User agusti roig - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:39:33Z http://mathoverflow.net/feeds/user/1246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5179/global-fibrations-of-simplicial-sheaves global fibrations of simplicial sheaves Agusti Roig 2009-11-12T09:47:54Z 2013-03-23T01:00:55Z <p>I'm reading the classical Brown-Gersten's paper "Algebraic K-theory as generalized sheaf cohomology" and I'm stuck with their choose of global fibrations. Namely, a morphism of simplicial sheaves $p : E \longrightarrow B$ is a global fibration if for every inclusion of open sets $U\subset V$ the natural map $E(V) \longrightarrow B(V) \times_{B(U)} E(U)$ is a (Kan) fibration of simplicial sets.</p> <p>My problem is: why these fibrations? As far as I can see, when they make use of this definition in constructing the factorizations of the model category structure, they could have chosen the fibrations to be defined open-wise: $p : E \longrightarrow B$ is a fibration if $p(V) : E(V) \longrightarrow B(V)$ is a (Kan) fibration of simplicial sets for every open set $V$ and apply as well the small object argument they use at this point.</p> <p>In other contexts I understand this kind of fibrations. For instance, for the model structure of the category of diagrams $C^I$ of a model category $C$ when $I$ is a 'very small' category (Dwyer-Spalinski, "Homotopy theories"), or a Reedy category. In this cases, this kind of fibrations ensures that you can extend your liftings by induction. But I don't see if this is their role with a category of sheaves, since no induction seems to be at hand.</p> <p>A colleague of mine has said to me thas this choice of fibrations is the consequence of choosing the cofibrations to be the monomorphism, following Joyal's "Letter to Grothendieck"; that is, these are precisely the fibrations if you choose monomorphisms as cofibrations and ask fibrations to have the RLP with respect to trivial cofibrations. But I couldn't find anywhere this famous Joyal's letter, so I would also be glad if someone could tell me where I can find it.</p> <p>Thanks in advance for any hints.</p> http://mathoverflow.net/questions/77094/a-model-structure-on-the-category-of-dualizing-maps/77126#77126 Answer by Agusti Roig for A model structure on the category of "dualizing maps" Agusti Roig 2011-10-04T12:45:29Z 2011-10-04T12:45:29Z <p>I'm not sure about your category of arrows, but it's certainly true the following result: giving a model category $\cal{C}$, you always have a model structure on the category of its arrows $\cal{C}^2$ with the following distinguished morphisms:</p> <ol> <li>Weak equivalences: pairs $(f, g)$ where both $f, g$ are weak equivalences.</li> <li>Fibrations: pairs $(f, g)$ where both $f, g$ are fibrations.</li> <li>Cofibrations: pairs $(f, g)$ where $f$ and the induced map $M'\sqcup_M A \longrightarrow A'$ are cofibrations </li> </ol> <p>Besides being probably folklore, this result can also be found in A. Roig, "Model category structures in bifibred categories", JPAA 95 (1994), corollary 7.3.</p> <p>So, in your case I guess you would need a substitute for the push-out $M'\sqcup_M A$, which perhaps it's not difficult to find.</p> <p>The above result for the category of arrows $\cal{C}^2$ is a particular case of a more general one about building model category structures for bifibred categories (op.cit., theorem 5.1). This theorem comes together with its dual (theorem 5.1$^o$), that you obtain exchanging fibrations and cofibrations. So, though it's not written explicitely in the paper, you have another "natural" model structure on $\cal{C}^2$, exchanging (2) and (3) above, and putting a pull-back in the place of the push-out.</p> http://mathoverflow.net/questions/56809/what-is-the-best-way-to-study-rational-homotopy-theory/56864#56864 Answer by Agusti Roig for What is the best way to study Rational Homotopy Theory Agusti Roig 2011-02-28T01:09:41Z 2011-02-28T01:09:41Z <p>After reading Griffiths-Morgan, Bott-Tu (not just the chapter on Rational Homotopy Theory, I would say) and Felix-Halperin-Thomas, maybe it wouldn't be a bad idea to be acquainted with:</p> <ol> <li>Halperin, Lectures on minimal models, Mémoires SMF 230 -aka "the bible": all technical details you won't find elsewhere.</li> <li>Bousfield, Gugenheim, On PL De Rham theory and rational homotopy type, Memoirs AMS 179 -the model category point of view; Sullivan's results can be stated as an equivalence of categories: find which.</li> <li>Lehman, Théorie homotopique des formes différentielles, Asterisque 45 -if you know French, this is a very nice introduction to the subject.</li> </ol> http://mathoverflow.net/questions/56166/do-homotopy-groups-always-commute-with-filtered-colimits Do homotopy groups "always" commute with filtered colimits? Agusti Roig 2011-02-21T11:58:26Z 2011-02-22T02:23:26Z <p>It is well-known that homotopy groups, of, say, simplicial sets, commute with filtered colimits.</p> <p>However, I could not find a reference for an analogous result for homotopy groups of spectra, or, under which hypothesis the same result would hold for an "arbitrary" simplicial model category.