After reading through these talks by John Baez by https://arxiv.org/pdf/math/0608420.pdf , specifically "The Power Of Negative Thinking", where the $n$-categorical point of view is extended to $-1$ and $-2$ categories, I wondered: Is there a version of Yoneda's Lemma for $0$-categories, $(-1)$-categories or $(-2)$-categories?
I have trouble coming up with the formulation. This is a careful restatement of YL for $1$-categories (size issues aside):
Preliminaries:
If $C$ is a 1-category, for any two objects there is a $0$-category of morphisms.
More precisely a $1-$Functor: $$\mathrm{Hom} : C^{\mathrm{op}} \times C \rightarrow 0-\mathrm{Cat} $$
to the $1$-category of all $0$-categories $0-\mathrm{Cat}$.
(1)-PreSheaves on $\mathcal{C}$ are $1$-Functors to $0-\mathrm{Cat}$, which form a $1$-Category again. The $1$-Yoneda Embedding is the $1$-Functor
$$h:c \mapsto h_c=\mathrm{Hom}(-,c)$$
Yoneda's Lemma for 1-categories:
The function $$\mathrm{Hom}(h_c,X)\rightarrow X(c), \quad \theta \mapsto \theta_c(1_c) $$
is an isomorphism of presheaves.
Best Answer
Yoneda for $0$-categories?
A $0$-category is a set $X$, a $0$-functor is a map of sets $f : X \to Y$, and a morphism $f \to g$ (natural transformation) between two $0$-functors exists, and is unique, if and only if $f=g$. A $0$-category is naturally enriched in the discrete monoidal category $\{0,1\}$ via* $$\hom(x,y) = \begin{cases} 1 & x = y \\ 0 & x \neq y \end{cases}.$$ So when looking at representable functors, we should look at maps $f : X \to \{0,1\}$, which correspond to subsets $A \subseteq X$ via $A = \{x \in X : f(x)=1\}$. The representable functor $\hom(x,-)$ corresponds to the singleton $\{x\} \subseteq X$.
In the full Yoneda Lemma, we talk about morphisms $\hom(x,-) \to f$, which here exist, and are then unique, if and only if $f = \hom(x,-)$, i.e. $A = \{x\}$ for the subset $A \subseteq X$ corresponding to $f$. On the other hand, we have the value $f(x) \in \{0,1\}$, which is $1$ iff $x \in A$. So it seems like the Yoneda Lemma would say that $A = \{x\}$ iff $x \in A$, which is absurd.
*As of writing this, I noticed that this is actually wrong. We don't have a morphism $\hom(x,y) \times \hom(y,z) \to \hom(x,z)$ in $\{0,1\}$, namely when $x = z \neq y$! So it seems like the whole approach using enriched categories does not work, and it's not clear how to formulate Yoneda otherwise.
Enriched categories
The conceptual reason for this is that the discrete monoidal category $\{0,1\}$ is not well-behaved. Generally speaking, in enriched category theory we need a monoidal category $\mathcal{V}$ that is complete to even construct the enriched functor category, namely here we define $$\hom(F,G) := \int^{x} \hom(F(x),G(x))$$ as an end in $\mathcal{V}$. (For details, see Kelly's classic book on enriched categories.)
But the discrete category $\{0,1\}$ does not even have a terminal object! This is the conceptual reason why functor categories are ill-behaved here.
Yoneda for preorders
To establish a $0$-dimensional Yoneda Lemma, we should instead take the poset $\{0< 1\}$ with the evident monoidal structure. It is complete so everything is fine. A category enriched in $\{0 < 1\}$ is then a $(0,1)$-category, or equivalently a preordered set. The correspondence is $$\hom(x,y) = \begin{cases} 1 & x \leq y \\ 0 & \text{else} \end{cases}.$$ An enriched functor is the same as an order-preserving map, and a morphism $f \to g$ between two order preserving maps exists and is unique iff $f(x) \leq g(x)$ for all $x$.
If $X$ is a preorder, a functor (=order-preserving map) $f : X \to \{0 < 1\}$ can equivalently be described by an upper set $A \subseteq X$. If $f,f' : X \to \{0,1\}$ are two functors, we have $f \to f'$ iff $A \subseteq A'$ for the corresponding upper sets.
Next, if $x \in X$, the representable functor $\hom(x,-) : X \to \{0 < 1\}$ corresponds to the upper set $X_{\geq x} := \{y \in X : x \leq y\}$.
Therefore, the Yoneda Lemma for $(0,1)$-categories states the following:
When you want to prove this (trivial) statement, the argument will be identical to the one for the $1$-dimensional Yoneda Lemma. In particular, the proof of $\implies$ uses the universal element $x \in X_{\geq x}$.
Yoneda for $0$-categories!
Every $0$-category (=set) is a $(0,1)$-category in particular. The partial order is just very trivial: $x \leq y$ iff $x = y$. Every subset is an upper set, and $X_{\geq x} = \{x\}$. Thus, the Yoneda Lemma for $0$-categories states:
The usual corollaries hold: A subset is determined by the singletons it contains, and two singletons are equal iff they have the same element.
PS: I have also thought a bit about the Yoneda Lemma for $(-1)$-categories, of which there are exactly two ($\bot,\top$), but I doubt that anything interesting will come out here.