I'm studying a small category $A$ and diagrams of based spaces or spectra indexed by $A$ (so let's say diagrams in a category $C$ that's closed symmetric monoidal, has a compatible model structure, etc.). I'm told that in this setting there's a projective model structure on diagrams $A \rightarrow C$, where the fibrations and weak equivalences are defined objectwise; and an injective model structure, where the cofibrations and weak equivalences are defined objectwise.

I want to take two diagrams $X,Y$ and form a mapping space $\text{Map}(X,Y)$ in a homotopy-invariant way. (The mapping space should be a subspace of the product of the mapping spaces for each level. In a more formal setting, I can pass to the twisted arrow category on $A$ and take an equalizer.) I should be able to accomplish this by taking a cofibrant replacement of $X$ and a fibrant replacement of $Y$, but there are two model structures available for doing this. This gives me two different mapping spaces; I want to show that they are equivalent. Any suggestions for approaches/tools/references?

(An intermediate step might be to show that they are both equivalent to the space I get by taking a projective-cofibrant replacement of $X$ and an injective-fibrant replacement of $Y$.)


Let me mention yet another approach using the simplicial localization due to Dwyer and Kan.

Given any category $\mathcal{C}$ equipped with a class of morphisms $W$ one can form a simplicial category $L(\mathcal{C}, W)$ which depends only on $\mathcal{C}$ and $W$ and not on any auxiliary structure like model structure. However, it is true that if $\mathcal{C}$ comes with a model structure (with weak equivalences $W$), then the mapping spaces of $L(\mathcal{C}, W)$ are weakly equivalent to the mapping spaces constructed using (co)simplicial resolutions coming from the model structure. In particular, if we have two model structures on $\mathcal{C}$ with the same weak equivalences $W$ (as in your situation), then the mapping spaces constructed from those model structures are weakly equivalent.

The precise reference is Proposition 4.4 of Function Complexes in Homotopical Algebra by Dwyer and Kan.

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  • $\begingroup$ +1: I like this answer best of all. $\endgroup$ – David White Feb 15 '12 at 16:30

(I deleted my first attempt at an answer, as I had right and left reversed and anyway I wanted to try to say it better.)

I would advocate the following broad and comparatively low-tech view of the matter.

You are asking why two homotopy invariant versions of Hom of diagrams should be equivalent.

In general if you are dealing with a functor $F:C\to D$, where both $C$ and $D$ have some maps called weak equivalences, you can define the notion of "right derived functor" of $F$ as follows: Take the category of all functors $C\to D$. Make the slice category of all functors $(G,T:F\to G)$ under $F$. In it, consider the full subcategory whose objects are those $(G,T)$ such that $G$ preserves weak equivalences. In this category call a morphism $(G_1,T_1)\to (G_2,T_2)$ a weak equivalence if the map $G_1\to G_2$ is a weak equivalence (objectwise in $C$). By definition a right derived functor of $F$ is a homotopically initial object. That is, it is an object $(G,T)$ that becomes initial when you invert these equivalences.

From this definition it is clear that any two right derived functors of $F$ are equivalent, in the sense that there is zigzag of weak equivalences connecting them.

You care about the case when $C=Fun(A,Top)^{op}\times Fun(A, Top)$ and $F:C\to Top$ is the straightforward Hom functor, taking $(X,Y)$ to the set of natural maps between diagrams, topologized as a subspace of a product of function spaces, one for each $A$-object. Now you need to know why the various homotopy invariant replacements that you can think of for $Hom$ are in fact derived functors in the sense above.

The usual way of making a right derived functor for $F$ goes like this: think up a class $C_0$ of special $C$-objects such that $F$ takes equivalences between such objects to equivalences. Think up an endofunctor $r$ of $C$ such that

  1. $r$ takes every object to an object in $C_0$,
  2. there is a natural weak equivalence $e:1\to r$ to $r$ from the identity.
  3. $r$ takes weak equivalences to weak equivalences

As derived functor of $F$ use the composition $F\circ r$, equipped with the map $F\to F\circ r$ induced by $e$. This is in fact homotopically initial in that category of homotopy-invariant functors under $F$, under extremely weak axioms about $C$ and $D$ and their weak equivalences.

In your case you ought to be able to make $r$ and $e:1\to r$ by using cofibrant replacement of $X$ and fibrant replacement of $Y$ in either the projective or the injective model structure. The point to check, either way, is that maps $X_2\to X_1$ and $Y_1\to Y_2$ of diagrams induce equivalences $Hom(X_1,Y_1)\to Hom(X_2,Y_2)$ if the $X_i$ are cofibrant and the $Y_i$ are fibrant.

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I think the best way to go about this is to go through the Reedy model category structure, but for that to work $A$ has to be a Reedy category. If you are in this situation, then Theorem 15.6.4 of Hirschhorn's book tells you that the projective model structure is Quillen equivalent to the Reedy one (this would imply what you need, right?). There is of course a dual statement saying the injective model structure is Quillen equivalent to the Reedy one. Of course, these equivalences don't need to be equalities.

This is discussed more on the nLab page for Reedy model category structure, which also points you to Example A.2.9.22 in Lurie's HTT. That example happens to be on page 666, so you know you've reached the true depths of evil when it comes to studying model categories. I wish you the best of luck in your journey.

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  • $\begingroup$ Link to Lurie's HTT, if you dare: arxiv.org/abs/math.CT/0608040 $\endgroup$ – David White Feb 14 '12 at 21:22
  • $\begingroup$ In all seriousness, I am not an expert on Reedy things. I don't know how far the average small category $A$ is from being Reedy or if you can somehow move from your situation into this situation. I should also mention that the functor which gives the Quillen equivalence is the identity, so Map($X,Y)_{Proj} \cong$ Map$(X,Y)_{Reedy} \cong$ Map$(X,Y)_{Inj}$ $\endgroup$ – David White Feb 14 '12 at 21:44
  • $\begingroup$ I need a weak homotopy equivalence between my two constructions of mapping spaces, and I think that's stronger than saying that the homotopy categories are equivalent. Also, my A is the category of based sets $\emptyset_+$, $\{1\}_+$, $\ldots$, $\{1,\ldots,n\}_+$ and based maps between them. I believe this is not a Reedy category but I could be mistaken. $\endgroup$ – Cary Feb 14 '12 at 21:46
  • $\begingroup$ If we make the degree of $\{1,\ldots,n\}_+$ equal to $n$, then $R_+$ can be all injective order-preserving maps, and $R_-$ can be all surjective maps. Then every map factors uniquely into a map in $R_-$ followed by a map in $R_+$. Unfortunately not every map in $R_-$ lowers degree, and not every map in $R_+$ raises degree. $\endgroup$ – Cary Feb 14 '12 at 22:46
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    $\begingroup$ @Cary: You're right, your category is not Reedy, but it is generalized Reedy: arxiv.org/pdf/0809.3341v1.pdf. Also Quillen equivalences will induce homotopy equivalences between Dwyer-Kan function complexes. The DK complex from $X$ to $Y$ will have the same homotopy type as the mapping space from a cofibrant replacement of $X$ to a fibrant replacement for $Y$ in any of the model structures if the model structure makes the category a simplicial/topological model category. $\endgroup$ – Justin Noel Feb 15 '12 at 17:27

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