TITLE: Why no morphisms from the contradictory proposition to the inconsistent context?
QUESTION [6 upvotes]: Consider Higher order predicate logic over dependent type theory (DPL) as defined in Chapter 11 of B. Jacobs's book "Categorical Logic and Type Theory" (though I think this question applies to first-order predicate logic too).
In the category $\mathbb{P}$ of propositions-in-dependent-type-contexts, an object is a well formed proposition $\Gamma \vdash \varphi:Prop$ and a morphism $(\Gamma \vdash \varphi:Prop) \rightarrow (\Delta \vdash \psi : Prop)$ consists of a context morphism $\vec{M} : \Gamma \rightarrow \Delta$ such that $\Gamma | \varphi \vdash \phi(\vec{M})$ is derivable.  This category $\mathbb{P}$ is fibred over the category of dependent type contexts $\mathbb{C}$ via $(\Gamma \vdash \varphi:Prop) \mapsto \Gamma$.
Consider the "contradictory proposition" object, $A$, in $\mathbb{P}$ which I define as $\emptyset \vdash \perp : Prop$ and the "inconsistent context" object, $B$, which I define as $x : 0 \vdash \top:Prop$ (where $0$ is the empty type).
I would expect $A$ and $B$ to be isomorphic in $\mathbb{P}$ since both objects seems equally void to me in the sense that neither appear to have any models, i.e. there are no morphisms from the terminal object, $1 := \emptyset \vdash \top : Prop$, to either $A$ or $B$. But, of course, $A$ and $B$ cannot be isomorphic since a morphism from $A$ to $B$ entails the existence of a context morphism $M : \emptyset \rightarrow x : 0$ which would imply that the void type is inhabited.
Can someone assuage my concerns about $A$ and $B$ being equally void yet not isomorphic?

REPLY [6 votes]: My understanding of the situation is that you have to correctly understand what the morphisms do. A geometric picture might be helpful.
Let us interpret a context $\Gamma = x_1 : A_1, \ldots, x_n : A_n$ as a cartesian product $|\Gamma|$ of topological spaces. A proposition in a context $\Gamma \vdash \phi$ is interpreted as a subspace of a cartesian product $|\phi| \subseteq |\Gamma|$. I think the confusion arises if you think of a morphism $M : (\Gamma \vdash \phi) \to (\Delta \vdash \psi)$ as a continuous map $|\phi| \to |\psi|$, which it is not. It is an ambient map, i.e, a continuous map $|\Gamma| \to |\Delta|$ which also happens to map $|\phi| \to |\psi|$. Now it is clear that even though $|\phi|$ and $|\psi|$ may be homeomorphic as spaces, the may be embedded in $|\Gamma|$ and $|\Delta|$ in wildly different way (the wild Cantor sets come to mind), so it won't be possible to extend a homeomorphism between them to an ambient map.
In your example, we are comparing $\emptyset$ as a subspace of $\emptyset$ versus $\emptyset$ as a subspace of $1$. Even though the two empty subspaces are homeomorphic, there is no ambient map taking the latter to the former.