The common definition of Reedy category seems to be this one that a Reedy category is a small category $R$ with two wide subcategories $R_+$ and $R_-$ and an ordinal-valued degree function on its objects such that

  • Every nonidentity morphism in $R_+$ raises degree,
  • Every nonidentity morphism in $R_-$ lowers degree, and
  • Every morphism $f$ in $R$ factors uniquely as a map in $R_-$ followed by a map in $R_+$.

However, in a few places, such as the DHKS book Homotopy Limit Functors on Model Categories and Homotopical Categories or Barwick's note On Reedy Model Categories, there is a slightly different definition in which the factorizations are only required to be functorial, rather than unique. Unique factorizations are functorial, but the converse is not generally true.

I think I can prove that a "Reedy category" with functorial factorizations is also a Reedy category with unique factorizations, but my proof is quite roundabout and involves (at least apparently) shrinking the subcategories $R_-$ and $R_+$. Are the definitions actually equivalent?

Edit: Now I think this claim is wrong; see my answer below.


2 Answers 2


Is this a counterexample?

$R$ is the poset category $1\to 0\to 2$.

Nonidentity maps in $R^+$: $0\to2$, $1\to 2$.

Nonidentity maps in $R^-$: $1\to 0$.

There are no other maps in $R$.

The map $1\to 2$ admits two distinct "Reedy factorizations": $1\to1\to 2$ and $1\to 0\to 2$. (This is the only map with more than one possible choice.)

It appears that $R$ has a functorial Reedy factorization $Fun([1],R)\to Fun([2],R)$, where the functorial factorization of $1\to 2$ is set to be $1\to 0\to 2$ (the other choice isn't functorial). Checking that this is well-defined involved looking at each of the $20$ possible commutative squares in $R$. I would not swear that I got it right.

  • 1
    $\begingroup$ Thanks! I think it isn't necessary to check all the squares by hand; it should be enough to observe that $R$ also admits a Reedy structure in the usual sense if you remove $1\to 2$ from $R^+$, and use the fact that unique factorizations are automatically functorial. So this is also an easy example of my claim that if you shrink $R^+$ and $R^-$ you can make the factorizations unique. $\endgroup$ Jul 28, 2014 at 6:23

If I'm not mistaken, here is an "even worse" counterexample than Charles'. Let $R$ be the walking isomorphism $(0\cong 1)$, let $R_+$ be $(0\to 1)$, and let $R_-$ be $(1\to 0)$. The conditions on degrees are satisfied, and since $R$ is a contractible groupoid, any factorization is automatically functorial. In this case there is no way at all to make $R$ a Reedy category with unique factorizations, and the usual theorems about inductive definition of objects and morphisms are not true. (So in particular, my claim at the end of the question is wrong.)

  • $\begingroup$ Hmmm. But it is equivalent to a Reedy category. An even worse counterexample at this stage would be one that isn't equivalent to a Reedy category. $\endgroup$
    – Zhen Lin
    Jul 28, 2014 at 18:15
  • $\begingroup$ @ZhenLin Feel free to come up with one... $\endgroup$ Jul 28, 2014 at 18:17
  • $\begingroup$ @ZhenLin Hmm, actually I may be able to modify my failed proof of the claim to show that there is always an equivalent full subcategory that is Reedy. $\endgroup$ Jul 28, 2014 at 19:20

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