# A Criterion for Categories on which every
Grothendieck Topology is Rigid

Jérémie Marquès

###### Abstract

Let $\cC$ be a Cauchy-complete category. The subtoposes of $[\cC^{\op},\cSet]$
are sometimes all of the form $[\cD^{\op},\cSet]$ where $\cD$ is a full
Cauchy-complete subcategory of $\cC$. This is the case for instance when $\cC$
is finite, an Artinian poset, or the simplex category. In order to unify these
situations, we give two formulations of a sufficient condition. The first
formulation involves a two-player game, and the second formulation combines
two “local” properties of $\cC$.

Let $\cC$ be a small Cauchy-complete category. If $\cD⊆\cC$ is a full Cauchy-
complete subcategory, then $[\cD^{\op},\cSet]$ is a subtopos of
$[\cC^{\op},\cSet]$. The Grothendieck topologies corresponding to these
subtoposes are called _rigid_. They can be characterized using the definitions
below, taken from [1, C2.2.18] which the reader can consult for more details.

  * ✦ Given a Grothendieck topology $J$ on $\cC$, an object $X∈\cC$ is called _$J$ -irreducible_ if the only covering sieve is the maximal sieve. ✦ We say that $J$ is _rigid_ if every object admits a $J$-covering by $J$-irreducible objects.

When $J$ is rigid, the subtopos $\Sh(\cC,J)$ is equivalent to
$[\cD^{\op},\cSet]$ where $\cD$ is the full category of $\cC$ on the
$J$-irreducible objects.

In some cases, every Grothendieck topology on $\cC$ is rigid. In [4], several
categories are shown to have this property, such as the categories of
simplicial and of semi-simplicial sets. At least two general criteria can be
found in the literature:

  * ✦ When the slice categories of $\cC$ are finite [1, C2.2.21] [2, Proposition 4.10]. ✦ When $\cC$ is an Artinian poset [3].

In order for every Grothendieck topology to be rigid, it is necessary that
$\cC$ is Cauchy-complete: Otherwise,
$[\cC^{\op},\cSet]≅[\ovl{\cC}^{\op},\cSet]$ where $\ovl{\cC}$ is the Cauchy-
completion of $\cC$, and there is $X∈\ovl{\cC}$ not in $\cC$. Then the closure
of $\set{X}$ under splitting of idempotents in $\ovl{\cC}$ yields a subtopos
of $[\cC^{\op},\cSet]$ not corresponding to a rigid topology on $\cC$. But
Cauchy-completeness isn’t sufficient, for instance [3] shows that when $\cC$
is a non-Artinian poset, there are non-rigid Grothendieck topologies.

In § 1, we present a condition under which every Grothendieck topology on
$\cC$ is rigid. In § 2, we show that this condition is equivalent to the
existence of a winning strategy for some two-player game on $\cC$. In § 3, we
give another formulation of this criterion as the combination of two “local”
properties of $\cC$ involving its endomorphism monoids and the poset
reflections of its slice categories.

### Notations

In this note, $\cC$ is a small category. The composition of two arrows
$f:X\textrightarrow Y$ and $g:Y\textrightarrow Z$ will be denoted $fg$.

### Acknowledgments

This research has been supported financially by the European Research Council
(ERC) under the European Union’s Horizon 2020 research and innovation program,
grant agreement #670624.

##  1 The Sufficient Condition

We will consider $\cC$ as sitting implicitly in $[\cC^{\op},\cSet]$ via the
Yoneda embedding. By the _image_ $\im(f)$ of a morphism $f$ in $\cC$, we mean
the image of the corresponding morphism in $[\cC^{\op},\cSet]$. With this
convention, the morphisms in $\cC$ which are “surjective” are the split
epimorphisms. The poset of subobjects of $X∈\cC$ of the form $\im(f)$ with
$f:Y\textrightarrow X$ in $\cC$ can be identified with the _poset reflection_
of $\cC_{/X}$. The poset reflection of a small category is obtained by
declaring that $X≤Y$ when there is a morphism $X\textrightarrow Y$, and by
quotienting by the induced equivalence relation.

A poset is _Artinian_ if there is no strictly decreasing sequence of elements.

