Call two functors two functors $H,H':S\longrightarrow T$ weakly equivalent, or equivalent up to a self-equivalence of the source category, iff there exists a self-equivalence of $s:S \longrightarrow S$ such that functors $H$ and $H'\circ s$ are equivalent.

Can this property "weakly equivalent" be nicely reformulated in the language of higher category theory? or maybe homotopy theory?

Are there theorems claiming that any two functors with certain properties (not involving choice) are weakly equivalent?

That is, for such a theorem to be interesting, the properties should not explicitly involve an arbitrary choice, e.g. it should not say: choose a topology, bijection, self-equivalence and then construct the functor in the following way. Rather, a functor should be described in terms of preserving some structure etc.

Below is the original question which was phrased very confusingly, it seems. I hope now it maybe is clearer.

Say a functor is "well-defined up to a self-equivalence of the source category" by certain properties/definition/construction iff, well, for any two functors $H,H':S\longrightarrow T$ with satisfying these properties/definition/obtained by this construction, there exists a self-equivalence of $s:S \longrightarrow S$ such that functors $H$ and $H'\circ s$ are equivalent.

Is there a nice way to reformulate this property "a functor unique up to self-equivalence of the source category", say in the language of 2-categories?

Are there any interesting examples of properties/definitions/constructions NOT involving arbitrary choice and yet such that the functor is well-defined up to a self-equivalence of source category ?

I am mostly interested to see an "algebraic" definition of a functor between "algebraic" categories which is well-defined up to self-equivalence but not well-defined.

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    $\begingroup$ A 'functor' which is not well defined is not a functor. $\endgroup$ – Fernando Muro Aug 17 '12 at 8:57
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    $\begingroup$ I don't think he means "well-defined" in the usual sense. On the other hand, "well-defined" here hasn't been exactly, um, well-defined -- the opening sentence is a bit slippery. But to give an example, I think he means that if a category $C$ has products but not chosen products, then "the product functor" $C \times C \to C$, while not specified, exists and is "defined well enough" since any two choices are canonically isomorphic. A property P of functors would be "well-defining" if any two functors satisfying P are isomorphic, e.g., a universal property P. (But the question needs work.) $\endgroup$ – Todd Trimble Aug 17 '12 at 10:54
  • $\begingroup$ Thanks for your comments. Then I shall just remove the opening sentence: it seems rather to confuse than clarify....Does the second paragraph seems unclear as well ? $\endgroup$ – o a Aug 17 '12 at 11:06
  • $\begingroup$ Todd Trimble: yes, that is what I mean....But I removed the sentence your comment refers to. $\endgroup$ – o a Aug 17 '12 at 11:24
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    $\begingroup$ One problem seems to be that every functor is well-defined (in your sense, which you erased, but you mean the special case where $s$ is the identity) by some property. Given a functor $F: C \to D$, define property P on functors $G: C \to D$ by the condition that $G = F$ identically. Any two functors satisfying property P are canonically isomorphic. This silly example indicates that you probably need to sharpen what you really intend by "well-defined". $\endgroup$ – Todd Trimble Aug 17 '12 at 15:45

In short, this is equivalence of objects in the weak slice 2-category $\mathbf{CAT}/T$.

First recall the classical notion of slice category: given any category $\mathcal{C}$ and object $C$, there's a category $\mathcal{C}/C$ (called $\mathcal{C}$ sliced over $C$) whose objects are the maps in $\mathcal{C}$ with codomain $C$ and whose maps are commutative triangles.

When $\mathcal{C}$ is a 2-category, you can make the same sort of definition (so that any object $C$ of $\mathcal{C}$ gives rise to a slice 2-category $\mathcal{C}/C$), but you now have a choice to make. The objects of $\mathcal{C}/C$ are defined as before, but you could ask for the 1-cells in $\mathcal{C}/C$ to be strictly commutative triangles, or triangles that commute up to a specified invertible 2-cell, or triangles that commute up to a specified not-necessarily-invertible 2-cell (and then you have to decided which way it points). In all cases, there's an obvious way to define the 2-cells of $\mathcal{C}/C$.

Let's take the "weak" or "pseudo" version of $\mathcal{C}/C$, in which the 1-cells are triangles that commute up to a specified invertible 2-cell. Take two objects of $\mathcal{C}/C$, say $h\colon D \to C$ and $h'\colon D' \to C$. In any 2-category, there's a notion of equivalence of objects. In this case, $h$ and $h'$ are equivalent in $\mathcal{C}/C$ if and only if there's an equivalence $s\colon D \to D'$ in $\mathcal{C}$ such that $h' \circ s \cong h$.

