0
$\begingroup$

Take a category $C$. Define an endofunctor $F:C \rightarrow C$ that is identity on objects. This maps morphisms to morphisms, preserving source and target. This suggests that the endofunctor endows the category $C$ with 2-category structure. Does this work? You might say $ \langle C, F \rangle $ is a 2-category.

I am just building on the idea that the 2-cells are given by $f \Rightarrow F(f)$. I haven't considered at all how the identities on morphisms are defined. Perhaps they can be trivially assumed. Furthermore, I have not considered composition. Perhaps there are special endofunctors that define composition.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ "This suggests that..." How exactly does it suggest that? Are you drawing an analogy to something else? $\endgroup$
    – S.C.
    Commented Aug 2, 2022 at 2:09
  • 3
    $\begingroup$ After rereading, I think I understand what you're getting at. You'd have a 2-cell $f \Rightarrow F(f)$. But what are the identities? Are you just freely adding them in? Composition? $\endgroup$
    – S.C.
    Commented Aug 2, 2022 at 2:18
  • $\begingroup$ @S.C. Thank you for considering this. I have updated the question. $\endgroup$
    – Ben Sprott
    Commented Aug 2, 2022 at 10:48
  • $\begingroup$ The category $Fun(C,C)$ is a $2$-category: there is one object, the $1$-cells are the functors $G:C\to C$ and the $2$-cells are natural transformations of endofunctors. We can take the $2$-full subcategory of the one above with $1$-cells the iterations of your endofunctor $F^n:C\to C$, $n\geq 0$. $\endgroup$
    – D.-C. Cisinski
    Commented Aug 2, 2022 at 13:09

1 Answer 1

Reset to default
1
$\begingroup$

Well, you can always take the "Free $2$-category generated by $C$ (as a category) and $2$-arrows of the form $f \Rightarrow F(f)$". So that the $2$-arrow will be "all the thing you can form by composing $2$-arrows of the form $f \to F(f)$". And any $2$-category looking something like what you are after will have a $2$-functor from this free object.

However, this free object does not appear to have a nice description - and the discusion below strongly suggest to me that no such $2$-category will have a nice description :

For example, because in a $2$-category you need to be able to (vertically) compose $2$-arrows you'll also have $2$-arrows of the form $f \Rightarrow F^2(f)$ and $f \Rightarrow F^3(f)$ etc... but then horizontal composition of $2$-arrows means that if an arrow $h$ can be written as a composite you'll also have $2$-arrows of the form

$$ h = f \circ g \Rightarrow F(f) \circ F^2(g) $$

Which you can't simply express this as a $h \Rightarrow F^k(h)$, so it follows that the existence of a $2$-arrow $h_1 \Rightarrow h_2$ is going to at least depends in a relatively complicated way on how $h_1$ and $h_2$ can be factored as composition.

And in facts things are even worst than this: because if $F(f) \circ F^2(g)$ can be itself written as a composite $u \circ v$ in a different way, then you'll have a further map

$$ f \circ g \Rightarrow F(f) \circ F^2(g) = u \circ v \Rightarrow F^4(u) \circ v $$

(for exemple, the $F^4$ is completely random) which cannot even be understand simply in terms of how $h = f \circ g$ factors only, it also involve factorization of $F(f) \circ F^2(g)$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .