User emily riehl - MathOverflow

Answer by Emily Riehl for Resolutions by Adapted Class of Objects and Model Categories

2011-12-23T17:30:39Z

(Disclaimer: I don't have a copy of Gelfand and Manin to hand, so I'm only conjecturing that what I'm about to say is relevant to your question.)

I think what you might be looking for is the construction of derived functors via deformations. This is a generalization of the construction that's used in the context of model categories to so-called homotopical categories, due originally to Dwyer, Kan, Hirschhorn, and Smith. A good summary can be found in the first few sections of this paper.

The main idea is the following: if $F \colon C \to D$ is a functor between categories equipped with some notion of weak equivalences satisfying the 2-of-3 property, to construct a left derived functor of $F$, you don't need a full model structure on $C$. Instead it suffices to have an endofunctor $Q$ equipped with a natural weak equivalence $Q \Rightarrow 1_C$ such that $F$ preserves all weak equivalences between objects in the image of $Q$.

The theorem is that in this case $LF = FQ$ together with the natural transformation $FQ \Rightarrow F$ is a (point-set) left derived functor of $F$ (meaning, if you compose with the localization functor $D \to Ho(D)$, this becomes a left derived functor in the usual sense). It's really easy to prove once you know it's true. I suggest it as an exercise.

The notation is meant to suggest cofibrant replacement. Assuming we have a functorial cofibration - trivial fibration factorization, we can factor maps $\emptyset \to X$ to obtain a cofibrant object $QX$ and a natural weak equivalence $QX \to X$. If $F$ is left Quillen (preserves (trivial) cofibrations and the initial object), then $F$ preserves all trivial cofibrations between cofibrant objects. By Ken Brown's lemma, $F$ then preserves all weak equivalences between cofibrant objects. So the left derived functor of $F$ can be constructed simply by precomposing with some cofibrant replacement.

But the point is it doesn't matter what sort of cofibrant replacement we use, or even that it is a cofibrant replacement for some model structure. Maybe this is what's going on with the ``class of objects adapted to $F$.''

Answer by Emily Riehl for How canonical is cofibrant replacement?

2011-02-06T19:34:37Z

I don't know the answer to your questions off the top of my head, but I think algebraic weak factorization systems (the new consensus terminology for what were originally called natural weak factorization systems) are the right context to search for the answer. I'll try to briefly explain why.

Loosely, an algebraic weak factorization system on a category $\mathcal{M}$ consists of a comonad $\mathbb{L}$ and monad $\mathbb{R}$ on the arrow category $\mathcal{M}^\bf{2}$ such that these fit together to form a functorial factorization (ie, a section of the composition functor $\mathcal{M}^\bf{3} \to \mathcal{M}^{\bf 2}$, $\bf{2}$ and $\bf{3}$ being the poset categories for these ordinals). A main point is that the arrows which admit coalgebra structures for $\mathbb{L}$ lift canonically against those arrows which admit algebra structures for $\mathbb{R}$. So the comonad-monad functorial factorization also algebraicizes the construction of lifts.

Here's why I suspect this is relevant to your question. A priori, the existence of pointwise lifts does not give rise to a natural transformation between two functorial factorizations for the same weak factorization system. But suppose we had some other functorial factorization $(L',R')$ for the underlying weak factorization system of $(\mathbb{L},\mathbb{R})$. Then if the functor $R' \colon \mathcal{M}^{\bf 2} \to \mathcal{M}^{\bf 2}$ factored through the category of algebras for $\mathbb{R}$ (as, for instance, $R$ tautologously does), then there would be a morphism $(L,R) \to (L',R')$ in the category you describe. Or dually, if $L'$ factored through the category of $\mathbb{L}$-coalgebras, then there would be a morphism $(L',R') \to (L,R)$ given by the canonical solutions described above to the lifting problems.

Garner's small object argument produces algebraic weak factorization systems for any cofibrantly generated ordinary weak factorization system (or model category), so this algebraic setting is a lot more common than you'd think. (Incidentally, the best paper of his to read is Understanding the small object argument.) Interestingly, more things are cofibrantly generated than were before, because his small object argument works for generating categories (ie, one can have morphisms in the form of squares between the arrows in the generating set; in other words, one can ask that the trivial fibrations lift "coherently" against the generating cofibrations). For example, the usual model structure on ${\bf \mathrm{Cat}}$ induces one on the functor category ${\bf \mathrm{Cat}}^{\mathcal{A}}$ where the fibrations and weak equivalences are defined representably. This isn't cofibrantly generated in the usual sense, but it is in algebraic context. I describe the generating categories in my paper.

I'm quite interested in the sort of question you posed and would be happy to talk more offline, if you'd like to get in touch.

Answer by Emily Riehl for What is the name for the following categorical property?

2009-12-10T00:30:20Z

This isn't quite the question you asked, but does address the notion of ''bijective'' morphisms in categories, so I hope you'll forgive this digression.