Applied to $\mathcal{C} = \mathbf{CAT}$, this gives the notion of "weak equivalence" you mention. (It just so happens that in your setting, the domains of $H$ and $H'$ are equal.)

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  • $\begingroup$ Yes. I wish I'd said that (instead of what I did say). $\endgroup$ – Colin McLarty Aug 18 '12 at 17:18
  • $\begingroup$ Thank you! I hoped for something like this. Now, the second part of my question: is there an example of something natural, an object of CAT/T, defined up to this equivalence ? That is, where it is important that you consider this equivalence rather then something stronger...But perhaps it is better stated as a separate question. $\endgroup$ – o a Aug 19 '12 at 17:15
  • $\begingroup$ Tom Leinster: in fact, I am confused. What definition of a 2-category you are using; is it in your survey of definitions of n-categories? E.g. what is the 'specified invertible 2-cell' in CAT/T ? I assume T is the codomain category T of H:S-->T. Sorry for bothering you about this. $\endgroup$ – o a Aug 19 '12 at 17:44
  • $\begingroup$ See for instance Kelly and Street's "Review of the elements of 2-categories" from 1974. If you're interested in learning about n-categories in general, that's an excellent place to start. $\endgroup$ – Tom Leinster Aug 19 '12 at 18:47

This is a mine observation (from old personal paper), may be can help. I hope my English is understandable

I thought about this question in correlation to proposition 2.2.1 p.31 of "Indexed categories and its Applications" LNM 661:

Give a functor $F:\mathcal{C}^{op}\to Cat$ (i.e. a strict indexed category) if its objets and morphisms presheaf are representable: $\mathcal{A}(C_0, -)\cong F_0$, $\mathcal{A}(C_1, -)\cong F_1$ by Yoneda lemma domain, codomain the transformations $d_0: F_1\Rightarrow F_0$, $d_1: F_1\Rightarrow F_0$ ecc. describe a internal category $\underline{C}$ of $\mathcal{A}$ and a isomorphism $F\cong [\mathcal{C} ]$ where $[\mathcal{C}]$ is the presheaf of (small) categories defined as $|[\mathcal{C}](A)|_0=\mathcal{A}(A, C_k)\ k=0, 1$ ecc.

But what if $F$ is a pseudo.functor? ($F_0$, $F_1$ aren't presheaves, for the composition we haven't the strict coherence, but accompanied by automorphisms on codomain..) The authors of the reference above consider $F_0(A)$ as the class of (canonically) isomorphic object of $|F(A)|_0$, and for $F_1(A)$ as the class of (canonically) isomorphic morphisms (as objects of $F(A)^{\underline{2}}$). At first reading I seem that this reduction to "pseudo-skeletons" avoid the trouble of the canonical isomorphisms. I remember that the skeleton of a (small) category is make by a chose of a object for any isomorphic call, and make the full subcategory. But how is related a small category $\mathcal{C}$ with its pseudo-skeleton $pS(\mathcal{C})$?

Of course we have the natural functor (make the quotient maps) $\pi: \mathcal{C}\to pS(\mathcal{C})$, now, try to define a reciprocal (equivalence we hope) functor $P$, merely a "chose functor": then $P$ map any object $\xi\in pS(\mathcal{C})$ to a its representant $X_{\xi}\in \mathcal{C}$ with $[X_\xi]=\xi$, and any morphism $\alpha: \xi\to \xi'$ to a its representant $f_\alpha: X_\xi\to X_{\xi'}$ with $[f_\alpha]=\alpha$,

(observe that a different chose $g_{\xi}$ is related to $f_\xi$ by automorphisms on domain and codomain $X_\xi$, $X_{\xi'}$ this is a generalization to the initial question)

we can chose $P(1_\xi)=1_{X_\xi}$ of course, but in general $f_{\gamma\circ\xi}\neq f_{\gamma}\circ f_\xi$ but these are related by automorphisms on domain and codomain.

Then I consider 2-category $G(\mathcal{C})$ make as follow: take the usual skeleton $Sk(\mathcal{C})$, then for $f, g: X\to Y$ define 2-cell $f\Rightarrow g$ as a couple of isomorhisms $s: X\to X,\ t: Y\to Y$ with $t\circ f=g\circ s$ (with natural composition and identities) then I have a correct 2-equivalences (make by a 2-functor and a pseudo-functor)) from $G(\mathcal{C})$ and its local skeleton (take skeleton on Hom-categories) that is just the pseudo-skeleton $pS(\mathcal{C})$ .

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