Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathcal C$ be a category, and consider a new category $\mathcal C'$ with

$Obj(\mathcal C') := \{$pairs $(X \in Obj(\mathcal C), T \in End_{\mathcal C}(X)) \}$

$Hom_{\mathcal C'}((X,T_X),(Y,T_Y)) := \{R \in Hom_{\mathcal C}(X,Y) : T_Y \circ R = R\circ T_X \}$

Does this $\mathcal C'$ have a name, or any non-formally-obvious properties?

I suppose that the more natural construction would let $T$ be an arbitrary morphism instead of an endomorphism. In my context $\mathcal C$ is additive and I require also that $T^2=0$.

share|improve this question
add comment

3 Answers

up vote 5 down vote accepted

I think what you have defined is just called the category of endomorphisms in $\mathcal{C}$. See for instance Marian Mrozek's 1992 paper Normal functors and retractors in categories of endomorphisms for some properties of this category.

The oldest mention that I have found in the literature so far is

Almkvist, Gert, The Grothendieck ring of the category of endomorphisms, J. of Algebra 28 (1974), 375–388.

Almkvist restricts attention to the case where $\mathcal{C}$ is the category of $A$-modules which admit a finite projective resolution. See also this question of mine about homotopy properties of iterating the endomorphism construction.

share|improve this answer
3  
Thanks for that Mrozek reference, Vidit. I'm very glad to discover it, for reasons that are probably quite different to Allen's (golem.ph.utexas.edu/category/2011/12/…). Mrozek, like me, uses $\text{Endo}(\mathcal{C})$ for Allen's category. Tiny nitpick: it's better to call it the category of endomorphisms in $\mathcal{C}$, not of $\mathcal{C}$, since the latter sounds like the category of endofunctors of $\mathcal{C}$. –  Tom Leinster Nov 15 '13 at 21:54
add comment

$\mathcal{C}'$ is isomorphic to a functor category, namely the category of functors $\mathbb{B} \mathbb{N} \to \mathcal{C}$, where $\mathbb{B} \mathbb{N}$ is the category freely generated by an endomorphism, i.e. the category with one object $*$ and $\mathrm{Hom}(*, *) = \mathbb{N}$.

If $\mathcal{C}$ is an $\mathbf{Ab}$-category and you also require $T^2 = 0$, then $\mathcal{C}'$ is isomorphic to the category of additive functors $\mathbb{B} A \to \mathcal{C}$, where $\mathbb{B} A$ is the $\mathbf{Ab}$-category with just one object $*$ and $\mathrm{Hom}(*, *) = A$, where $A$ is the ring $\mathbb{Z} [T] / (T^2)$. As always, $\mathcal{C}'$ is an $\mathbf{Ab}$-category.

share|improve this answer
2  
$C'$ can also be described as the subcategory of the category $\text{Arr}(C)$ of arrows in $C$ with the same source and target, or equivalently as the equalizer of the two projections $\text{Arr}(C) \to C$. This reflects the fact that $B\mathbb{N}$ itself can be described as the coequalizer of the two inclusions of the terminal category $\text{pt}$ into the interval category $I$. This is a categorical version of the construction of the circle by gluing together an interval at its endpoints, and so one can think of $C'$ as a categorical version of the (free) loop space. –  Qiaochu Yuan Nov 15 '13 at 19:08
6  
$C'$ is not the full subcategory of $\mathrm{Arr}(C)$ on arrows with the same source and target. In that full subcategory, a morphism $(X, f:X \to X) \to (Y, g:Y \to Y)$ is a pair of morphisms $p,q : X \to Y$ such that $g \circ p = q \circ f$, whereas in the actual endomorphism category the morphisms are just those where $p=q$, @QiaochuYuan. The equalizer description is correct, however (being an equalizer also forces $p=q$). –  Omar Antolín-Camarena Nov 15 '13 at 19:48
add comment

I would call it the category of representations of the loop quiver in $C$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.