What is the determinant of a set?

Question

Cross-posted from MSE where the question didn't get much attention. The question is related to and has a similar motivation as this MO question.

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Let$\newcommand{\from}{\colon}\newcommand{\sgn}{\mathrm{sgn}}\newcommand{\defby}{:=}\newcommand{\after}{\circ}\newcommand{\Vec}{\mathbf{Vec}}\newcommand{\Hom}{\mathrm{Hom}}\newcommand{\L}{\mathbf{L}}\newcommand{\Set}{\mathbf{Set}}$ $X$ be a finite set and let $\sigma \from X \to X$ be a map. When $\sigma$ is bijective, the definition of $\sgn(\sigma) \in \{\pm1\}$ is well-known, I propose to define furthermore $\sgn(\sigma) \defby 0$ if $\sigma$ is not bijective. Then the law $\sgn(\sigma \after \tau) = \sgn(\sigma)\sgn(\tau)$ remains true. Also, we still have $\sgn(\sigma) = \det(L_\sigma)$ where $L_\sigma \from \mathbb{R}^X \to \mathbb{R}^ X$ is the induced map given by acting on the basis vectors according to $\sigma$.

Speaking of determinants, in linear algebra you can do a little bit more than defining determinants of an endomorphism: You have a functor $\det \from \Vec_n \to \Vec_1$ from the category of $n$-dimensional vector spaces to the category of $1$-dimensional vector spaces which associates to every $n$-dimensional vector space $V$ its determinant line $\det(V) \defby \bigwedge^n V$ and to a morphism $f \from V \to W$ of $n$-dimensional vector spaces the map $\det(f) \defby \bigwedge^n f \from \det(V) \to \det(W)$. If $V = W$ then the composition $\Hom(V, V) \to \Hom(\det(V), \det(V)) \to \mathbb{R}$ recovers the usual definition of the determinant of an endomorphism. Here the isomorphism $\Hom(L, L) \to \mathbb{R}$ for a line $L$ is given by $f \mapsto f(e) / e$ where $\{e\}$ is some basis for $L$, the choice doesn't matter.

Given an arbitrary morphism $f \from V \to W$ in $\Vec_n$ it is still true that $f$ is an isomorphism if and only if $\det(f)$ is an isomorphism. Moreover, even though $\det(f)$ is not a number, given two morphisms $f$ and $g$ (and, say, $g$ an isomorphism) we can still consider the quotient $\det(f) / \det(g) \in \mathbb{R}$ and determine if $f$ and $g$ have "the same orientation" or more generally, if they increase the signed volume of some parallelepiped by the same amount.

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I wonder if the same also works for sets: Is there some category $\L$ (the category $\Set_1$ of one-element sets won't work) together with a functor $\det \from \Set_n \to \L$ such that it makes sense to speak about the determinant of an arbitrary map of sets $X \to Y$ of cardinality $n$ such that we can recover the sign in the case of an endomorphism. In fact, the answer is yes by 2. below, but I would like to have a more basic combinatorial description of this functor.

Of course, it is enough for $\L$ to consist of a single isomorphism class. Moreover, the unique object $L$ (up to isomorphism) of $\L$ should have the property that $\Hom(L, L) \cong \{-1, 0, 1\}$.

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Some remarks:

1. If we were to restrict our attention to the core $(\mathbf{Set}_n)_{\cong}$ (having only bijections as morphisms), then we could choose $\L$ to be the core of $\Set_2$: Objects are two-element sets and morphisms are bijections. We have $\Hom(L, L) \cong \{\pm1\}$ for an object $L$ of this category. A construction of a reasonable functor $\det \from (\Set_n)_{\cong} \to (\Set_2)_{\cong}$ is indicated e.g. in this MO answer: One can define $\det(X) \defby T_X / {\sim}$ with $T_X$ and $\sim$ defined as in the linked answer. However, I don't see how to make this work also for non-isomorphisms.

2. A definition which does work is $\L \defby \Vec(\mathbb{F}_3)_1$, the category of $1$-dimensional $\mathbb{F}_3$-vector spaces. We have $\Hom(L, L) \cong \{-1, 0, 1\}$ in this category as required. Furthermore, we can use the functor $X \mapsto \det(\mathbb{F}_3^X)$, the determinant of an $\mathbb{F}_3$-vector space being defined simiarly as in the real case above. What I don't like here is that there is so much linear algebra involved for a basically set-theoretic question. Moreover the entrance of the field $\mathbb{F}_3$ seems very ad-hoc.

3. An equivalent and more elementary way to describe the category $\L$ in 2. is as the category of free-and-transitive $M$-sets for the monoid $M = \{-1, 0, 1\}$. What would be nice then, would be a linear-algebra-free construction of the $\det$ functor.

4. I've read that $\mathbb{F}_1$-geometry might be relevant to such questions, but unfortunately, I know nothing about it.

Answer 1

Consider the category of three-element sets with a marked zero element, where a morphism $A \to B$ is a function $B \to A$ that sends the zero element to the zero element and, if it sends any nonzero element to a zero elements, sends every nonzero element to the zero element. Note that I have reversed the order of the arrows.

Then morphisms $A \to A$ are naturally in bijection with $\{1,0,-1\}$ with $0$ sending everything to zero, $1$ the identity, and $-1$ swapping the two nonzero elements.

For a finite set $S$, one can define an equivalence relation on the class of preorders of $S$ where every two preorders that are not total orders are equivalent, and two total orders are equivalent if and only if the number of pairs of elements of the set where the two total orders disagree is even.

For $S \to T$ a map of $n$-element sets, a preorder on $T$ gives by pullback a preorder on $S$. It is straightforward to check that pullback respects the equivalence relation and so induces a map on equivalence classes. One can furthermore check that, if the map is not a bijection, it sends every order to a non-total-order, and if the map is a bijection, it sends total orders to total orders and non-total-orders to non-total-orders. Hence $S\to T$ induces a map in the category of three-element sets above, giving the desired functor.