</p> <p>More precisely, let $\cal{M}$ be a simplicial model category. For a fibrant object $X \in \cal{M}$ its <em>homotopy groups</em> with coefficients in a cofibrant object $W\in \cal{M}$ may be defined as</p> <p>$$\pi_n (X; W) = [\Sigma^nW, X] = \pi_n \mathrm{map}(W,X) \ ,$$</p> <p>where $\mathrm{map}$ denotes the simplicial mapping space from $W$ to $X$.</p> <blockquote> <p>So my first question is: which hypothesis do I have to assume for $W$ to obtain an isomorphism</p> <p>$$\mathrm{colim}_i \pi_n (X_i;W) = \pi_n (\mathrm{colim}_i X_i;W) \ ?$$</p> <p>And the second one: in which kind of model category such an isomorphism holds <em>for every</em> cofibrant object $W$ -or, at least, for "sufficiently" many cofibrant objects $W$? </p> </blockquote> <p>The reason behind my question is the following (and explains the meaning of that "sufficiently"): I have a filtered category $I$, functors $X_\bullet, Y_\bullet : I \longrightarrow {\cal M}_f$ and a natural transformation $f_\bullet : X_\bullet \longrightarrow Y_\bullet$, such that, for every cofibrant object $W$, $f_\bullet$ induces isomorphisms</p> <p>$$\mathrm{colim}_i \pi_n (X_i ; W) = \mathrm{colim}_i \pi_n (Y_i ; W) \ ,$$</p> <p>for every $n$. And I want to conclude that the induced map between the colimits</p> <p>$$\mathrm{colim}_i X_i \longrightarrow \mathrm{colim}_i Y_i$$</p> <p>is a weak equivalence. Which would be true if</p> <ol> <li><p>I could commute colimits and homotopy groups, at least for</p></li> <li><p>"enough" cofibrant objects $W$ -in case of simplicial sets, one point $W = *$ is enough.</p></li> </ol> <p>I suspect the answer involves words like "smallness / compactness" and "cellular model category". For instance an answer like: "You can do that in no matter what simplicial cellular model category" -in which every cofibrant object is compact- would be fine. Nevertheless, as long as I can understand, commutations like</p> <p>$$\mathrm{colim}_i {\cal M} (W, X_i ) = {\cal M} (W, \mathrm{colim}_i X_i)$$</p> <p>hold for $W$ small and $\lambda$-<em>sequences</em>; that is, when the domain of the functor $X : \lambda \longrightarrow \cal{M}$ is an <em>ordinal</em>; in particular, a <em>totally ordered</em> set, which my filtered $I$ needs not to be.</p> <p>So any references of a result along these lines, even just for spectra, are welcome.</p> http://mathoverflow.net/questions/28972/colimits-of-spectral-sequences colimits of spectral sequences Agusti Roig 2010-06-21T17:55:56Z 2011-02-21T19:12:59Z <p>I'm looking for some references about colimits of spectral sequences.</p> <p>More precisely: let $X : I \longrightarrow \cal{C}$ be a functor from a filtered category $I$ to the category of double cochain complexes of an abelian category $\cal{C}$, in which filtered colimits exist and commute with cohomology.</p> <p>Let $E_2(X_i)$ be the second page of the first filtration ss associated to $X_i$. Assuming that the $X_i$ are right-half plane double complexes, it weakly converges to $H^*(\mbox{Tot}^\prod X_i)$ for all $i$ (Weibel, "An introduction to homological algebra", page 142):</p> <p>$$E_2(X_i) \Longrightarrow H^*(\mbox{Tot}^\prod X_i)\ ,$$</p> <p>where $\mbox{Tot}^\prod$ is the <em>total product complex</em>,</p> <p>$$(\mbox{Tot}^\prod X)^n = \prod_{p+q=n} X^{pq} \ .$$</p> <p>For the same reason:</p> <p>$$E_2(\underset{i}{\lim_\longrightarrow} X_i) \Longrightarrow H^*(\mbox{Tot}^\prod \underset{i}{\lim_\longrightarrow} X_i )\ .$$</p> <p>Then, because of the exactness of $\displaystyle \lim_\longrightarrow$, we have</p> <p>$$\underset{i}{\lim_\longrightarrow} E_2 (X_i) = E_2(\underset{i}{\lim_\longrightarrow} X_i) \ .$$</p> <p>Then my question is: under which conditions can I assure that I have a comparison theorem like</p> <p>$$\underset{i}{\lim_\longrightarrow} H^* (\mbox{Tot}^\prod X_i) = H^*(\mbox{Tot}^\prod \underset{i}{\lim_\longrightarrow} X_i) \quad \mbox{?}$$</p> <p>Any hints or references will be appreciated. </p> http://mathoverflow.net/questions/28263/tot-and-colimits Tot and colimits Agusti Roig 2010-06-15T15:36:06Z 2010-11-14T12:52:47Z <p>This must be a well-known exercise with spectral sequences, but I don't know a reference for it. I'm trying to figure out when does $Tot$ commute with colimits.</p> <p>More precisely, let $X$ be a double cochain complex of, say, $R$-modules, $R$ a commutative ring with unit, or, more generally, a double complex in an abelian category. Let $\cal{C}$ denote the category of these double cochain complexes.</p> <p>We have two different <em>total functors</em>, $\mbox{Tot}^\prod$ and $\mbox{Tot}^{\bigoplus}$, from the category of double complexes to the category of cochain complexes:</p> <p>$$\mbox{Tot}^{\prod}(X)^n = \prod_{p+q=n}X^{p,q} \qquad \mbox{and} \qquad \mbox{Tot}^{\bigoplus}(X)^n = \bigoplus_{p+q=n}X^{p,q} \quad .$$</p> <p>Let $\mbox{Tot}$ denote anyone of them and let $I$ be a (filtered) category, and $X: I \longrightarrow \cal{C}$ a functor. We have a natural morphism</p> <p>$$\theta: \varinjlim_i \mbox{Tot} (X_i) \longrightarrow \mbox{Tot} (\varinjlim_i X_i) \quad .