The common definition of Reedy category seems to be [this one](http://ncatlab.org/nlab/show/Reedy+category) 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](http://ncatlab.org/nlab/show/Homotopy+Limit+Functors+on+Model+Categories+and+Homotopical+Categories) or Barwick's note [On Reedy Model Categories](http://arxiv.org/abs/0708.2832), 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.



The definition of Reedy category
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. 