The examples you've mentioned - Set, Gp, Top - are all concrete, meaning they are equipped with a forgetful functor U to Set. We say a morphism f in a concrete category C is injective if its image Uf is injective, i.e., monic in the category Set. Dually, f is surjective if Uf is surjective. One usually thinks of concrete categories as "sets with structure", so these definitions coincide with the common use of such terminology: e.g., we call a map of spaces surjective when the underlying map of sets is.

So we have four adjectives to use for arrows in C: monic, epic, injective, surjective. It's an easy exercise to see that all injections are monic and all surjections are epic. The converse is not true in general, but finding examples of monos that aren't injective and epis that aren't surjective can be tricky, and here's why.

Often, particularly in ''algebraic'' examples, the functor U : C → Set has a left adjoint F. When this is the case, it is an easy exercise to see that every mono must be injective. Dually, if U has a right adjoint, then every epi is surjective. So for example, the forgetful functor U : Top → Set has both adjoints, and hence for spaces the notions injective/surjective and monic/epic coincide, at which point Tom's post answers your question.

Here are some examples of concrete categories where these concepts differ, all of which can be found in Francis Borceux's Handbook of Categorical Algebra (I think). In the category of divisible abelian groups, the quotient map $\mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z}$ is monic, though it's clearly not injective. In the category of monoids, the inclusion $\mathbb{N} \rightarrow \mathbb{Z}$ is epic, though not surjective. In the category of Hausdorff spaces, the epis are continuous functions with dense image, so also need not be surjective.

Answer by Emily Riehl for Joins of simplicial sets

2009-11-30T05:00:58Z

It might be helpful to work through some simple examples. You probably know that Δⁿ ★ Δ^k = Δ^n+k+1. This has to do with the ordinal sum: one way of defining joins is as a restriction of the monoidal structure on augmented simplicial sets, which are contravarient functors from the category Δ₊ of all finite ordinals (including the empty ordinal) into sets. The category Δ₊ has a monoidal structure given by ordinary addition with ∅ as the unit, and this induces the aforementioned monoidal structure on augmented simplicial sets. The thing we call n when we are talking about simplicial sets is really the ordinal n+1, so the formula above holds because

(n+1) + (k+1) = (n+k+1)+1.

Of course, this example doesn't illustrate the asymmetry you asked about, but this one will:

∂Δⁿ ★ Δ⁰ = Λⁿ⁺¹[n+1] while Δ⁰ ★ ∂Δⁿ = Λ⁰[n+1].

To work out the details, you'll need to understand how the face maps of S★T are defined, as alluded to above. Here's my notation: (S★T)_n = S_n ∪ T_n ∪ (∪_{j+k = n+1} S_j × T_k ).

The i-th boundary map d_i : (S★T)_n → (S★T)_n-1 is defined on S_n and T_n using the i-th boundary map on S and T. Given σ∈S_j and τ∈T_k , we have:

d_i (σ, τ) = (d_i σ,τ) if i ≤ j, j ≠ 0.
d_i (σ, τ) = (σ,d_i-j-1 τ) if i > j, k ≠ 0.

If j = 0, d₀(σ, τ) = τ ∈ T_n-1 ⊂ (S★T)_n-1. If k = 0, d_n(σ, τ) = σ ∈S_n-1 ⊂ (S★T)_n-1 .

Try this out for n = 1 or 2 first, to get a feel for things. While these sorts of computations can be quite annoying, I find they do really help me develop my intuition. Best of luck!

Comment by Emily Riehl

2012-01-10T23:29:30Z

The standard definition of "directed" colimits requires that the indexing category is a preorder, so coequalizers aren't examples. Indeed, coequalizers don't commute with products in Set (though reflexive coequalizers, well-known "sifted colimits", do). For example, consider two copies of the diagram that includes the one element set into the two element set in two different ways. Taking coequalizers first, yields a 1-element set; taking products, a 3-element set. (Of course, the functor that forms the product with a fixed set is a left adjoint and preserves all colimits.)

Comment by Emily Riehl

2011-02-06T22:08:55Z

No, I meant the one with weak equivalences the categorical equivalences, fibrations the isofibrations, and cofibrations functors injective on objects. I often call this the ``folk'' model structure, but I seem to remember that Steve Lack, who has written on this sort of thing, doesn't like that name for some reason.

Comment by Emily Riehl

2011-02-06T20:04:35Z

I should have said, you technically only need $R'$ to lift through the category of algebras for the pointed endofunctor part of the monad $\mathbb{R}$ (meaning, you can drop the coherence condition for the multiplication). This functor exists if and only if the arrows in the image of $R'$ lift naturally against their left factors with respect to the functorial factorization of $(\mathbb{L},\mathbb{R})$. And dually of course.

Comment by Emily Riehl

2009-12-01T05:19:48Z

Sorry. Both notations are fairly standard. I chose the former because it looked better in html. But when writing $\Lambda^k[j]$, one typically means for $j$ to be the dimension and $0 \leq k \leq j$ to indicate which face is missing. (So this is $\Lambda^j_k$ in your notation.)