5
$\begingroup$

EDIT: Following the lines of some suggestions in the comments below, I try to add something more to explain the problem better. A map $\text{hocolim}Y\rightarrow\bar{Y}$ in $\text{Ho}(\mathbf{M})$ is the same as a map $Y\rightarrow c_{\bar{Y}}$ in $\text{Ho}(\mathbf{M^{I}})$. As pointed out by მამუკაჯიბლაძე in the comments, "there is a canonical $Y\rightarrow c_{\bar{Y}}$, the morphism $Y(i)\rightarrow \bar{Y}$ being the homotopy class of the projection map $X(i)\times_{\bar{X}}^{h}\bar{Y}$ (which is defined up to homotopy anyway)".

My problem comes up exactly here. The morphisms $Y(i)\rightarrow \bar{Y}$, being homotopy classes of maps, do not define a map $Y\rightarrow c_{\bar{Y}}$ in $\text{Ho}(\mathbf{M}^{I})$ but rather in $\text{Ho}(\mathbf{M})^{I}$ (!). To get a map from $Y$ to $c_{\bar{Y}}$ in $\text{Ho}(\mathbf{M}^{I})$, I should find actual functors $I\rightarrow \mathbf{M}$, say $A$ and $B$ and a zig-zag of maps among them (where those pointing backward are weak equivalences) such that the image of such a zig-zag in $\text{Ho}(\mathbf{M}^{I})$ gives me, up to isomorphisms, a map $Y\rightarrow c_{\bar{Y}}$ there. A similar remark applies also to $X_{i}\times_{\bar{X}}^{h}\bar{Y}$. Let me try to explain this: consider the shape category of pullbacks, $J=\bullet\rightarrow \bullet\leftarrow\bullet$. Now, in general I can find a cospan $X_{i}\rightarrow \bar{X}=\text{hocolim}X\leftarrow \bar{Y}$ but as an object in $\text{Ho}(\mathbf{M})^{J}$ and not in $\text{Ho}(\mathbf{M}^{J})$ (where I can apply the homotopy pullback functor). Indeed, I can lift $\text{hocolim}$ to a functor $\mathbf{M}^{I}\rightarrow \mathbf{M}$ as $\text{colim}\circ Q$, where $Q$ is a homotopical functor $\mathbf{M}^{I}\rightarrow (\mathbf{M}^{I})_{0}$ which is part of a left $\text{colim}-$deformation retract (see DHKS). For example, one can take a functorial cofibrant replacement in the projective model structure on $\mathbf{M}^{I}$, if that is available. Then of course I get a cospan $(QX)_{i}\rightarrow \text{colim}QX\leftarrow \bar{Y}$ in $\mathbf{M}$ (i.e an object in $\mathbf{M}^{J}$), hence also an object $(QX)_{i}\rightarrow \text{colim}QX\leftarrow \bar{Y}$ in $\text{Ho}(\mathbf{M}^{J})$, call it $C_{i}$. Under the canonical functor $\text{Ho}(\mathbf{M}^{J})\rightarrow \text{Ho}(\mathbf{M})^{J}$, such an object $C_{i}$ becomes isomorphic to a cospan $X_{i}\rightarrow \bar{X}=\text{hocolim}X\leftarrow \bar{Y}$ in $\text{Ho}(\mathbf{M})$, i.e. to an object of $\text{Ho}(\mathbf{M})^{J}$, because there is a weak equivalence $(QX)_{i}\rightarrow X_{i}$ in $\mathbf{M}$. However, I can apply the hopullback functor only to $C_{i}$ and not to $X_{i}\rightarrow \bar{X}=\text{hocolim}X\leftarrow \bar{Y}$, simply because this latter cospan is not an object of $\text{Ho}(\mathbf{M}^{J})$ (in general). Applying the hopullback to $C_{i}$, I do get a map $(QX)_{i}\times_{\text{colim}QX}^{h}\bar{Y}\rightarrow \bar{Y}$ in $\text{Ho}(\mathbf{M})$, for each $i$, which, I guess, could model $X_{i}\times_{\bar{X}}^{h}\bar{Y}\rightarrow \bar{Y}$, whatever this may mean. Lifting hopllbck to a functor from $\mathbf{M}^{I}$ as $\text{lim}_{J}\circ R$, where $R$ is (part of) a right $\text{lim}_{J}-$deformation retract, I can lift these maps $(QX)_{i}\times_{\text{colim}QX}^{h}\bar{Y}\rightarrow \bar{Y}$ in $\text{Ho}(\mathbf{M})$ to actual maps in $\mathbf{M}$ and such maps do form a natural transformation between functors $A,B\colon I\rightarrow\mathbf{M}$ which, in the homotopy category $\text{Ho}(\mathbf{M^{I}})$, gives, up to isomorphisms, something as a map $Y\rightarrow c_{\bar{Y}}$. This is essentially the construction I already pointed out when I first posted the question here (see below).