######  Theorem 1.

main Suppose that $\cC$ satisfies the following conditions:

  1. ✦ The poset reflection of $\cC_{/X}$ is Artinian, for each $X∈\cC$. ✦ For each $f:Y\textrightarrow X$, there is $f^{\prime}:Y^{\prime}\textrightarrow X$ with $\im(f)=\im(f^{\prime})$ and such that for every $r:Z\textrightarrow Y^{\prime}$, if $\im(f^{\prime})=\im(rf^{\prime})$ then $\im(r)=Y^{\prime}$.

Then every Grothendieck topology on $\cC$ is rigid.

Let $J$ be a Grothendieck topology on $\cC$ and let $X∈\cC$. We show that for
all $f:Y\textrightarrow X$, the image $\im(f)$ of $f$ is $J$-covered by
$J$-irreducible objects of $\cC$. We proceed by Artinian induction on
$\im(f)$. Taking $f$ the identity of $X$, this will show that $X$ is
$J$-covered by $J$-irreducible objects.

Let $f:Y\textrightarrow X$ and suppose that the property holds for all
$g:W\textrightarrow X$ with $\im(g)<\im(f)$. Thanks to the second hypothesis
on $\cC$, we can suppose that for all $r:Z\textrightarrow Y$, if
$\im(rf)=\im(f)$, then $\im(r)=Y$. If $Y$ is $J$-irreducible, then $\im(f)$ is
covered by $Y$ and we are done. Otherwise, there is some $J$-covering $S$ of
$Y$ which doesn’t generate the maximal sieve, hence which contains no split
epimorphism. We thus have $\im(rf)<\im(f)$ for every $r∈S$, and by induction
there is a $J$-covering of $\im(rf)$ consisting of $J$-irreducible objects.
This transfers to $\im(f)$ by composition of $J$-coverings.

In particular, the conditions of Theorem LABEL:thm:main imply that $\cC$ is
Cauchy complete, since it is necessary in order for every Grothendieck
topology to be rigid. Cauchy-completeness is actually a special case of the
second condition, when $f$ is idempotent.

##  2 A Game of Split Epimorphisms

The _game of split epimorphisms_ on $\cC$ is defined as follows. There are two
players _Cleaner_ and _Reducer_. The state of the game is determined by a
morphism $f$ in $\cC$ whose codomain never changes. A play by Cleaner consists
in factorizing $f$ through a split epimorphism. More explicitly, Cleaner
chooses a factorization $f=rf^{\prime}$ where $r$ is a split epimorphism, and
the state of the game changes from $f$ to $f^{\prime}$. A play by Reducer
consists in precomposing $f$ by a morphism which is _not_ a split epimorphism.
Cleaner wins if Reducer cannot make a valid move, and Reducer wins if the game
continues forever.

We could allow more generally Cleaner to factorize $f$ through _any_ morphism
$r$. However, if $r$ is not a split epimorphism, this move is useless since
Reducer can just cancel it by precomposing by $r$.

Alternatively, a move by Cleaner when the game is in position
$f:Y\textrightarrow X$ is determined by an idempotent $e:Y\textrightarrow Y$
such that $ef=f$. We split $e=rs$ and the position of the game becomes $sf$.
This corresponds to the factorization $f=r(sf)$. Note that Cleaner cannot
precompose $f$ by an _arbitrary_ split monomorphism, because of the condition
$ef=f$.

Cleaner never modifies the image of $f$ and Reducer always makes it smaller,
though not necessarily strictly smaller. We think of a game position
$f:Y\textrightarrow X$ as a “presentation” of $\im(f)⊆X$. The goal of Reducer
is to continuously make $\im(f)$ smaller, or at least _pretend to_. The goal
of Cleaner is to “clean up” that presentation, so as to prevent Reducer from
lying.

######  Proposition 2.

game The conditions of Theorem LABEL:thm:main are satisfied if and only if
Cleaner has a winning strategy.

Suppose that the conditions of Theorem LABEL:thm:main are satisfied. The
winning strategy of Cleaner in position $f:Y\textrightarrow X$ is to answer
with the morphism $f^{\prime}:Y^{\prime}\textrightarrow X$ given by the second
condition. Since $\im(f)=\im(f^{\prime})$, there is a morphism
$r:Y^{\prime}\textrightarrow Y$ such that $f=rf^{\prime}$, and $r$ is a split
epimorphism because we must have $\im(r)=Y^{\prime}$. In the next move,
Reducer is forced to produce a morphism
$f^{\prime\prime}:Y^{\prime\prime}\textrightarrow X$ with
$\im(f^{\prime\prime})<\im(f^{\prime})$. Since $\cC_{/X}$ is Artinian, the
process must terminate.

