\section{Choice for finite sets\titledagger}

\label{sec:choicefin}

This section is a short overview of how group theory is involved in

relating different choice principles for families of finite sets. A

paradigmatic case is that if we have choice for all families of

$2$-element sets, then we have choice for all families of $4$-element

sets.%

\footnote{This is due to Tarski,

see~\citeauthor{Jech-AC}\footnotemark{}, p.~107.}\footcitetext{Jech-AC}

The axiom of choice is a principle that we may add to our type theory

(it holds in the standard model), but there are many models where it doesn't hold.

\begin{principle}[The Axiom of Choice]\label{pri:ac}

For every set $X$ and every family of \emph{non-empty} sets

$P : X \to \Set_{\ne\emptyset}$,

there exists an dependent function of type $\prod_{x:X}P(x)$.

In other terms, for any set $X$ and any family of sets $P:X\to\Set$,

we have

\begin{equation}\label{eq:ac-impl}

\prod_{x:X}\Trunc{P(x)} \to \Trunc[\bigg]{\prod_{x:X}P(x)}.\qedhere

\end{equation}

\end{principle}

\begin{remark}

We have an equivalence between the Pi-type $\prod_{x:X}P(x)$ and the

type of sections of the projection map $\prj_1 : \sum_{x:X}P(x) \to X$,

under which families of non-empty sets correspond to surjections between sets

(using that $X$ is a set).

Thus, the axiom of choice equivalently says that any surjection

between sets admits a section.

Because of this equivalence, we'll sometimes also call elements of the

Pi-type \emph{sections}.

\end{remark}

The following is usually called Diaconescu's theorem\footcite{Diaconescu} or the Goodman--Myhill theorem\footcite{Goodman-Myhill}, but it was first observed in a problem in Bishop's book on constructive analysis~\footcite{Bishop}.

\begin{theorem}

The axiom of choice implies the law of the excluded middle, \cref{pri:lem}.

\end{theorem}

\begin{proof}

Let $P$ be a proposition, and consider the quotient map $q : \bn 2 \to \bn 2/\sim$,

where $\sim$ is the equivalence relation on $\bn 2$ satisfying $(0 \sim 1) = P$.

Like any quotient map, $q$ is surjective, so by the axiom of choice,

and because our goal is a proposition,

it has a section $s : \bn 2/\sim \to \bn 2$.

That is, we also have $q\circ s = \id$.

Using decidable equality in $\bn 2$, check whether $s([0])$ and $s([1])$ are equal

or not.

If they are, then we get the chain of identifications

$[0] = q(s([0])) = q(s([1])) = [1]$, so $P$ holds.

If they aren't, then assuming $P$ leads to a contradiction, meaning $\lnot P$ holds.

\end{proof}

We'll now define some restricted variants of the axiom of choice,

that however are not always true,

and our goal is to see how they relate to each other and to other principles.

\begin{definition}

Let $\AC$ denote the full axiom of choice, as in~\cref{pri:ac}.

If we fix the set $X$, and consider \eqref{eq:ac-impl} for arbitrary families $P:X\to\Set$, we call this the \emph{$X$-local axiom of choice}, denoted $\lAC{X}$.

If we restrict $P$ to take values in $n$-element sets, for some $n:\NN$,

we denote the resulting principle $\AC(n)$.

(That is, here we consider families $P : X \to \BSG_n$.)

If we both fix $X$ and restrict to families of $n$-element sets,

we denote the resulting principle $\lAC{X}(n)$.

\end{definition}

\begin{xca}

Show that $\lAC{X}$ is always true whenever $X$ is a finite set.

\end{xca}

\begin{lemma}\label{lem:ac-impl-triv-coh-sets}

If $\lAC{X}$ holds for a set $X$,\marginnote{%

In fancier language, this says that the axiom of choice

implies that all cohomology sets $\constant{H}^1(X,G)$ are trivial.}

then $\Trunc{X \to \BG}_0$ is contractible for any group $G$.

