# Is assigning the endomorphism object in some sense functorial?

Let $\mathcal V$ be a monoidal category and let $\mathcal C$ be a $\mathcal V$-category. Let's denote the $\mathcal V$-valued hom-functor $[-,-]$. Now for every object $X\in\mathcal C$ we have it's endomorphism object $\mathcal End(X):=[X,X]$ - it is actually a monoid in $\mathcal V$. Can the assignment $$X\mapsto \mathcal End(X)$$ be considered functorial in some way? Is there a language that captures the relations between $\mathcal End(X)$ when $X$ varies? What happens on the level of module categories $\mathcal V^{\mathcal End(X)}$?

I've allready thought about this a bit but i don't want to reinvent the wheel.  So here's what i've been thinking of:

For every object $X\in\mathcal C$ we get a functor $[X,-]:\mathcal C\to\mathcal End(X)-\operatorname{mod}$. I think these functors are connected in a vaguely functorial way - hopefully by adjunctions between the module categories (adjunction in the 2-category of categories under $\mathcal C$).

My idea:

1. Let $f:X\to Y$ be a morphism in $\mathcal C_0$. On the one hand $[Y,X]$ is a bimodule from $\mathcal End(Y)$ to $\mathcal End(X)$. On the other hand $[Y,X]$ becomes a bimodule in the other direction - from $\mathcal End(X)$ $\mathcal End(Y)$ - by pre- and postcomposition with $f$ i.e. pulling back the module structure along $[f,f]:[Y,X]\to[X,Y]$. So assuming $\mathcal V$ is nice enough we have two functors $[Y,X]\otimes_{\mathcal End(X)}$ and $[Y,X]\otimes_{\mathcal End(Y)}$. However i can think of no canditate for a unit of a supposed adjunction between these two.
2. As in (2) $[Y,X]$ also becomes a semigroup object in $\mathcal V$ that has a (left/right) unit - and thus is a monoid - precisely when $f$ has an (left/right) inverse.

My vague guess is that the framework where this question could be handled is that of extranatural transformations.

• $End(X)$ is always a monoid regardless of enrichment. – Harry Gindi Apr 14 '10 at 10:25
• I'm posting this in a comment because I'm not positive, but it doesn't appear to work in a naive way. You can transport endomorphisms by looking at pairs of morphisms $X\to Y$ and $Y\to X$. For example, you can't transport endomorphisms between the $0$ and any other commutative ring simply because $Hom(0,X)$ is empty for all $X\neq 0$. If you have a category $C$ where the morphisms are pairs of morphisms in another category, and you retain the same objects, it might work, but it doesn't seem like this will be very useful in general. – Harry Gindi Apr 14 '10 at 10:42
• If you've already thought about this you should include a bit about what you've been able to figure out and what you haven't. Otherwise it feels like you're fishing and trying to get others to do your work for you. – Chris Schommer-Pries Apr 14 '10 at 11:16
• @Chris: Yes. I'm hoping someone has allready done the work and/or could fix me up with a reference. The question seems totally natural and thus not very original to me. I guess a lot of people did allready think about this. – Gerrit Begher Apr 14 '10 at 12:46
• Not exactly what you are looking for but might be related: arxiv.org/abs/1307.5956 – Marcel Bischoff Sep 1 '15 at 12:57

## 2 Answers

It's hard to answer this question on this abstract level in any other way that by saying "no, it's not a functor". Of course it is a bifunctor $\mathcal{C}^{op} \times \mathcal{C} \to \mathcal{V}$, and the language of ends and coends deals with such functors (Mac Lane's Categories for the working mathematician, or for the enriched version, Kelly's Basic concepts of enriched category theory).

You'll get functoriality if you restrict your category $\mathcal{C}$ to those morphisms $f\colon X \to Y$ such that $[Y,X] \to [X,X]$ is an iso (then the endomorphisms become a covariant functor) or those such that $[X,X] \to [X,Y]$ is an iso (then it's contravariant). The latter is related to the concept of centric maps in topology. It has been used to study realizations of diagrams in the homotopy category (Dwyer-Kan, Centric maps and realization of diagrams in the homotopy category, Proceedings of the AMS 114(2), 1992).

This reminds me of the discussion on pages 264-265 of Eilenberg and Mac Lane's General Theory of Natural Equivalences of the "degree of invariance" of various constructions on the category of groups; see in particular the table on the top of page 265.

For any $\mathcal{V}$-category $\mathcal{C}$, End defines a functor from the underlying groupoid iso$\mathcal{C}_0$ of $\mathcal{C}_0$ to the category of monoids in $\mathcal{V}$, where an isomorphism $f \in \mathcal{C}_0$ acts by conjugation $f \circ - \circ f^{-1}$.

One way to understand this construction is as a restriction of the hom bifunctor $\mathcal{C}_0^{\text{op}} \times \mathcal{C}_0 \to \mathcal{V}$ using the diagonal and the functor $(-)^{-1} \colon \text{iso}\mathcal{C}_0^{\text{op}} \to \mathcal{C}_0$. Depending on which $\mathcal{C}$ and which $\mathcal{V}$ you have in mind, there might be other (partially-defined) functors $\mathcal{C}^{\text{op}} \to \mathcal{C}$ (e.g., "transpose" or "adjoint") though they might give rise to morphisms in $\mathcal{V}$ that are not necessarily monoid homomorphisms.