Here's a statement of Yoneda's lemma for n-category.

Soient C a n-category and $C^{\wedge}=[C^o,n-1Cat]$ the n-category of presheaves on C.

$C^o$ is the opposite n-category of C and $n-1Cat$ is the n-category of (n-1)-categories.

We have the Yeneda embedding:

$h_C:C \rightarrow C^{\wedge}$ , X being sent to the hom-presheaf $hom_C(-,X)$.

Now Yoneda's lemma says:

1. For $X \in C $ and $A \in C^{\wedge}$,

$hom_{C^{\wedge}}(h_C(X),A) \approx A(X) $ in the n-category $n-1Cat$ up to n-equivalence;

2. For $A \in C^{\wedge}$,

$hom_{C^{\wedge}}(h_C(-),A) \approx A $ in the n-category $C^{\wedge}$ up to n-equivalence.

I want to test the Yoneda in the case n=0.

Now a 0-category is, supposedly, an ensemble and a (-1)-category is either 1 or 0 (truth values). So the 0-category of (-1)-categories is

$$-1Cat=\{0,1\}$$.

Now soit C a 0-category ; $C^{\wedge}=[C^o,-1Cat]$ is nothing but the power set of C since a function of C in {0,1} defines a subset of C.

The Yoneda embedding is

$h_C:C \rightarrow C^{\wedge}$ , x being sent to the hom-presheaf $hom_C(-,x)=\{x\}$ (since $hom_C(y,x)=0$ if $y \neq x$), that is, x being sent to the singleton {x}.

So the Yoneda lemma gives:

$hom_{C^{\wedge}}(h_C(x),A) \approx A(x) = 0$ if $x \notin A$

and

$hom_{C^{\wedge}}(h_C(x),A) \approx A(x) = 1$ if $x \in A$

in the 0-category $-1Cat$ up to 0-equivalence which is equality.

Thus

$hom_{C^{\wedge}}(h_C(x),A) = 1$ as long as $x \in A$,

so this indeed makes $C^{\wedge}$ not just a set but a set with equivalence relation, also called setoid in nLab.

I don't actually expect this: Yoneda's lemma forces us to consider setoids instead of plain sets as 0-categories.

So now my question is: Given two 0-categories C and D,

[C, D] (functions of C in D) should be a 0-category, i.e., a setoid.

So what is the equivalence on [C,D]?

In the case of [C,{0,1}], a subset A and a singleton {x} are equivalent ifif x is an element of A. Two functions A and B (i.e. two sunsets of C) are equivalent ifif their intersection is not empty; indeed any subset is either equivalent to the empty set or the whole set C itself (if the 0-category C is strictly a set).

This is indeed rather confusing. I hope someone can perhaps clarify.

Or perhaps the Yoneda should not be applied in this rather trivial case at all?

Notes:

1. I use words like "soit" and "soient"; they're French words for "let be". I think you all know this. I just don't like seperate "let" and "be".

2. I use "ifif" instead of "iff".

Sorry for these bizzare usage.

Here's a statement of Yoneda's lemma for n-category.

Let C be a n-category and $C^{\wedge}=[C^o,n-1Cat]$ be the n-category of presheaves on C.

$C^o$ is the opposite n-category of C and $n-1Cat$ is the n-category of (n-1)-categories.

We have the Yeneda embedding:

$h_C:C \rightarrow C^{\wedge}$ , X being sent to the hom-presheaf $hom_C(-,X)$.

Now Yoneda's lemma says:

1. For $X \in C $ and $A \in C^{\wedge}$,

$hom_{C^{\wedge}}(h_C(X),A) \approx A(X) $ in the n-category $n-1Cat$ up to n-equivalence;

2. For $A \in C^{\wedge}$,

$hom_{C^{\wedge}}(h_C(-),A) \approx A $ in the n-category $C^{\wedge}$ up to n-equivalence.

I want to test the Yoneda in the case n=0.

Now a 0-category is, supposedly, an ensemble and a (-1)-category is either 1 or 0 (truth values). So the 0-category of (-1)-categories is

$$-1Cat=\{0,1\}$$.

Now let C be a 0-category ; $C^{\wedge}=[C^o,-1Cat]$ is nothing but the power set of C since a function of C in {0,1} defines a subset of C.

The Yoneda embedding is

$h_C:C \rightarrow C^{\wedge}$ , x being sent to the hom-presheaf $hom_C(-,x)=\{x\}$ (since $hom_C(y,x)=0$ if $y \neq x$), that is, x being sent to the singleton {x}.

So the Yoneda lemma gives:

$hom_{C^{\wedge}}(h_C(x),A) \approx A(x) = 0$ if $x \notin A$

and

$hom_{C^{\wedge}}(h_C(x),A) \approx A(x) = 1$ if $x \in A$

in the 0-category $-1Cat$ up to 0-equivalence which is equality.

Thus

$hom_{C^{\wedge}}(h_C(x),A) = 1$ as long as $x \in A$,

so this indeed makes $C^{\wedge}$ not just a set but a set with equivalence relation, also called setoid in nLab.

I don't actually expect this: Yoneda's lemma forces us to consider setoids instead of plain sets as 0-categories.

So now my question is: Given two 0-categories C and D,

[C, D] (functions of C in D) should be a 0-category, i.e., a setoid.

So what is the equivalence on [C,D]?

In the case of [C,{0,1}], a subset A and a singleton {x} are equivalent iff x is an element of A. Two functions A and B (i.e. two sunsets of C) are equivalent iff their intersection is not empty; indeed any subset is either equivalent to the empty set or the whole set C itself (if the 0-category C is strictly a set).

This is indeed rather confusing. I hope someone can perhaps clarify.

Or perhaps the Yoneda should not be applied in this rather trivial case at all?