\end{lemma}

\begin{proof}

Suppose we have a map $f : X \to \BG$.

We need to show that $f$ is merely equal to the constant map.

Consider the corresponding family of sets

consisting of the underlying sets of the $G$-torsors represented by

$f(x) : BG$, for $x:X$.

That is, define $P : X \to \Set$ by setting $P(x) \defeq (\shape_G = f(x))$.

Since $\BG$ is connected, this is a family of non-empty sets,

so by the axiom of choice for families over $X$,

there exists a section.

Since we're proving a proposition, let $s : \prod_{x:X}(\shape_G = f(x))$

be a section.

Then $s$ identifies $f$ with the constant map, as desired.

\end{proof}

We might wonder what happens if we consider general \inftygps $G$

in~\cref{lem:ac-impl-triv-coh-sets}.

Then the underlying type of a $G$-torsor is no longer a set, but can be any type.

Correspondingly, we need an even stronger version of the axiom of choice,

where the family $P$ is allowed to be arbitrary.

Let $\AC_\infty$ denote this untruncated axiom of choice,

and let $\lAC{X}_\infty$ denote let local version, fixing a set $X$.

This is connected to another principle, which is much more constructive,

yet still not true in all models.

\begin{principle}[Sets Cover]\label{pri:sc}

For any type $A$, there exists a set $X$ together with a surjection $X \to A$.

\end{principle}

We abbreviate this as $\constant{SC}$.

\begin{xca}

Prove that the untruncated axiom of choice, $\AC_\infty$,

is equivalent to the conjunction of the standard axiom of choice, $\AC$,

and the principle that sets cover, $\constant{SC}$.

\end{xca}

\begin{xca}

Prove that we cannot relax the requirement that $X$ is a set

in the axiom of choice.

Specifically, prove that $\lAC{\Sc}(2)$ is false

\end{xca}

We now come to the analogue of~\cref{lem:ac-impl-triv-coh-sets}

for arbitrary \inftygps.

\begin{xca}

Prove that if the untruncated $X$-local axiom of choice, $\lAC{X}_\infty$,

holds for a set $X$,

then $\Trunc{X \to \BG}_0$ is contractible for all \inftygps $G$.

\end{xca}

We now discuss two partial converses to~\cref{lem:ac-impl-triv-coh-sets},

both due to \citeauthor{Blass-Finite-Choice}\footcite{Blass-Finite-Choice}.

\begin{theorem}[Blass]\label{thm:Blass-1}

Let $X$ be a set such that $\Trunc{X \to \BG}_0$ is contractible

for all groups $G$.

Then every family of non-empty sets over $X$, $P : X \to \Set$,

that factors through a connected component of $\Set$,

merely admits a section.

\end{theorem}

\begin{proof}

We suppose $P : X \to \Set$ is such that all the sets $P(x)$

have the same size, \ie the function $P$ factors through

$\BAut(S)$ for some non-empty set $S$.

This in turn means that we have a function $h : X \to \BG$,

where $G \defeq \Aut(S)$, with $P = \prj_1 \circ h$,

where $\prj_1 : \BAut(S) = \sum_{A : \Set}\Trunc{S \simeq A} \to \Set$

is the projection.

By assumption, $h$ is merely equal to the constant family.

But since we are proving a proposition, we may assume that $h$

\emph{is} constant, so $P$ is the constant family at $S$.

And this has a section since $S$ is non-empty.

\end{proof}

Obviously, the same argument works if we consider all \inftygps $G$

and families of types that are all equivalent.

For the second partial converse, we look at decidable sets.

\begin{theorem}[Blass]\label{thm:Blass-2}

Let $X$ be a decidable set such that $\Trunc{X \to \BG}_0$ is contractible

for all groups $G$.

