How are expressions like $(Df)(x)$ treated in categorial set-theories (in a foundational context)?

Question

In the categorial set-theory approach to the foundations (like ETCS), an element of a set $X$ is usually taken to be a function $x : 1 \rightarrow X,$ where $1$ is a singleton set. This is analogous to dealing with $\{x\}$ rather than $x$. Anyway, this basically means that, given a function $f : X \rightarrow Y$, we should technically write $f \circ x$ as opposed to $f(x)$.

This works fine when there's only two "levels" to contend with, like elements (read: singleton subobjects) and functions, but how about when there's three? For example, what if $D$ is the differentiation operator, $f$ is a differentiable function $f : \mathbb{R} \rightarrow \mathbb{R}$, and $x$ is a real number? We want to distinguish the expression $(Df)(x)$ from $D(fx),$ since the latter is undefined, or perhaps everywhere $0$ if we view $fx$ as a constant function. Anyway, the naive approach interprets the first expression as $(D \circ f) \circ x$, and the second expression as $D \circ (f \circ x)$, which by associativity are equal, which is emphatically not what we want.

So I'm just wondering, how are expressions like $(Df)(x)$ treated?

Answer 1

Let $S \subseteq \mathbb{R}^\mathbb{R}$ be the subobject of differentiable functions.

Then $D : S \to S$.

If you have any generalized element $f \in S$: that is, a morphism $f : A \to S$, then

$$Df = D \circ f : A \to S$$

is again a generalized element of $S$.

Recall that we have a "function evaluation" morphism $\theta : \mathbb{R}^\mathbb{R} \times \mathbb{R} \to \mathbb{R}$

If you have a generalized element $x \in \mathbb{R}$: that is, $x : B \to \mathbb{R}$, then

$$(Df)(x) = \theta(D \circ f, x) : A \times B \to \mathbb{R} $$

is a generalized element of $\mathbb{R}$.

The other way,

$$ f(x) = \theta(f, x) : A \times B \to \mathbb{R} $$

is not a generalized element of $S$. Thus $D(fx)$ is nonsense.

---

That said, my instincts are that this is not the right way to deal with differentiation. Instead, you want to work in the topos $\mathscr{E} := \mathcal{Sh}(\mathbb{R})$. Then, there is a subobject $S$ of $\mathbb{R}_\mathscr{E}$ corresponding to the set of "continuously differentiable numbers" (i.e. the set of continuously differentiable scalar fields), another object $\Omega (\cong \mathbb{R}_\mathscr{E})$ corresponding to the the set of continuous differential forms. Then the differential is $d : S \to \Omega$.

$f$ fits into this picture as being an element of $S$. $x$ fits into this picture as being a geometric morphism $\mathbf{Set} \to \mathscr{E}$: so "evaluation at $x$" means pulling back from $\mathscr{E}$ to $\mathbf{Set}$ by the inverse image part of the geometric morphism.

Of course, this is just differential geometry in disguise.

Answer 2

The domain of $D$ consists of functions on $\mathbb R$. Since $fx$ is not such a function but rather an element of $\mathbb R$ (i.e., a function $1\to\mathbb R$, not a function $\mathbb R\to\mathbb R$), $D(fx)$ is, strictly speaking, undefined. Your assertion that it is everywhere $0$ tells me that you are probably identifying the number $fx$ with the constant function on $\mathbb R$ having that value. (In other words, you are forming the composite of the unique function $\mathbb R\to1$ and $fx:1\to\mathbb R$ to get a constant function $\mathbb R\to\mathbb R$, whose derivative is indeed $0$.) In that case, I would write this as either $D(f\circ x\circ *)$, where $*$ denotes the map from $\mathbb R$ to $1$, or $D(y\mapsto fx)$ (or, for people who like lambda-calculus notation) $D(\lambda y.fx)$.