}.
$$

Therefore $F$ does not preserve this coequalizer.

However, Barr–Beck does not require preservation of all coequalizers, only those which become **split coequalizers** after applying $F$.

We now verify that this pair is indeed $F$-split.

===== 4. The coequalizer is split in $\mathrm{Set}$ =====

Recall that a split coequalizer consists of

$$
A
\underset{g}{\overset{f}{\rightrightarrows}}
B
\xrightarrow{q} Q
$$

together with maps

$$
s:Q\to B,
\qquad
t:B\to A,
$$

satisfying

$$
qs=\mathrm{id}_Q,
\qquad
ft=\mathrm{id}_B,
\qquad
gt=sq.
$$

These maps are not required to preserve any additional algebraic structure if we work in $\mathrm{Set}$.

In our example, let

$$
q:\mathbb Z\to\{0,1\}
$$

be reduction modulo $2$.

Define a section

$$
s:\{0,1\}\to\mathbb Z,
\qquad
s(0)=0,\quad s(1)=1.
$$

Set $r(n)=s(q(n))$, so that $r(n)\in\{0,1\}$ is the parity representative of $n$.

Now define

$$
t:\mathbb Z\to\mathbb Z^2,
\qquad
t(n)=
\left(
r(n),
\frac{n-r(n)}{2}
\right).
$$

We check the identities:

$$
\begin{aligned}
q(s(i))&=i,\cr
f(t(n))&=r(n)+2\frac{n-r(n)}2=n,\cr
g(t(n))&=r(n)=s(q(n)).
\end{aligned}
$$

Consequently,

$$
qs=\mathrm{id},
\qquad
ft=\mathrm{id},
\qquad
gt=sq.
$$

This proves that the coequalizer is split in $\mathrm{Set}$.

**Conclusion.** We have found an $F$-split pair in $\mathcal C$ whose coequalizer is not preserved by $F$.

Thus the second condition of the Barr–Beck theorem genuinely fails.

The essential point is that the splitting maps $s,t$ are set maps, not homomorphisms of abelian groups.

===== 5. An explicit $F$-split simplicial object =====

We can express the same obstruction using geometric realizations of simplicial objects, as in the higher-categorical formulation of Barr–Beck.

Consider the simplicial object $X_\bullet$ in $\mathrm{Ab}_{\mathrm{tf}}$ defined by

$$
X_n=
\left\{
(a_0,\ldots,a_n)\in\mathbb Z^{n+1}
\ \middle|
a_i\equiv a_j\pmod 2
\text{ for all }i,j
\right\}.
$$

Each $X_n$ is a subgroup of $\mathbb Z^{n+1}$ and hence is torsion-free.

The face maps delete coordinates, and the degeneracy maps repeat coordinates.

In low degrees,

$$
X_0=\mathbb Z,
$$

$$
X_1=
\{(a,b)\in\mathbb Z^2:a\equiv b\pmod 2\}.
$$

The maps $d_0,d_1:X_1\rightrightarrows X_0$ are the two coordinate projections (in either order).

**After applying $F$.**

The simplicial set $F(X_\bullet)$ is the nerve of the equivalence relation

$$
a\sim b
\quad\Longleftrightarrow\quad
a\equiv b\pmod 2.
$$

Equivalently, it is the nerve of a groupoid with two connected components, consisting of even and odd integers, with a unique arrow between any two objects in the same component.

Its geometric realization in $\mathrm{Set}$ is

$$
{|}F(X_\bullet){|}_{\mathrm{Set}}
=
\{0,1\}.
$$

Moreover, the augmentation

$$
F(X_\bullet)\longrightarrow\{0,1\}
$$

is split as an augmented simplicial object.

Indeed, we can choose the representatives $0$ and $1$ in the two components. An extra degeneracy is given by

$$
s_{-1}(a_0,\ldots,a_n)
=
(r(a_0),a_0,\ldots,a_n),
$$

where $r(a_0)$ is the parity representative.

This gives an explicit simplicial contraction over the two-element set.

**Before applying $F$.**

The geometric realization in $\mathrm{Ab}_{\mathrm{tf}}$ is the coequalizer of the two face maps

$$
X_1\rightrightarrows X_0.
$$

Their difference has image $2\mathbb Z$, so the coequalizer in ordinary abelian groups is $\mathbb Z/2\mathbb Z$.

Passing to the torsion-free category kills this quotient. Therefore,

$$
{|}X_\bullet{|}_{\mathrm{Ab}_{\mathrm{tf}}}=0.
$$

It follows that

$$
F({|}X_\bullet{|})
\not\cong
{|}F(X_\bullet){|}.
$$

We have thus exhibited an explicit $F$-split simplicial object whose geometric realization is not preserved by $F$.

===== 6. Conceptual interpretation =====

For an adjunction

$$
F^L:\mathcal D
\rightleftarrows
\mathcal C:F,
$$

the monad $T=FF^L$ describes how free objects in $\mathcal C$ interact.

Monadic reconstruction attempts to recover an arbitrary object $A\in\mathcal C$ from a simplicial resolution by free objects, of the form

$$
\cdots
\rightrightarrows
F^L T^2 F(A)
\rightrightarrows
F^L T F(A)
\rightrightarrows
F^L F(A).
$$

After applying $F$, the augmented resolution is split, and its realization recovers $F(A)$.

To recover $A$ itself, one needs $F$ to preserve the relevant geometric realization. Conservativity can then detect that the resulting map is an isomorphism.

Thus the two Barr–Beck conditions have different roles:

  * **Conservativity:** $F$ detects whether a morphism is an isomorphism.
  * **Preservation of $F$-split realizations:** The colimit relations used in reconstructing objects from free objects remain valid after applying $F$.

In the torsion-free abelian group example, a quotient of free objects may create torsion. The category $\mathrm{Ab}_{\mathrm{tf}}$ cannot retain that torsion and instead replaces the quotient by its torsion-free reflection.

However, the monad $FF^L$ on $\mathrm{Set}$ is the ordinary free abelian group monad, which remembers no such torsion-free restriction.

Consequently,

$$
\operatorname{Alg}_{FF^L}(\mathrm{Set})
\simeq\mathrm{Ab}
$$

rather than $\mathrm{Ab}_{\mathrm{tf}}$.

This is a concrete example showing why conservativity and the existence of a left adjoint do not suffice for monadic reconstruction.