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Motivation: In his utterly famous paper, Rezk (here, (pag. 7)) defines a structure called "Quillen ring". I'm wearing my algebraist's hat today, so I was wondering if this definition is chosen to suggest that somewhere a "true" ring structure is hidden: now I'm in particular interested in dualizing the definition to obtain the "pushout-product" arrow, which I would like to taxonomize here with your kind help. [What I'm kindly asking here is: does the "pullback-exponential" operation give a "ring operation", in some reasonable sense? Does its dual give a co-ring operation?]

Let me start from the beginning. For the moment I don't even need a model structure, as I am not interested in cofibration-preservation properties (which can be used to define in a compact way the notion of "left Quillen bifunctor", as I learned about an hour ago).

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Take your favourite finitely bicomplete category $\cal C$ (in particular I'll ask for pullbacks and pushouts).

Define a binary operation on the set (of isomorphism classes) of arrows in $\cal C$, taking $f\colon A\to B$ and $g\colon C\to D$ and sending them to the arrow $$\newcommand{\diam}[4]{\left(#1 \times #4\right)\coprod_{#1 \times #3}\left( #2 \times #3 \right)} f\diamond g\colon \quad \diam{A}{B}{C}{D}\longrightarrow B\times D$$ I would like to unravel the algebraic properties of this operation $\diamond \colon {\rm Mor}({\cal C})\times {\rm Mor}({\cal C})\to {\rm Mor}({\cal C})$.

  1. Is it associative?
  2. Is it commutative?
  3. It seems to have a neutral element (the arrow $\varnothing\to 1$).
  4. Is there an operation on which $\diamond$ distributes over?

Associativity seems painful, as it is linked to simplification properties which I'm not able to deduce from easy arguments.

This is the scary form of the isomorphism which proves the associativity $f(gh) \cong (fg)h$ for $f\colon A\to B,g\colon C\to D, h\colon V\to W$ $$ \diam{\diam{A}{B}{C}{D}}{B\times D}{V}{W} \quad\cong \quad \diam{A}{B}{\diam{C}{D}{V}{W}}{D\times W} $$ Great honour to whom will be able to simplify it!

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3  
You see? When I read "utterly famous paper [of Rezk]" I thought of another one! Rezk is really a great mathematician and has several famous papers ;-) –  Fernando Muro Oct 28 '13 at 16:55
    
:-) one of his many, many famous papers –  tetrapharmakon Oct 28 '13 at 17:05
    
The terminology "Quillen ring" was invented, I believe, by Dan Kan. (That's who I heard it from, anyway.) "Ring" suggests symmetric monoidal, and "Quillen" means things play nice with the model catogory structure. –  Charles Rezk Oct 29 '13 at 0:08

1 Answer 1

up vote 10 down vote accepted

Whenever you start with a biclosed monoidal category $(\mathcal C, \otimes, I)$, the push-out product is a (biclosed) monoidal structure on the category of arrows $\operatorname{Mor}(\mathcal C)$. The trick is to think of $\operatorname{Mor}(\mathcal C)$ as the category of functors $\mathbf 2\rightarrow\mathcal C$, where $\mathbf 2$ is the poset $\{0<1\}$. Two morphisms $f$ and $g$ give rise to a functor $f\otimes g\colon \mathbf 2\times \mathbf 2\rightarrow\mathcal C$ and their push-out product can be expressed as

$$f\diamond g=\operatorname{colim}\limits_{(\mathbf 2\times \mathbf 2)\setminus \{(1,1)\}}f\otimes g\longrightarrow \operatorname{colim}\limits_{\mathbf 2\times \mathbf 2}f\otimes g$$

The map is induced by the inclusion of posets. As for the target, notice that $\mathbf 2\times \mathbf 2$ has a final object, $(1,1)$, so the colimit is the value at the final object.

A more or less elementary computation shows that if $\otimes$ preserves push-outs in each variable (and this happens if $\mathcal C$ is biclosed), then the iterated push-out product of morphisms $f_1,\dots,f_n$ is simply

$$f_1\diamond\cdots\diamond f_n=\operatorname{colim}\limits_{\mathbf 2^n\setminus \{(1,\dots,1)\}}f_1\otimes\cdots\otimes f_n\longrightarrow \operatorname{colim}\limits_{\mathbf 2^n}f_1\otimes\cdots\otimes f_n.$$ This is the way of checking associativity.

As you point out, it is easy to see that $\varnothing\rightarrow I$ is a tensor unit. You can use that $X\otimes\varnothing\cong \varnothing\cong\varnothing\otimes X$ for any object $X$ since $\mathcal C$ is biclosed.

If $\mathcal C$ is symmetric, hence so is the category of arrows. This is really easy and does not require the property of being biclosed.

The pentagon and other coherence axioms follow from the corresponding axioms for $\mathcal C$ and the universal property of colimits.

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Incredibly neat! Thank you. –  tetrapharmakon Oct 28 '13 at 17:57
    
P.S.: "I don't know what you mean by coassociative or cocommutative" ...In the end, nothing. Let me edit! –  tetrapharmakon Oct 28 '13 at 18:00
1  
I'm grateful for your appreciation :-) –  Fernando Muro Oct 28 '13 at 18:05

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