Is assigning the endomorphism object in some sense functorial? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:23:22Z http://mathoverflow.net/feeds/question/21318 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21318/is-assigning-the-endomorphism-object-in-some-sense-functorial Is assigning the endomorphism object in some sense functorial? Garlef Wegart 2010-04-14T10:15:39Z 2010-04-14T12:38:06Z <p>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 $\mathcal V^{\mathcal End(X)}$?</p> <p>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:</p> <p>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 $\mathcal C$).</p> <p>My idea:</p> <ol> <li>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.</li> <li>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.</li> </ol> <p>My vague guess is that the framework where this question could be handled is that of extranatural transformations.</p> http://mathoverflow.net/questions/21318/is-assigning-the-endomorphism-object-in-some-sense-functorial/21329#21329 Answer by Tilman for Is assigning the endomorphism object in some sense functorial? Tilman 2010-04-14T12:04:52Z 2010-04-14T12:04:52Z <p>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 <em>Categories for the working mathematician</em>, or for the enriched version, Kelly's <em>Basic concepts of enriched category theory</em>).</p> <p>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 <em>centric maps</em> in topology. It has been used to study realizations of diagrams in the homotopy category (Dwyer-Kan, <em>Centric maps and realization of diagrams in the homotopy category</em>, Proceedings of the AMS 114(2), 1992).</p>