$$</p> <p>When dealing with $\mbox{Tot}^\bigoplus$, this $\theta$ is an isomorphism, because a direct sum is a colimit and colimits commute with colimits.</p> <p>What happens when we take $\mbox{Tot}^\prod$? Is $\theta$ at least a quasi-isomorphism (a morphism inducing an isomorphism in cohomology)? In which cases? Do we need some extra hypothesis on the abelian category (AB...)? Is the hypothesis "filtered" really needed, or we can deal with arbitrary colimits in general?</p> <p>Of course, if our double complex has finite diagonals, then $\mbox{Tot}^\prod = \mbox{Tot}^\bigoplus$, and we are done. But what happens without this hypothesis?</p> <p>I'm mainly interested in the case of a right half-plane double complex, that is $X^{p,q} = 0$ if $p&lt;0$, but I'll be glad to learn about all possible cases.</p> <p>Any references or hints will be welcome.</p> http://mathoverflow.net/questions/8543/learning-to-think-categorically/8586#8586 Answer by Agusti Roig for Learning to Think Categorically Agusti Roig 2009-12-11T18:10:36Z 2010-11-02T04:31:08Z <p>Though most people seem to advise against reading Mac Lane's "Categories for the working mathematician", and neither did I read it from start to finish, looking at it, I find that it is highly worth trying to read some parts because it's extremely very well written: to the best of my knowledge this book is second to none in his field (of course, nowadays you have also Borceux's book, but its scope is something different). So if your mathematical interests force you to use categories, you'll have to consult Mac Lane's again and again. Starting to read it directly is a way to get an idea of where to find things when you need them.</p> <p>Ok, there is a problem because, for instance, it begins with "metacategories" (chapter I, section 1), so when you arrive to <em>real</em> categories (section 2) you may be already completely lost. Hence, I asked myself: what parts of Mac Lane did I really have used in my own work and find them useful, worth reading or consulting, or are unavoidable in the language of categories? The following is my own selection of some chapters and sections of Mac Lane's book, based only on my personal tastes and biases (the selection is from the first edition, but, if I'm not wrong, the only difference is that the second one has an extra chapter on monoidal and braided categories and functors near the end):</p> <p>Chapter I: 2, 3, 4, 5, 8. Chapter II: 2, 3. Chapter III: 1, 2, 3, 4, 5 Chapter IV: 1, 2, 4. Chapter V: 1, 4, 5. Chapter VI: 1, 2. Chapter VII: 1, 3, 4, 5, 6. Chapter VIII: 1, 2, 3, 4. Chapter IX: 1, 2.</p> <p>In general, I've tried to avoid both too abstract issues, logical foundations and too specific or specialized matters.</p> <p>This kind of reading, of course, raises the problem of encountering terms you haven't seen defined before, or results you haven't studied. But in this cases, I think there's no harm in going to the index and find where the term is defined, or taking the result you haven't seen before on faith. Trying to do some exercises is of course necessary and the historical notes at the end of the chapters are interesting too.</p> http://mathoverflow.net/questions/42755/non-standard-model-category-structure-on-cochain-complexes Non standard (?) model category structure on (co)chain complexes. Agusti Roig 2010-10-19T08:25:23Z 2010-10-19T10:14:05Z <p>Let $\cal{A}$ be an abelian category with enough projectives and $\mathbf{C}_+ (\cal{A})$ the category of bounded below chain complexes.</p> <p>Since Quillen (Homotopical algebra, 1.2, examples B), there is a well-known "standard" model category structure on $\mathbf{C}_+ (\cal{A})$ taking as weak equivalences the maps inducing isomorphisms on homology, as fibrations the degree-wise epimorphisms in $\cal{A}$ and with cofibrations maps $i$ which are injective and such that $\mathrm{cok}\ i$ is a complex having a projective object of $\cal{A}$ in each degree.</p> <p>More recently, Hovey (Model categories), proved an analogous result for the category of <em>unbounded</em> chain complexes, but with ${\cal A} = R$-modules, $R$ a ring (but cofibrations are not so easy to characterise). Finally, it's folklore (at least, I don't know if it is published somewhere) that the same holds for $\cal{A}$ an abelian category with a projective generator -the fact that allows the small object argument to work, as Eric Wofsey points out in his answer to <a href="http://mathoverflow.net/questions/141/model-category-structures-on-categories-of-complexes-in-abelian-categories" rel="nofollow">this MO question</a>.</p> <blockquote> <p>I'm interested in the following variant of this problem: is there a model structure on $\mathbf{C}_+ (R)$ taking as weak equivalences the <strong>homotopy</strong> equivalences?