Then every family of non-empty decidable sets over $X$

merely admits a section.\footnote{%

We might call this conclusion $\lAC{X}^{\mathrm{dec}}$.}

\end{theorem}

\begin{proof}

Equivalently, consider a surjection $p : Y \to X$,

where $X$ and $Y$ are decidable sets,

and let $C$ be the higher inductive type with constructors

$c : C$, $f : X \to C$, and $k : \prod_{y:Y}(c = f(p(y))$.\footnote{%

This kind of higher inductive type is also known as a pushout,

and its constructors fit together to give a commutative square:

\[

\begin{tikzcd}[ampersand replacement=\&]

Y \ar[d]\ar[r,"p"] \& X \ar[d,"f"] \\

\bn 1 \ar[r,"c"'] \& C

\end{tikzcd}

\]}

Using the same kind of argument as in~\cref{lem:wedgeofgpoidisgpoid}

and~\cref{thm:free-group-elements}, we can show,

using decidability of equality in $X$ and $Y$,

that the identity type $c =_C f(x)$ is equivalent to a type of reduced words

over $Y \coprod Y$.

In particular, $C$ is a groupoid, and it's easy to check that it's connected.

Hence we can form the group $G \defeq \mkgroup(C,c)$.

By assumption, the map $f$ is merely equal to the constant map,

so since we're proving a proposition,

we may assume we have a family of elements $h(x) : c = f(x)$, for $x:X$.

Taking for each $x$ the last $y$ in the corresponding reduced word,

we get a family of elements $s(x) : Y$

such that $p(s(x)) = x$,

but this is precisely the section we wanted.

\end{proof}

It seems to be an open problem, whether we can do without the decidability assumption,

\ie whether the converse of~\cref{lem:ac-impl-triv-coh-sets} holds generally.

Now we turn as promised to the connections between the various local choice principles

$\lAC{X}(n)$.

The simplest example is the following.

\begin{theorem}\label{thm:lAC-2-3-4}

Let $X$ be any set. Then $\lAC{X}(4)$ follows from $\lAC{X}(2)$ and $\lAC{X}(3)$.

\end{theorem}

\begin{proof}

Let $P : X \to \BSG_4$ be a family of $4$-element sets over $X$.

Consider the map $\Bf : \BSG_4 \to \BSG_3$ that maps a $4$-element set

to the $3$-element set of its $2+2$ partitions.

Choose a section of $\Bf \circ P$ by $\lAC{X}(3)$.

Now use $\lAC{X}(2)$ twice to choose for each chosen partition

first one of the $2$-element parts, and secondly one of the $2$

elements in each chosen part.

\end{proof}

We now look a bit more closely at what happened in this proof,

so as to better understand the general theorem.

The key idea is the concept of ``reduction of the structure group''.

[TODO, Elaborate: For a family of $n$-element sets over a base type $X$, $P : X

\to \BSG_n$, there is a section if and only if there is a

`` to a subgroup of $\SG_n$,

whose action on the standard $n$-element set, $\bn n$, has a fixed point.]

Now we return to the local case, and we give the general

sufficient condition that ensures that $\lAC{X}(n)$ follows from $\lAC{X}(z)$ for each $z:Z$, where $Z$ is a finite subset of $\NN$.

\begin{definition}

The condition $L(Z,n)$ is that for every finite subgroup $G$ of $\SG_n$

that acts on $\bn n$ without fixed points,

there exists finitely many proper, finite subgroups $K_1,\cdots,K_r$ of

$G$ such that the sum of the indices,

\[

|G \idxcolon H_1| + \cdots + |G \idxcolon H_r|,

\]

lies in $Z$.

\end{definition}

We now turn to the global case, where we can change the base set.

Here the basic case is Tarski's result alluded to above,

which shows that we don't need choice for $3$-element sets,

in contrast to the local case, \cref{thm:lAC-2-3-4}.

\begin{theorem}

$\AC(2)$ implies $\AC(4)$.

\end{theorem}

\begin{proof}

Let $P: X \to \BSG_4$ be a family of $4$-element sets indexed by a set $X$.