For the other direction, suppose that the poset reflection of $\cC_{/X}$ is
not Artinian for some $X∈\cC$. We produce a winning strategy for Reducer for
the game starting in position $\id_{X}:X\textrightarrow X$. Let
$(f_{i}:Y_{i}\textrightarrow X)_{i∈ℕ}$ be such that $\im(f_{i+1})<\im(f_{i})$
for all $i∈ℕ$. Then Reducer can win by playing $f_{i}$ as its $i$th move.
Cleaner answers with $f^{\prime}_{i}:Y^{\prime}_{i}\textrightarrow X$ such
that $\im(f^{\prime}_{i})=\im(f_{i})$. Since
$\im(f_{i+1})<\im(f_{i})=\im(f^{\prime}_{i})$, there is an arrow
$r:Y_{i+1}\textrightarrow Y^{\prime}_{i}$ with $rf^{\prime}_{i}=f_{i+1}$. The
inclusion being strict, $r$ is not a split epimorphism. Hence Reducer can play
$f_{i+1}$ as its next move, and so on.

Now, suppose that the second condition of Theorem LABEL:thm:main is not
satisfied. This means that there is some $f:Y\textrightarrow X$ such that for
every $f^{\prime}:Y^{\prime}\textrightarrow X$ satisfying
$\im(f)=\im(f^{\prime})$, there is an $r:Z\textrightarrow Y^{\prime}$ such
that $\im(r)<Y^{\prime}$ but $\im(rf^{\prime})=\im(f^{\prime})$. We will again
produce a winning strategy for Reducer starting from the identity
$\id_{X}:X\textrightarrow X$. The first move of Reducer is used to put the
game in position $f:Y\textrightarrow X$. This is possible since $\im(f)<X$,
otherwise we could take $f^{\prime}=\id_{X}$. Afterward, the game will always
be in a position $f^{\prime}:Y^{\prime}\textrightarrow X$ such that
$\im(f^{\prime})=\im(f)$. Cleaner cannot change this. Reducer can then use the
$r:Z\textrightarrow Y^{\prime}$ given by the hypothesis and put the game in
state $rf^{\prime}$. This is a valid move since $r$ is not a split
epimorphism.

As an example, we show Corollary LABEL:cor:degree-criterion that applies to
the simplex category.

######  Corollary 3.

degree-criterion Let $\cC$ be a small category equipped with an ordinal-valued
degree function $d:\abs{\cC}\textrightarrow\Ord$ such that every morphism
factors as a split epimorphism followed by either an isomorphism or a morphism
$f:A\textrightarrow B$ with $d(A)<d(B)$. Then every Grothendieck topology on
$\cC$ is rigid.

The winning strategy of Cleaner is to react to the moves of Reducer as
follows. Suppose that the previous move of Reducer was to precompose
$f:Y\textrightarrow X$ by $h:Z\textrightarrow Y$. Then Cleaner factorizes
$h=rg$ with $r$ the split epimorphism given by the hypothesis on the degree
function.

$\tikzcd Z\ar[r,"h"]\ar[d,->>,"r"]&Y\ar[r,"f"]X\\\
Y^{\prime}\ar[ru,"g"^{\prime}]$

The morphism $g$ cannot be invertible as it would imply that $h$ is a split
epimorphism. Hence $d(Y^{\prime})<d(Y)$. As the degree strictly decreases, the
game cannot continue forever.

##  3 Two Local Conditions

In this section, we reformulate the second condition of Theorem LABEL:thm:main
so as to only involve the endomorphism monoids of $\cC$.

We say that a monoid $M$ has _enough idempotents on the left_ if for every
$f∈M$, either $f$ has a left inverse or there is a non identity idempotent $e$
and some $n≥1$ satisfying $ef^{n}=f^{n}$. Every finite monoid has enough
idempotents on the left, by taking $e=f^{ω}$ the only idempotent power of $f$.

######  Proposition 4.

The conditions of Theorem LABEL:thm:main are equivalent to the conjunction of:

  1. ✦ $\cC$ is Cauchy-complete. ✦ The poset reflection of $\cC_{/X}$ is Artinian, for each $X∈\cC$. ✦ $\End(X)$ has enough idempotents on the left, for each $X∈\cC$.