</p> </blockquote> <p>If it's true, I think this should be easy to verify: just taking a look to the classical proof and seing if you can change "<strong>homology</strong> equivalences" everywhere by "<strong>homotopy</strong> equivalences". I'm willingly going to do it, but, prior to start, I would like to know if it is already done, much as in the case of topological spaces where, together with the "standard" (Quillen too) model structure with <strong>weak homotopy</strong> equivalences as weak equivalences, there is the Strom model structure (The homotopy category is a homotopy category), with <strong>homotopy</strong> equivalences as weak equivalences.</p> http://mathoverflow.net/questions/32173/mitchells-embedding-theorem Mitchell's embedding theorem Agusti Roig 2010-07-16T14:04:45Z 2010-09-25T04:15:41Z <p>Mitchell's embedding theorem <a href="http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem</a> tells us that every small abelian category ${\cal A}$ has a full, faithful and exact embedding $V : {\cal A} \longrightarrow \mathbf{Mod}_R$ in a category of $R$-modules, for some ring $R$.</p> <p>Now, $V$ being exact is the same as saying that it preserves all <em>finite</em> limits and colimits.</p> <p>I would be glad to know if Mitchell's embedding theorem could be improved in order to have that $V$ preserves also:</p> <p>(a) arbitrary products, and (b) filtered colimits.</p> <p>Or, alternatively,</p> <p>(c) injective objects.</p> <p>Or, which conditions on the abelian category ${\cal A}$ would guarantee (a) and (b), or (c)? Are there any results in these directions?</p> <p>The reason behind my question is the following: I realized that my answer to my previous question <a href="http://mathoverflow.net/questions/29883/vanishing-theorems" rel="nofollow">http://mathoverflow.net/questions/29883/vanishing-theorems</a> is wrong: sheaf cohomology is not defined uniquely in terms of exact sequences, so the fact that $V$ is exact doesn't guarantee that $H^n(X; {\cal F}) = H^n(X; V({\cal F}))$ as I claimed. But, if I had (a) and (b), I could say that $V$ preserves Godement resolutions. And if I had (c), $V$ would preserve injective resolutions.</p> http://mathoverflow.net/questions/36494/is-the-first-filtration-hausdorff/36681#36681 Answer by Agusti Roig for Is the first filtration Hausdorff? Agusti Roig 2010-08-25T16:14:16Z 2010-08-25T16:14:16Z <p>[This should be a comment, but went too long and somewhat involved to write.]</p> <p>This seems a perfect counter-example: thank you very much. Still, I have some doubts.</p> <p>If I understand you correctly, you are saying that, since every term in the "superdiagonal" becomes zero eventually at some page of the spectral sequence, then the corresponding "superdiagonal" in the $E_\infty$ page will be zero. So we will have $H^1(\mathbf{Tot}X) = F^0 = F^1 = F^2 = \dots$</p> <p>But isn't that already true at the $E_2$ page? I mean: aren't the differentials $d_1 : E_1^{p,-p} \longrightarrow E_1^{p+1,-p}$ simply the projections $A_p \longrightarrow A_p/\mathrm{im}\alpha_{p+1}$, where $\alpha_p : A_p \longrightarrow A_{p-1}$ are the morphisms of the inverse system?</p> <p>If this is so, we're already killing the "superdiagonal" at the $E_2$ page, aren't we?</p> http://mathoverflow.net/questions/36494/is-the-first-filtration-hausdorff Is the first filtration Hausdorff? Agusti Roig 2010-08-23T20:40:07Z 2010-08-25T16:14:16Z <p>Maybe this is too technical and elementary, but I cannot make up my mind, nor find a reference.</p> <p>The situation is the following: let $X$ be a double cochain (right half-plane) complex of abelian groups and let </p> <p>$$(\mathbf{Tot}^{\prod} X)^n = \prod_{p+q = n}X^{pq}$$</p> <p>denote its total-product complex. The first filtration on $X$,</p> <p>$${}_I F^s(X) = \begin{cases} X^{pq} &amp; \ \text{if} \ p \geq s , \\ 0 &amp; \ \text{otherwise} \end{cases}$$</p> <p>gives you the filtration on $\mathbf{Tot}^\prod X$:</p> <p>$$(F^s \mathbf{Tot}^\prod X)^n = \prod_{p+q=n , p\geq s} X^{pq}$$</p> <p>and you have, as with any filtered differential complex $(A, F, d)$, an induced filtration in cohomology:</p> <p>$$F^pHA = \mathbf{im} (HF^pA \longrightarrow HA) \ .$$</p> <p>My question is the following: is the filtration induced by ${}_I F$ on $H(\mathbf{Tot}^\prod X)$ Hausdorff? That is, </p> <p>$$\bigcap_p F^pH(\mathbf{Tot}^\prod X ) = 0 \ ?$$</p> <p>I couldn't find an answer in the literature. Weibel's book says that in this situation the spectral sequence arising from the first filtration is "convergent". Unfortunately, for Weibel this only means that you have an isomorphism $E_0 HA = E_\infty A$. Cartan-Eilenberg "Homological Algebra" doesn't work with the total-product complex, but with the total-sum one:</p> <p>$$(\mathbf{Tot}^{\bigoplus} X)^n = \bigoplus_{p+q = n}X^{pq}$$</p> <p>For this one, I think, the answer is "yes": if I had some cohomology class</p> <p>$$[x] \in \bigcap_p F^pH^n(\mathbf{Tot}^\bigoplus X ) \quad \Longleftrightarrow \quad [x] \in F^pH^n(\mathbf{Tot}^\bigoplus X ) \ \text{for all} \ p$$</p> <p>then I could find representatives for $[x]$ like</p> <p>$$(0, \stackrel{p-1}{\dots}, 0, x^{p,n-q}, x^{p+1, n-q-1}, \dots )$$</p> <p>for all $p \geq 0$. Since there is only a finite number of $x^{pq} \neq 0$ for each element of $(\mathbf{Tot}^\bigoplus X)^n$, in a finite number of steps, I can be sure that I can find a representative for $[x]$ which is zero, so $[x]=0$.