# Lecture 7: Step functions and exponential modalities

Teacher: Paul-André Melliès


## Step functions

$D_{A ⊸ B} = \underbrace{D_A ⊸ D_B}_{\text{domain of linear functions}}$

NB: this theory is heavily inspired by functional analysis, and in particular step functions.

$A ⊸ B = (A ⊗ B^⊥)^⊥ = A^⊥ ⅋ B$

Every pair $(a,b) ∈ \vert A \vert × \vert B \vert$ defines a linear function:

$δ_{(a,b)} = u ⟼ \begin{cases} \lbrace b \rbrace &&\text{ if } a ∈ u \\ ∅ &&\text{ else} \end{cases}$ $\Tr δ_{(a,b)} = \lbrace (\lbrace a \rbrace, b)\rbrace$

Think of $δ_{(a,b)}$ as a “step function”.

Analogously, for $u ∈ !A, \, b ∈ B$:

$δ_{(u,b)} = v ⟼ \begin{cases} \lbrace b \rbrace &&\text{ if } v ⊆ u \\ ∅ &&\text{ else} \end{cases}$

Exercise: Show that $δ_{(u,b)}: D_A ⟶ D_B$ is a stable function.

$\Tr δ_{(u,b)} = \lbrace (u, b)\rbrace$

Fact: $f$ is the union of $δ_{(u,b)}$ for $(u,b) ∈ \Tr f$ whenever $f: D_A ⟶ D_B$ is a stable function:

$f: x ⟼ \bigcup \lbrace δ_{(u,b)} \; \mid \; (u,b) ∈ \Tr f\rbrace (x) \\ = \bigcup \lbrace δ_{(u,b)}(x) \; \mid \; (u,b) ∈ \Tr f\rbrace\\ = \lbrace b \; \mid \; ∃ u ∈ !A, u ⊆ x; (u,b) ∈ \Tr f\rbrace$

In the case of linear functions $D_A ⟶ D_B$: when can we define the union of $δ_{(a,b)}$ and $δ_{(a’,b’)}$? You want to map $\lbrace a, a’\rbrace$ to $\lbrace b, b’\rbrace$, if $a \sim_A b$ and $b \sim_B b’$: this precisely defines $A ⊸ B$!

## Exponential modalities

Linear decomposition of the intuitionistic implication:

$A ⇒ B \; = \; !A ⊸ B$ $u \sim_{!A} v ⟺ \text{the union } u ∪ v \text{ is a finite clique }\\ ⟺ u ↑ v \text{ (compatible)}$ $?A = (!A^⊥)^⊥$

If $f: D_A ⟶ Σ = D_1 = D_⊥$ is a stable function determines and is determined by a set of pairwise incompatible elements of $!A$

So a stable function $D_A ⟶ Σ = D_1 = D_⊥$ is the same thing as an anticlique of $!A$, hence the same as a clique of $!A ⊸ ⊥$

More generally, a stable function $D_A ⟶ D_B$ is the same thing as a clique of $A ⊸ B$ (every stable function is the union of step functions $δ_{(u,b)}$ compatible with one another!)

$!(A \& B) \; ≅ \; !A ⊗ !B \qquad !⊤ = 1$

Categorical properties of the exponential:

$Δ_A : A ⊸ A \& A\\ Δ_A = \lbrace (a, \texttt{inl } a) \; \mid \; a ∈ \vert A \vert\rbrace ∪ \lbrace (a, \texttt{inr } a) \; \mid \; a ∈ \vert A \vert\rbrace$ $\underbrace{!A ⟶ !(A \& A) \overset{≅}{⟶} !A ⊗ !A}_{\text{comonoid structure → contraction}}$

Using the $!$, we can turn the cartesian product (which is not closed!) into some kind of diagonal for the tensor

### Monoids (category theory)

A monoid in $Set$:

$M × M \overset{m}{⟶} M$

that is associative (the evident rectangle commutes)

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