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In 1-category theory a representation of a 1-functor $F:C\to Set$ is a 0-cell $X$ in $C$ together with a universal element $u\in FX$ such that the transformation $C(X,-)\to F$ is an isomorphism (=a 1-equality) in the 1-category of functors $Fun(C,Set)$. It is well known how unique a representation $(X,u)$ is: unique up to a unique compatible isomorphism.

To see the uniqueness of the representation better, one can unfold $F$ into its category of elements $el(F)$. The representations of $F$ are then precisely the initial objects of $el(F)$. When we consider only the full subcategory $rep(F)$ of representations of $F$ in $el(F)$, and if we consider the unique functor $rep(F)\to \mathbf 1$, the one point category, then we see that this functor is an equivalence. This means that $rep(F)$ is contractible and the representation is unique up to contractibility.

Can we mimic this in the (weak) 2-categorical setting?

All my $2$-categories, functors and transformations are per default weak (but not lax). A representation of a $2$-functor $F:C\to Cat$ is a pair $(X,u)$, where $X$ is a $0$-cell of $C$ and $u$ an object of $F X$, such that the transformation of 2-functors $$C(X,-)\to F$$ induced by the 2-Yoneda lemma is an equivalence in the $2$-category $2Cat(C,Cat)$. How unique is a representation of $F$?

Is it possible to unfold $F$ into a (weak) 2-category $el(F)$ such hat the representations are precisely (weak) 2-initial objects in that 2-category, and such that the full sub-2-category $rep(F)$ of representations is contractible? By this I mean that the 2-functor $rep(F)\to \mathbf 1$ into the one-point 2-category is a 2-equivalence.

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    $\begingroup$ I think a better strategy is to use the Yoneda lemma: a representation of a 1-functor $F$ is the same thing as a natural isomorphism $h^X\cong F$, and if you have two of these, say $h^X\cong F$ and $h^Y\cong F$, then $h^X\cong h^Y$ too, and the Yoneda lemma then gives an isomorphism $X\cong Y$. And this time this argument generalises well to bicategories: we just need to replace the Yoneda lemma with the bicategorical Yoneda lemma =) $\endgroup$
    – Emily
    Commented Nov 9, 2022 at 17:59
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    $\begingroup$ @Emily Yes I know that. But I want to see that the representation is not just unique up to an equivalence, but that the equivalence is in some sense canonical (like the "unique up to a unique compatible isomorphism" in 1-categories =). $\endgroup$
    – Nico
    Commented Nov 9, 2022 at 18:16
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    $\begingroup$ Doesn't it follow from the fact that if $X$ is any (higher) groupoid, and $x\in X$, then $X_{/x}$ is contractible ? Here you're talking about some groupoid of representables, and its slice over $F$ : if $F$ is representable, then this groupoid is equivalent to a slice over a represented object (by the $2$-Yoneda lemma), and so fits into the previous result. $\endgroup$
    – Maxime Ramzi
    Commented Nov 10, 2022 at 15:49
  • $\begingroup$ @MaximeRamzi I have not studied higher categories yet, so if you could describe the slice category explicitly for me, I'd be very happy! $\endgroup$
    – Nico
    Commented Nov 10, 2022 at 15:56

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All my 2-categories, 2-functors, 2-(co)limits etc. are per default weak (or pseudo- or bi- if you so like).

One can unfold a 2-functor $M: K\to Cat$ into the 2-category $el(M)$ of its elements, and any representation of $M$ will be a 2-initial object in that 2-category. Unfortunately, and in contrast to the 1-categorical case, a 2-initial object in $el(M)$ need not be a representation of $M$. Still, the full sub-2-category of representations in $el(M)$ will be 2-contractible (if a representation exists), so that a 2-representation is unique up to a 2-contractible choice.

Definition 1. The 2-category $el(M)$ has the following cells.

  • The 0-cells are pairs $(C,m)$, where $m\in M(C)$.