However, now, I should prove that this construction provides a map at the level of the homotopy category which does not depend (up to isomorphisms) upon the thousands choices that I had to make (specific liftings for hocolim and hopllbck, specifing choices of $Q$ and $R$...), at least if I want to have some hopes to get a formulation of property (P1) which is somehow handy (at the end, when it comes down to prove that property (P1) -in one formulation or another - holds for some specific model categories, say simplicial set, it seems to me that one do uses some concrete models for hocolims and hopllbcks, at least for the proofs I have seen of it). Of course, at the level of the homotopy category, all hocolims and hopllbcks are naturally isomorphic (they are all adjoints to the same functor!), but I do not see how this can help me to give a rigorous proof of the fact that two maps $\text{hocolim}Y\rightarrow \bar{Y}$ built up using different choices in the above construction correspond each other under some isomorphism.

I am sure there is something, probably something obvious, that I am missing, but I do not see how the problems I explained above could be circumvented or, even more radically, how they are not problems at all.


Here is the original question.

I have some troubles in trying to give a meaningful interpretation to the following property which is stated in this preprint by professor Rezk (see Definition 6.5) as part of the requirement for a model category $\mathbf{M}$ to have what the author calls descent. Here is the problematic condition:

(P1) Let $X\colon I\longrightarrow \mathbf{M}$ be a functor from a small category $I$ to a model category $\mathbf{M}$. Set $$\bar{X}:=\text{hocolim}X$$ and consider any map $f\colon\bar{Y}\rightarrow\bar{X}$ in $\mathbf{M}$. Form a functor $Y\colon I\longrightarrow\mathbf{M}$ by setting, for $i\in I$, $$Y(i):=X(i)\times^{h}_{\bar{X}}\bar{Y}.$$ Then the evident map $$\text{hocolim Y}\rightarrow \bar{Y}$$ is a weak equivalence.

(Here $(-)\times^{h}_{\bar{X}}(-)$ denotes a homotopy pullback over $\bar{X}$, I guess).

My doubts arise as I do not see a clear way to understand the construction made. Here are some points that are obscure to me:

  1. To have a chance to construct something as $Y(i)$ or the "evident" map $\text{hocolim}Y\rightarrow\bar{Y}$, I should be able to get maps $X(i)\rightarrow \bar{X}$ and $Y(i)\rightarrow\bar{Y}$. The latter family of maps should then induce the alleged arrow $\text{hocolim}Y\rightarrow\bar{Y}$, or at least this is what I believe. The point now is: are there some models for the homotopy colimits and the homotopy pullbacks (for which they form functors from the appropriate diagram categories) that allow me to get maps $X(i)\rightarrow \text{hocolim}X$ and $Y(i)\rightarrow\bar{Y}$ (as for ordinary colimits and pullbacks)? Can these maps be built so as to form a (co)cone (over $\text{hocolim}X$ or over $\bar{Y}$) either in a strict sense or in some homotopically relaxed version (to be defined), so that I can use, for example, the maps $Y(i)\rightarrow\bar{Y}$ to get $\text{hocolim}Y\rightarrow\bar{Y}$?

  2. Is the evidence of the map $\text{hocolim}Y\rightarrow\bar{Y}$ really so evident?

  3. Are there some canonical properties of the homotopy colimits or of the homotopy pullbacks which allow me to make the construction done by Rezk (and to answer the questions in 1.) in a way which is independent of the chosen model to realize those homotopy colimits and pullbacks?

Note: Rezk considers arbitrary model categories when formulating the property (P1) above, but I would be very happy to get an answer to my questions (which ultimately boil down to the final question: "How should I interpret property (P1)?") in the case where $\mathbf{M}$ is (at most) a combinatorial model category (but not simplicial, if possible).