Consider the new set $Y$ consisting of all $2$-element subsets

of $P(x)$, as $x$ runs over $X$,

\[

Y \defeq \sum_{x:X}[P(x)]^2.

\]

The set $Y$ carries a canonical family of $2$-element sets,

so we may choose an element of each.

In other words, we have chosen an element of each of the $6$

different $2$-element subsets of each of the $4$-element sets

$P(x)$.

For every $a : P(x)$, let $q_x(a)$ be the number of $2$-element

subsets $\set{a,b}$ of $P(x)$ with $b\ne a$ for which $a$ is the

chosen element.

Define the sets $B(x) \defeq \setof{a:P(x)}{\text{$q_x(a)$ is a

minimum of $q_x$}}$, and remember that they are subsets of $P(x)$.

This determines a decomposition of $X$ into three parts $X = X_1 +

X_2 + X_3$, where

\[

X_i \defeq \sum_{x:X}(\text{$B(x)$ has cardinality $i$}),

\quad i = 1,2,3.

\]

Note that $B(x)$ can't be all of $P(x)$,

since that would mean that $q_x$ is constant,

and that is impossible, since the sum of $q_x$ over the $4$-element $P(x)$ is $6$.

Over $X_1$, we get a section of $P$ by picking the unique element

in $B(x)$.

Over $X_3$, we get a section of $P$ by picking the unique element

\emph{not} in $B(x)$.

Over $X_2$, we get a section of $P$ by picking the already chosen

element of the $2$-element set $B(x)$.

\end{proof}

The following appears as Theorem~6 in Blass\footcite{Blass-Finite-Choice}.

\begin{theorem}

Assume $\Trunc{X \to \BCG_n}_0$ is contractible for all sets $X$ and

positive integers $n$. Then $\AC(n)$ holds for all $n$.

\end{theorem}

\begin{proof}

We use well-founded induction on $n$, the case $n\jdeq 1$ being trivial.

Let $P : X \to \BSG_n$ be a family of $n$-element sets,

and let $Y \defeq \sum_{x:X}P(x)$ be the domain set of this \covering.

Consider the family $Q : Y \to \BSG_{n-1}$ defined by

\[

Q((x,y)) \defeq \setof{y' : P(x)}{y \ne y'} = P(x) \setminus \set{y},

\]

where we use the fact that $P(x)$ is an $n$-element set

and thus has decidable equality,

so we can form the $(n-1)$-element complement $P(x) \setminus

\set{y}$.

By induction hypothesis, we get a section of $Q$, which we can

express as a family of functions

\[

f : \prod_{x:X}\bigl(P(x) \to P(x)\bigr)

\]

where $f_x(y) \ne y$ for all $x,y$.

Since $P(x)$ is an $n$-element set, we can decide whether $f_x$

is a permutation or not, and if so, whether it is a cyclic

permutation.

We have thus obtained a partition $X = X_1 + X_2 + X_3$,

where

\begin{align*}

X_1 &\defeq \setof{x:X}{\text{$f_x$ is not a permutation}}, \\

X_2 &\defeq \setof{x:X}{\text{$f_x$ is a non-cyclic permutation}}, \\

X_3 &\defeq \setof{x:X}{\text{$f_x$ is a cyclic permutation}}.

\end{align*}

We get a section of $P$ over $X_1$ by induction hypothesis

by considering the family of the images of $f_x$.

We get a section of $P$ over $X_2$ by first choosing a cycle of $f_x$

(there are fewer then $n$ cycles because there are no $1$-cycles),

and then choosing an element of the chosen cycle.

We get a section of $P$ over $X_3$ by the assumption applied

to the map $X_3 \to \BCG_n$ induced by equipping each $P(x)$ with

the cyclic order determined by the cyclic permutation $f_x$.

\end{proof}

[TODO: State the general positive result due to

Mostowski\footcite{Mostowski-Finite-Choice}, maybe as an exercise

and give references to the negative results, due to Gauntt (unpublished).]

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