  • A 1-cell $(C,M)\to (D,n)$ consists of a 1-cell $f:C\to D$ in $K$ together with an isomorphism $f_\ast m \cong n$. Here $f_\ast m$ is an abbreviation for $(Mf)(m)$. I will not give extra notation the isomorphism $f_\ast m\cong n$, although it is part of the data of the 1-cell.

  • A 2-cell $\theta:f\to g$ in $el(M)$ is a $2$-cell $\theta:f\to g$ in $K$ such that $\theta_\ast m$ is the composite $f_\ast m\cong n=n\cong g_\ast m$.

The composition of 2-cells is just that of $K$. Given 1-cells $f:(C,m)\to (D,n)$ and $g:(D,n)\to (E,l)$ in $el(M)$, we let their composite be the morphism $gf$ together with the isomorphism $$(gf)_\ast m \cong g_\ast(f_\ast m)\cong g_\ast n\cong l$$The coherence cells of $el(M)$ are those of $K$. I am skipping over a lot of details here, but it feels like it should work out (this is always a dangerous thing to say).

Definition 2. A 2-initial object in a 2-category $E$ is an object $I$ such that for each other $0$-cell $X$ the 1-category $E(I,X)$ is 1-contractible. More explicitly, this means that there is one and precisely one invertible $2$-cell between any two $1$-cells $I\to X$, and $E(I,X)$ is not empty.

Claim 1. If $(X,u)$ is a representation of $M$, then $(C,u)$ is 2-initial in $el(M)$.

Proof. Let $(A,m)$ be any other 0-cell in $el(M)$. The fact that "evaluation at $u$" is an equivalence $$K(X,A)\to M(A)$$ means that I can find some $f:X\to A$ together with an isomorphism $f_\ast u \cong m$. This means that there is at least one 1-cell $(X,u)\to (A,m)$. Assume that $g$ is a second such 1-cell in $el(M)$. Then $g$ comes equipped with a choice of an isomorphism $g_\ast u \cong m$. A 2-cell $f\to g$ in $el(M)$ consists of a 2-cell $\theta: f\to g$ in $K$ such that $\theta_\ast m$ is the composite $f_\ast u \cong m \cong g_\ast u$.The fact that $K(X,A)\to M(A)$ is an equivalence of 1-categories tells me also that there is one and precisely one such $\theta$. $\square$

Let us denote by $rep(M)$ the full 2-subcategory of $el(M)$ with 0-cells the 2-representations of $M$.

Claim 2. The 2-category $rep(M)$ is 2-contractible. This means that the 2-functor $rep(M) \to pt$ is a 2-equivalence.

Proof. What this means in elementary terms is that each hom-1-category in $el(M)$ must be 1-contractible. That is, whenever there are two representations $(A,a)$ and $(B,b)$, then there must be at least one 1-cell $(A,a)\to (B,b)$, and whenever there is another one then the two 1-cells are comparable by a unique 2-cell which is also invertible. But we see immediately that this is the case, because we have already shown that all the representations are 2-initial objects. $\square$

Claim 3. Unfortunately, being 2-initial in $el(M)$ doesn't automatically mean that you are a 2-representation of $M$. So while $el(M)$ answers the question how unique representations are, it does not characterize them.

Proof. This is in length discussed in the paper Bi-initial objects and bi-representations are not so different by tslil clingman AND LYNE MOSER, which by accident asks precisely my question in the middle of page 2. :) I only found the text weeks after I have asked the question. It also contains some positive results. $\square$

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    $\begingroup$ "Weak initial object" is perhaps not the best choice of terminology here, since the meaning of "object with a not-necessarily-unique map to every object is already taken." For that matter, I don't understand all this use of "weak 2-category" when the word "bicategory" already exists. $\endgroup$
    – Kevin Carlson
    Commented Jan 19, 2023 at 1:23
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    $\begingroup$ @KevinArlin It should be better now. "Weak 2-initial" probably would have also been less confusing. $\endgroup$
    – Nico
    Commented Jan 19, 2023 at 8:51

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