</p> <p>But this reasoning doesn't work with $\mathbf{Tot}^\prod X$: you can only state with certainty that, for every $p$ there is some $x_p \in F^p$ and $b_p$ such that </p> <p>$$x - x_p = db_p \quad \Longleftrightarrow \quad x- db_p \in F^p \ .$$</p> <p>These equations have a nice interpretation: if you topologize $\mathbf{Tot}^\prod X$ taking as basic open sets $x + F^p$ for all $x \in \mathbf{Tot}^\prod X$ and $p$, they read:</p> <p>$$(db_p) \longrightarrow x \ .$$</p> <p>That is, $x$ is a limit of coboundaries. But, unless the filtration is finite, this doesn't imply that $x$ itself <em>is</em> a coboundary, does it?</p> <p>So I could ask my question this way: in this situation, is the set of coboundaries closed?</p> http://mathoverflow.net/questions/35094/colimit-of-intersections Colimit of intersections Agusti Roig 2010-08-10T10:08:00Z 2010-08-10T11:35:13Z <p>Let $B_i^p$ be a family of sets, where $p\in \mathbb{N}$ and $i \in I$, $I$ being a directed set, and such that, for every $i$, we have a descending chain of inclusions</p> <p>$$\dots \supset B_i^{p-1} \supset B_i^p \supset B_i^{p+1} \supset \dots$$</p> <p>Question: is the following implication true?</p> <p>$$\bigcap_p B_i^p = \emptyset, \ \text{for all} \ i \quad \Longrightarrow \quad \bigcap_p \varinjlim_i B_i^p = \emptyset \ .$$</p> <p>Since $\bigcap_{}$ is a limit, this seems a problem of an interchange of limits and filtered colimits and indeed there is a universal map</p> <p>$$\varphi: \varinjlim_i \bigcap_p B_i^p \ \longrightarrow \ \bigcap_p \varinjlim_i B_i^p$$</p> <p>If $\varphi$ was a bijection, then my implication would be true with no doubts, but, since the intersection is not finite, I can not say that $\varphi$ is a bijection. Nevertheless, could my implication still be true, without $\varphi$ being a bijection?</p> <p>The reason behind my question is the following: let $(A_i, F_i)$ be a directed family of filtered sets (or abelian groups, or modules; in fact, in my problem they are cochain complexes). Since filtered colimits (direct limits) are exact, you can define a filtration on the colimit like this:</p> <p>$$F^p\varinjlim_i A_i = \varinjlim_iF_i^pA_i \ .$$</p> <p>Now assume all the filtrations $F_i$ are Hausdorff; that is, $\bigcap_p F_i^pA_i = 0$ for all $i$. Is it then necessarily true that the filtration $F$ on $\varinjlim_iA_i$ is Hausdorff too?</p> <p>This question is a sequel to my previous question <a href="http://mathoverflow.net/questions/32762/convergence-of-right-half-plane-spectral-sequence-bounded-on-the-right" rel="nofollow">http://mathoverflow.net/questions/32762/convergence-of-right-half-plane-spectral-sequence-bounded-on-the-right</a> . Despite Tilman's counterexemple to my guess there, I think I've managed almost to prove it because my spectral sequences are right half-plane and this is the final detail I need.</p> http://mathoverflow.net/questions/3398/mayer-vietoris-homotopy-groups-sequence-of-a-pull-back-of-a-fibration Mayer-Vietoris homotopy groups sequence of a pull-back of a fibration Agusti Roig 2009-10-30T04:43:17Z 2010-07-23T17:01:20Z <p>This must be an elementary question: could somebody tell me a reference for the Mayer-Vietoris <em>homotopy</em> groups sequence of a pull-back of a fibration?</p> <p>I'm working in the category of pointed simplicial sets. So I've a pull-back of a (Kan) fibration of pointed simplicial sets, and I've read that in this situation you have an associated Mayer-Vietoris sequence relating the homotopy groups of the simplicial sets of the pull-back that looks like the classical Mayer-Vietoris sequence for the singular <em>homology</em> of a pair of open sets covering a topological space.</p> <p>I've been searching in May's "Simplicial objects in Algebraic Topology" and Goerss-Jardine's "Simplicial Homotopy Theory", but I couldn't find it.</p> http://mathoverflow.net/questions/32762/convergence-of-right-half-plane-spectral-sequence-bounded-on-the-right Convergence of right half-plane spectral sequence bounded on the right Agusti Roig 2010-07-21T08:42:14Z 2010-07-21T20:08:38Z <p>This is a sequel to my previous question <a href="http://mathoverflow.net/questions/28972/colimits-of-spectral-sequences" rel="nofollow">http://mathoverflow.net/questions/28972/colimits-of-spectral-sequences</a> .</p> <p>I think I've found the answer in S.A. Mitchell's paper "Hypercohomology spectra and Thomason's descent theorem". There the author states a "colimit lemma" (page 42) for ss of homotopy groups of spectra, which I think can be literally translated with no harm for cohomology groups of cochain complexes and it's exactly the result I was looking for.</p> <p>However, my problem is more basic and shameful. Previously, but in the same page, Mitchell says that, having a right half-plane cohomology spectral sequence (coming, say, from a double complex, or a filtered complex), that is</p> <p>$$E_2^{pq} = 0 \qquad \mbox{if} \quad p &lt; 0 \ , \qquad \qquad \qquad [1]$$</p> <p>then it <em>converges</em> if it is, for instance, <em>bounded on the right</em>, that is, if there exists $d$ such that</p> <p>$$E_2^{pq} = 0 \qquad \mbox{if} \quad p > d . \qquad \qquad \qquad [2]$$</p> <p>I assume the author uses the term "converge" in the sense of Cartan-Eilenberg (otherwise, I don't understand the proof of his "colimit lemma" at all), that is:</p> <p>(a) We have an <em>exhaustive</em> filtration $F$ on the limit $G$, $\bigcup_p F^PG = G$, and also isomorphisms $E_\infty^p = F^pG / F^{p+1}G$, and</p> <p>(b) The filtration on $G$ is <em>Hausdorff</em>, that is $\bigcap_p F^pG = 0$.