Let $w$ be an element of the free group $F_n$ on the generators $\{x_i\}_{i=0}^{n-1}$. Define a function $\theta: F_n\to F_n$ by $x_i\mapsto x_{i+1}$ (and extend this to all elements of $F_n$), where the subscripts are taken modulo $n$. Consider the cyclic group presentation

$$\mathscr{P}_n(w):=\langle x_0, \dots , x_{n-1}\mid w, \theta(w), \dots , \theta^{n-1}(w)\rangle.$$

Then the group $\mathscr{G}_n(w)$ defined by $\mathscr{P}_n(w)$ is called a cyclically presented group.

Similarly, one can construct a cyclically presented semigroup.

The Question:

Are there such things as "cyclically presented (arrows-only) categories"?


It might be necessary to restrict ourselves to arrows-only categories to make sense of the analogue of the "shift" $\theta$ in the case of group presentations.

I think I understand the definition of a presentation of a category:

Let $G$ be a directed graph, and let $R$ be a function that assigns to each pair $a,b$ of objects of the free category $F(G)$ a binary relation $R_{a,b}$ on the hom-set $F(G)(a,b)$. The category with generators $G$ and relations $R$ is the quotient category. For a category $C$, an isomorphism $C\to F(G)/R$ is called a presentation of $C$.

This definition seems to require that the binary relations, which in groups define the relators of the presentation, need to be "in two generators only", for lack of a better way of putting it. That might prevent us from making sense of a "cyclically presented (arrows-only) category".

Based on what it says about (the definition of) quotient categories being "evil" (which I don't fully understand), perhaps a cyclically presented category would, too, be "evil".

Please help :)


It's not necessarily the case that $x_n\mapsto x_{n+1}$ gives a well defined function on the relations. You need a condition that if $x_mx_n$ is defined then so is $x_{n+1}x_{m+1}$. Given that, there's no issue.

It's not at all the case that relations can involve only two generators. They can involve only two words in the generators, but that's also true for groups.

The problem of evil is not relevant here.

  • 1
    $\begingroup$ Thank you, @KevinCarlson. I took the liberty of moving the final sentence of this answer to a paragraph of its own. It's a separate issue from the paragraph it was in previously. $\endgroup$ – Shaun Nov 12 '18 at 3:42

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