There is a somehow related question on MO which suggests, in the case where $\mathbf{M}$ is a category of spaces (say, simplicial sets), to feel free to cofibrantly and fibrantly replace everything (objects, arrows, diagrams) as long as this may be needed to give a meaning to constructions like the one in (P1) above. I have then tried to see if I could work out something with this underlying philosophy.

Here are my trials: in what follows, I assume $\mathbf{M}$ to be a cofibrantly generated category and keep the same notations as in (P1) above.

Let $Q$ be the functorial cofibrant replacement in the projective model structure on $\mathbf{M}^{I}$ and let $R$ be the fibrant replacement functor for the Reedy model structure on the category of cospans in $\mathbf{M}$. Given a cospan $$ A\rightarrow B\leftarrow C $$ in $\mathbf{M}$, I will adopt a little abuse of notation and denote the image of this cospan under $R$ by $$ RA\rightarrow RB\leftarrow RC. $$

Now, a model for the homotopy colimit of $X$ is given by $\bar{X}=\text{colim}\ QX$. In this case, I thus get a cocone $$ (QX)(i)\rightarrow \text{hocolim} X $$ and, for each $i\in I$, I can consider the cospan in $\mathbf{M}$ $$ (QX)(i)\rightarrow \text{hocolim} X \leftarrow \bar{Y} $$ A model for the homotopy pullback of this cospan is given by doing the ordinary pullback of $$ R((QX)(i))\rightarrow R(\bar{X})\leftarrow R(\bar{Y}) $$ Thus, my (corrected) functor $Y$ is given by $$ Y(i):=R((QX)(i))\times_{R(\bar{X})}R(\bar{Y}) $$ and I have an evident natural transformation $$ Y\Rightarrow c(R(\bar{Y})), $$ where $c$ is the constant functor from $I$ at $R(\bar{Y})$. Thus, I get a map $$ \text{hocolim}(Y)=\text{colim}(QY)\rightarrow \text{hocolim}(c(R(\bar{Y})))=\text{colim}(Q(c(R(\bar{Y})))), $$ hence also a map $$ t\colon\text{hocolim}(Y)\rightarrow R(\bar{Y}) $$ because I have a natural transformation $Q(c(R(\bar{Y})))\Rightarrow c(R(\bar{Y}))$ and the colimit of the RHS (EDIT) has a canonical map to $R(\bar{Y})$. I may then require this $t$ to be a weak equivalence.

Question: Does this make any kind of sense?

I would really appreciate any kind of partial answer to any of my questions above or even some general suggestions about how to interpret these kinds of constructions with homotopy colimits, pullbacks and so on.

Thank you in advance!

$\endgroup$
  • 2
    $\begingroup$ Reposting from math.stackexchange.com/questions/1211666/…, as I realized the question may suit MO as well. I hope this is indeed the case. $\endgroup$ – Marco Vergura Mar 31 '15 at 16:16
  • 2
    $\begingroup$ I think everything you said is correct except for the fact that $\mathrm{colim}c(R(\bar{Y}))$ is $R(\bar{Y})$. It is not always true that the colimit of $c(X)$ is $X$ (think about coproducts). However you always have a map $\mathrm{colim}c(X)\to X$ which is enough for your purpose. $\endgroup$ – Geoffroy Horel Apr 2 '15 at 8:41
  • $\begingroup$ @GeoffroyHorel Yes, you are right of course, I have just edited the text of the question...So, in the end, do you think that the proposed solution may be the right way (even if not so handy, I would say) to interpret the construction made in the preprint, at least as long as cofibrantly generation is available for our model category $\mathbf{M}$? $\endgroup$ – Marco Vergura Apr 2 '15 at 8:56
  • 3
    $\begingroup$ Yes I think your proposed solution is correct. These cofibrant-fibrant replacement are a bit annoying and people tend to not write too much about them. They trust that the reader will be able to fill in the details which you did apparently :) $\endgroup$ – Geoffroy Horel Apr 2 '15 at 9:32
  • $\begingroup$ Sorry I don't quite see where is the problem. I believe the canonical morphisms in homotopy categories provided by universal properties of the left adjoint $\textrm{hocolim}:\mathrm{Ho}({\mathbf M}^I)\to\mathrm{Ho}({\mathbf M})$ and the similar right adjoint in case of homotopy pullbacks suffice to produce an explicit morphism in the homotopy category. Does not it suffice to show that the latter morphism is an isomorphism in the homotopy category? $\endgroup$ – მამუკა ჯიბლაძე Apr 12 '15 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.