</p> <p>Now, I have no problem in assuming that in my particular ss the original filtration of my filtered complex is already exhaustive and hence so it is the induced one on the "limit" $G$ and the isomorphisms for $E_\infty$ (as it is the case for both filtrations of a double complex) and I think Mitchell is assuming this too implicitely, because these conditions seems "for free".</p> <p>Also, the boundness conditions [1] and [2] will imply that the filtration on $G$ is in fact <em>finite</em>. So, if we had (b), the converge would be in the strong sense. Great! :-)</p> <p>So the point is the Hausdorff condition of the filtration on $G$: why would [1] and [2] imply that the filtration on $G$ should also be Hausdorff?</p> http://mathoverflow.net/questions/29883/vanishing-theorems vanishing theorems Agusti Roig 2010-06-29T08:53:01Z 2010-07-06T10:39:05Z <p>I would be glad to know about possible generalizations of the following results:</p> <p>1) (Grothendieck) Let $X$ be a noetherian topological space of dimension $n$. Then for all $i>n$ and all sheaves of abelian groups $\cal{F}$ on $X$, we have $H^i(X; \cal{F})=$ 0. [See Hartshorne, Algebraic Geometry, III.2.7.]</p> <p>2) Let $X$ be an $n$-dimensional $C^0$-manifold. Then for all $i>n$ and all sheaves of abelian groups $\cal{F}$ on $X$, we have $H^i(X; \cal{F})=$ 0 . [See Kashiwara-Schapira, Sheaves on manifolds, III.3.2.2]</p> <p>More precisely, I'm interested in dropping the "abelian groups" hypothesis: could I take sheaves in any, say, AB5 abelian category?</p> <p>Apparently, in Grothendieck's theorem, the "abelian groups" hypothesis is necessary -at least in Hartshorne's proof-, because at the end you see a big constant sheaf $\mathbf{Z}$. But what happens if we talk about sheaves of $R$-modules, with $R$ any commutative ring with unit, for instance?</p> <p>Are those generalizations trivial ones? False for trivial reasons?</p> <p>Any hints or references will be welcome.</p> http://mathoverflow.net/questions/29883/vanishing-theorems/30756#30756 Answer by Agusti Roig for vanishing theorems Agusti Roig 2010-07-06T10:39:05Z 2010-07-06T10:39:05Z <p>Well, I think I can answer my question, thanks to Boyarsky's remark.</p> <p>The point is that, since the theorem is also true for sheaves of $R$-modules, given a sheaf $\cal{F}$ with values in an abelian category $\cal{A}$, with the help of Mitchell's embedding theorem, <a href="http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem</a>, we can consider it as a sheaf of $R$-modules, for some ring $R$. Moreover, the embedding $V: {\cal A} \longrightarrow \mathbf{Mod}_R$ is full, faithful, and <em>exact</em>. That is to say, $V$ sends exact sequences to exact sequences. So $H^n(X;\cal{F})$ = $H^n(X;V(\cal{F}))$.</p> <p>Hence, both vanishing theorems are also (trivially) true for sheaves with values in any abelian category.</p> http://mathoverflow.net/questions/20214/principle-when-limits-level-by-level-dont-commute-with-simplicial-structure/20229#20229 Answer by Agusti Roig for Principle when limits level by level don't commute with simplicial structure Agusti Roig 2010-04-03T11:18:46Z 2010-04-03T13:35:06Z <p>Limits and colimits can be computed point-wise in functor categories, such as the category of simplicial objects of any category. That is, if you have a functor category $\mathcal{C}^I$ ($I = \Delta$ in your case) and a functor $F: P \longrightarrow \mathcal{C}^I$, in particular, for every object $p$ in $P$ you have a functor $F_p : I \longrightarrow \mathcal{C}$. Then $\varinjlim_p F_p$ is an object of $\mathcal{C}^I$; that is, a functor $I \longrightarrow \mathcal{C}$, whose value on objects $i$ in $I$ is</p> <p>$$(\varinjlim_p F_p)(i) = \varinjlim_p (F_p(i)) \ .$$</p> <p>This is true as far as the colimit on the right exists for every object $i$ in $I$. An analogous statement applies for limits in functor categories.</p> <p>For limits, you can find the result in MacLane, "Categories for the working mathematician", chapter V, section 3, theorem 1 (see at the end of the proof also). For colimits, see "Sheaves in geometry and logic: a first introduction to topos theory", by Saunders Mac Lane,Ieke Moerdijk, p.40.</p> http://mathoverflow.net/questions/19446/sheaves-on-discrete-spaces sheaves on discrete spaces Agusti Roig 2010-03-26T17:19:19Z 2010-03-26T20:17:54Z <p>Sorry if this question is too broad, but if I explain why I'm <em>really</em> interested in it, then it would be too technical and specific. So, let me try to ask it this way.