What are the subfunctors of $\operatorname{Hom}_{\Bbb Q}(q,-)$

Question

Let $\Bbb Q$ be the linearly ordered set of rational numbers, and $q\in\Bbb Q$ . I think the functor $\operatorname{Hom}_{\Bbb Q}(q,-)$ is in bijection with $A=\{t : t\in \Bbb Q, t>q\}$. I want to see if it is possible to show using Dedekind's that the subfunctors of $\operatorname{Hom}_{\Bbb Q}(q,-)$ are in bijection with $\{r: r\in\text{ extended real numbers}, r>q\}?$

(I need this to prove in category $Sets^Q$ theory, the subobject classifier $S$ has $S (q)=\{r: r\in\text{ extended real numbers}, r>q\}$ ). Please help....

Answer 1

For the purpose of this answer, a Dedekind cut is a subset $D$ of $ℚ$ that is upwards closed: for all $x ∈ D$ and $y ∈ ℚ$ with $x ≤ y$ we have also $y ∈ D$. The linearly ordered set of extended real numbers is given by $\overline{ℝ} = \{ -∞ \} ∪ ℝ ∪ \{ ∞ \} = [-∞, ∞]$.

Dedekind cuts and extended real numbers

We recall:

• For $r ∈ \overline{ℝ}$, the set $\{ q ∈ ℚ \mid q ≥ r \}$ is a Dedekind cut.

• Every Dedekind cut admits an infimum in $\overline{ℝ}.

These two constructions are mutually inverse, and give a bijection between the set of Dedekind cuts and $\overline{ℝ}$. Given two Dedekind cuts $D$ and $D'$ with associated extended real numbers $r$ and $r'$, we have $D ⊆ D'$ if and only if $r' ≤ r$.

Dedekind cuts and functors

Let $F$ be a functor from $ℚ$ to $\newcommand{\Set}{\mathsf{Set}}\mathbf{Set}$.

• The set $$ \newcommand{\supp}{\operatorname{supp}} \supp(F) ≔ \{ q ∈ ℚ \mid \text{$F(q)$ is non-empty} \} \,. $$ is a Dedekind cut. To see this, let $x ∈ \supp(F)$ and $y ∈ ℚ$ with $x ≤ y$. There exists a (unique) morphism $i \colon x \to y$ in $ℚ$. The induced map $F(i)$ goes from the set $F(x)$ to the set $F(y)$. Since the set $F(x)$ is non-empty, the set $F(y)$ must again be non-empty. Therefore, $y ∈ \supp(F)$.

Intuitively speaking, this tells us there is a clear point $r ∈ \overline{ℝ}$ at which the functor $F$ start doing something: $F$ is trivial below $r$, but never vanishes above $r$.

• For every Dedekind cut $D$ we can form a subfunctor $F|_D$ of $F$ by cutting off everything below $D$. More explicitly, $$ F|_D(q') ≔ \begin{cases} F(q') & \text{for $q' ∈ D$}, \\ ∅ & \text{otherwise}, \end{cases} $$ for every $q' ∈ ℚ$. We have $\supp(F|_D) = \supp(F) ∩ D$, whence $\supp(F) ⊆ D$ if and only if $F|_D = F$.

• For every subfunctor $F'$ of $F$ we have $\supp F' ⊆ \supp F$.

For the arbitrary functor $F$ we have no control about what it does outside of its support. The situation is different for representable functors:

Dedekind cuts and $\newcommand{Hom}{\operatorname{Hom}} \Hom$-functors

Let $q$ be a rational number and let us consider the functor $\Hom_ℚ(q, -)$ from $ℚ$ to $\mathbf{Set}$.

• For every $q' ∈ ℚ$ the set $\Hom_ℚ(q, q')$ consists of at most one element. Any subfunctor $F'$ of $\Hom_ℚ(q, -)$ is therefore uniquely determined by its Dedekind cut $\supp(F')$. Indeed, for $q' ∈ \supp(F')$ we have $F'(q') = ∅$, and for $q' ∉ \supp(F')$ we have $F'(q') = \Hom_ℚ(q, q')$.

Let $D ≔ \supp \Hom_ℚ(q, -)$. Every subfunctor of $\Hom_ℚ(q, -)$ is uniquely determined by its support, which is a Dedekind cut contained in $D$. Conversely, every Dedekind cut $D'$ with $D' ⊆ D$ arises in this way, since it is the support of $\Hom_ℚ(q, -)|_{D'}$.

We hence find that the assignment $\supp$ gives us a bijection between subfunctors of $\Hom_ℚ(q, -)$ and Dedekind cuts $D'$ with $D' ⊆ D$. These Dedekind cuts correspond precisely to those extended real numbers $r'$ with $r' ≥ q$.