We already saw that the conditions of Theorem LABEL:thm:main imply that $\cC$
is Cauchy-complete. Suppose that $\cC$ is Cauchy-complete and that the poset
reflections of $\cC_{/X}$ are Artinian. We must show that the endomorphism
monoids of $\cC$ have enough idempotents on the left if and only if:

$\stk{\text{\normalsize Foreachf:Y\textrightarrow
X,thereissomef^{\prime}:Y^{\prime}\textrightarrow
Xwith\im(f)=\im(f^{\prime})}\\\ \text{\normalsize
andsuchthatforeveryr:Z\textrightarrow
Y^{\prime},if\im(f^{\prime})=\im(rf^{\prime})then\im(r)=Z.}}\tag{\star}$ (1)

Suppose that the endomorphism monoids have enough idempotents on the left. Let
$f:Y\textrightarrow X$. If there is a non-identity idempotent
$e_{1}:Y\textrightarrow Y$ such that $e_{1}f=f$, we split $e_{1}$ as
$r_{1}s_{1}$ and we define $f_{1}=s_{1}f$. Then $\im(f)=\im(f_{1})$, because
$\im(f)=\im(r_{1}s_{1}f)⊆\im(s_{1}f)=\im(f_{1})$. Repeat the process as long
as there is a non-identity idempotent $e_{i+1}$ with $e_{i+1}f_{i}=f_{i}$. The
process must terminate, otherwise $\im(s_{1}),\im(s_{2}s_{1}),…$ is an
infinite strictly decreasing sequence of subobjects of $X$. Let
$f^{\prime}:Y^{\prime}\textrightarrow X$ be the morphism obtained at the last
step, so that the only idempotent $e:Y^{\prime}\textrightarrow Y^{\prime}$
verifying $ef^{\prime}=f^{\prime}$ is the identity. Let $r:Z\textrightarrow
Y^{\prime}$ such that $\im(rf^{\prime})=\im(f^{\prime})$. This means that
there is some $s:Y^{\prime}\textrightarrow Z$ such that
$srf^{\prime}=f^{\prime}$. We wish to show that $\im(r)=Z$, i.e., that $r$ has
a left inverse. Since $\End(Y^{\prime})$ has enough idempotents on the left,
there are two possibilities. Suppose there is a non identity idempotent
$e:Y^{\prime}\textrightarrow Y^{\prime}$ and $n≥1$ such that
$e(sr)^{n}=(sr)^{n}$. That would imply that
$ef^{\prime}=e(sr)^{n}f^{\prime}=(sr)^{n}f^{\prime}=f^{\prime}$, but this is
impossible by the hypothesis on $f^{\prime}$. Hence $sr$ has a left inverse,
and $r$ too.

Reciprocally, suppose that condition-2 is satisfied. As an intermediate step,
we show:

$\stk{\text{\normalsize Foreachf:Y\textrightarrow Xandg:Y\textrightarrow
Ysuchthatgf=f,eithergisleft-invertible}\\\ \text{\normalsize
orthereisanonidentityidempotente:Y\textrightarrow
Ywithef=f.}}\tag{\star\star}$ (2)

Let $f:Y\textrightarrow X$ and $g:Y\textrightarrow Y$ such that $gf=f$. Apply
condition-2 to $f$ to obtain the factorization below.

$\tikzcd X\\\
Y\ar[u,"f"]\ar[r,->>,"r",shiftleft=0.3em]&Y^{\prime}\ar[lu,"f^{\prime}"^{\prime}]\ar[l,shiftleft=0.3em,>->,"s"]$

We have $\im(grf^{\prime})=\im(f)=\im(f^{\prime})$, hence
$\im(gr)=X^{\prime}$. If $rs=\id_{X}$, this implies that $\im(g)=X$, hence
that $g$ is left-invertible. Otherwise, $rs$ is a non identity idempotent
verifying $(rs)f=f$.

We can now show that the monoids of endomorphisms of $\cC$ have enough
idempotents on the left. Let $f:X\textrightarrow X$ be an endomorphism.
Consider the sequence $f,f^{2},f^{3},…$ We have $\im(f^{i+1})⊆\im(f^{i})$. The
poset reflection of $\cC_{/X}$ being Artinian, there are $n,k≥1$ and
$g:X\textrightarrow X$ such that $gf^{k}f^{n}=f^{n}$. Applying condition-3, we
obtain either that $gf^{k}$ is left-invertible, hence $f$ too, or an
idempotent $e$ such that $ef=f$.

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