</p> <p>I would like to know properties of sheaves on discrete spaces. That is, let $X_{\mathrm{dis}}$ be a topological space with the discrete topology. What can be said about sheaves on $X_{\mathrm{dis}}$? -With values in, say, the category of abelian groups, complexes of an abelian category, your prefered category, an abstract category...</p> <p>For instance, I've already done part of my homework, and found, that, if I'm not wrong: </p> <ul> <li>Such sheaves are determined by its values on points $\mathcal{F}(x)$: for any $U \subset X_{\mathrm{dis}}$, $\mathcal{F} (U) = \prod_{x \in U} \mathcal{F}(x)$.</li> <li>For any $x \in X_{\mathrm{dis}}$, $\mathcal{F}_x = \mathcal{F}(x)$.</li> <li>Every such a sheaf is <em>flasque</em> (flabby). So, for instance, there is no great deal in deriving the direct images functor: for any map $f: X_{\mathrm{dis}} \longrightarrow Y\$, $f_*$ is an exact functor.</li> </ul> <p>What else can be said? Any reference for these toy sheaves?</p> <p>EDIT. As I've been told, sheaves on discrete spaces appear in different contexts. For instance, you can describe the cosimplicial Godement resolution with them: let </p> <p>$$I = id : X_{\mathrm{dis}} \longrightarrow X$$</p> <p>be the identity map between $X$ with the discrete topology and any topological space $X$. Then the cosimplicial Godement resolution can be defined as follows: $I$ induces a couple of adjoint functors (direct and inverse images)</p> <p>$$I_* : Sh(X_{\mathrm{dis}}) \leftrightarrows Sh(X): I^* \ ,$$</p> <p>you have a triple associated</p> <p>$$T = I_* I^* : Sh(X) \longrightarrow Sh(X)$$</p> <p>and the cosimplicial Godement resolution may be defined, using the standard construction associated to a triple (see McLane's "Categories for the working mathematician) as</p> <p>$$C^p (\mathcal{F}) = T^{p+1} (\mathcal{F})\ .$$</p> <p>Ok. Any other contexts where sheaves on discrete spaces appear?</p> http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/19444#19444 Answer by Agusti Roig for Why is a topology made up of 'open' sets? Agusti Roig 2010-03-26T16:51:40Z 2010-03-26T16:51:40Z <p>My answer will not be of philosophical nature, neither historical, but perhaps pedagogical.</p> <p>I find Munkress' Topology a great book. Among other merits, because its introduction, which I summarize as follows:</p> <ol> <li><p>You recall what a metric space is. Define open balls and subsequently, open sets. Prove that, in a metric space:</p> <p>1.1 The empty set and the total space are open sets.</p> <p>1.2 The union of an arbitrary number of open sets is an open set.</p> <p>1.3 The intersection of a finite number of open sets is an open set.</p></li> <li><p>Recall what a continuous map between metric spaces is (the $\epsilon$-$\delta$ definition). Prove the theorem that says that a map between metric spaces $f: X \longrightarrow Y$ is continuous if and only if $f^{-1} (U) \subset X$ is an open set for every open set $U \subset Y$.</p></li> </ol> <p>And you have a motivation for the definition of topological space and continuous map as well.</p> <p>Of course this is not an historical explanation of how topological spaces arised, nor does it justify why you chose <em>these</em> properties of open sets in metric spaces and not others: "experience" has told us that <em>these</em> are the good ones. (For instance, if I'm not wrong, when Hausdorff first defined topological spaces included the property of being... Hausdorff among the axioms. "Experience" -and not an a priori argument- showed us that it could be interesting to work with non-Hausdorff topological spaces.)</p> http://mathoverflow.net/questions/13995/nontrivial-isomorphisms-of-categories/14019#14019 Answer by Agusti Roig for nontrivial isomorphisms of categories Agusti Roig 2010-02-03T18:16:25Z 2010-02-03T19:51:30Z <p>I don't know if the following example may be considered as trivial, but it's quite useful.</p> <p>Let $\cal{C}$ be a category, $\cal{S}$ a class of morphism in $\cal{C}$.</p> <p>Assume that, for instance, $\cal{S}$ is a class of <em>homotopy equivalences</em>. By which I mean that you have a cylinder (or path object) for every object in $\cal{C}$ -for example, because it is a Quillen model category-, and $\cal{S}$ is the class of morphism which are invertible up to the homotopy relation $\sim$ generated by these path or cylinder objects.</p> <p>Then, on one hand, you can consider the <em>quotient category</em> $\cal{C}/\sim$, whose objects are those of $\cal{C}$ and whose morphisms are the homotopy classes of morphisms.</p> <p>On the other hand, you can consider the <em>localized category</em> $\mathrm{Ho}\cal{C}$, with the same objects, but inverting the morphisms of $\cal{S}$.</p> <p>Well, at least when your homotopy relation $\sim$ is generated by a cylinder or path object, these two categories are canonically isomorphic.</p> <p><em>Remark.</em> Do not confuse my statement with Quillen's <em>equivalence of categories</em>. I'm sorry for the notation $\mathrm{Ho}\cal{C}$, but I don't know how to write square brackets here.</p> http://mathoverflow.net/questions/12920/stokes-theorem-for-manifolds-with-corners Stokes theorem for manifolds with corners? Agusti Roig 2010-01-25T09:57:23Z 2010-02-01T09:47:21Z <p>Maybe this is an elementary question, but I'm unable to find the appropriate reference for it. The <em>Stokes theorem</em> tells us that, for a $n+1$-dimensional manifold $M$ with boundary $\partial M$ and any differentiable $n$-form $\omega$ on $M$, we have $\int_{\partial M} \omega = \int_M d\omega$.</p> <p>But Stokes theorem is also true, say, for a cone <code>$M = \{(x,y,z) \in \mathbb{R}^3 \ \vert\ \ x^2 + y^2 = z^2, 0\leq z \leq 1 \}$</code>, or a square in the plane, <code>$M =\{(x,y) \in \mathbb{R}^2 \ \vert\ 0 \leq x, y\leq 1 \}$</code> which are not manifolds. So my questions are:</p> <ol> <li>Are these cone and square examples of what I think are called "manifold with corners"?</li> <li>If this is so, where can I find a reference for a version of Stokes' theorem for manifolds with corners?</li> <li>If "manifold with corners" is not, which is the appropriate setting (and a reference) for a Stokes' theorem that includes those examples?</li> </ol> <p>Any hints will be appreciated.</p> <p>EDIT: Since thanking individually everyone would be too long, let me edit my question to acknowledge all of your answers. Thank you very much: I've found what I was looking for and more.</p> http://mathoverflow.net/questions/13005/what-is-formal/13063#13063 Answer by Agusti Roig for What is 'formal' ? Agusti Roig 2010-01-26T19:11:21Z 2010-02-01T08:41:00Z <p>Paraphrasing Groucho Marx: if you don't like my first answer..., well I have another one. :-)</p> <p>Here it is: let $X$ be a simply connected differentiable manifold.</p> <p>Rational homotopy theory tells us that the <em>rational homotopy type</em> of $X$ (that is, its homotopy type modulo torsion) is contained in its <em>minimal model</em>, <code>$M_X$</code>, which is a <em>commutative</em> differential graded (cdg) algebra.</p> <p>By definition, this means that you have a quasi-isomorphism (<em>quis</em>, a morphism of cdg algebras inducing an isomorphism in cohomology)</p> <p>$$M_X \longrightarrow \Omega^*(X) \ .$$</p> <p>Here, $\Omega^* (X)$ is the algebra of differential forms of $X$ and the <em>minimality</em> of $M_X$ means that, in a certain, but precise, sense, it is the smallest cdg algebra for which such a quis exists.</p> <p>The fact that $M_X$ <em>contains</em> the rational homotopy type of $X$ implies, for instance, that you can obtain the ranks of the homotopy groups of $X$ from it:</p> <blockquote> <p>rank $\pi_n(X) =$ number of degree n generators (as an algebra) of $M_X$, for $n \geq 2$.</p> </blockquote> <p>Nice, isn't it? :-)</p> <p>The problem is that the algebra $\Omega^*(X)$ is, in general, not computable, so you can not obtain from it the minimal model $M_X$. And here is where formality comes to help you.</p> <p>Almost by definition, $X$ is a <em>formal</em> space if there exists two quis</p> <p>$$\Omega^*(X) \longleftarrow M_X \longrightarrow H^*(X;\mathbb{Q})<br />$$</p> <p>Hence, if $X$ is formal you can compute its minimal model $M_X$, and hence its rational homotopy type, directly from the cohomology algebra <code>$H^*(X; \mathbb{Q})$</code>, which is nicer (smaller, more computable) than $\Omega^*(X)$.</p> <p>And the final point is that there are plenty of examples of spaces which are known to be formal.</p> <p>(Final remark: Actually, you'd have to put <code>$A_{PL}^*(X;\mathbb{Q})$</code> instead of $\Omega^*(X)$ to work over the rationals, but this you can find it explained in the references we have provided for you.)</p> http://mathoverflow.net/questions/13005/what-is-formal/13028#13028 Answer by Agusti Roig for What is 'formal' ? Agusti Roig 2010-01-26T09:17:44Z 2010-01-26T09:17:44Z <p>Maybe you could take a look at</p> <p>Y. Félix, J. Oprea, D. Tanré; Algebraic models in Geometry, Oxford Graduate Text in Math. 17 (2008)</p> <p>where they talk about formality in the context of rational homotopy theory, RHT, (for instance, in sections 2.7 and 3.1.4). Also the more classical, but excellent little book</p> <p>D. Lehmann; Théorie homotopique des formes différentielles, Astérisque 45</p> <p>is worth reading (section V.9).</p> <p>As for formality in the context of operads, allow me a little self-promotion :-) :</p> <p>F. Guillén, V. Navarro, P. Pascual, Agustí Roig, Moduli spaces and formal operads; Duke Math. J. 129, 2 (2005).</p> <p>In this work, we translate some classical results concerning formality in RHT to chain operads. For instance, the Deligne-Griffiths-Morgan-Sullivan theorem about formality of Kähler manifolds, formality's independence of the ground field... And extend them also to modular operads.</p> http://mathoverflow.net/questions/9490/what-are-the-fibrant-objects-in-the-injective-model-structure/9502#9502 Answer by Agusti Roig for What are the fibrant objects in the injective model structure? Agusti Roig 2009-12-21T20:09:35Z 2009-12-21T20:09:35Z <p>I think you can find an answer in my question about global fibrations of simplicial sheaves: <a href="http://mathoverflow.net/questions/5179/global-fibrations-of-simplicial-sheaves" rel="nofollow">http://mathoverflow.net/questions/5179/global-fibrations-of-simplicial-sheaves</a> .</p> <p>There, Andreas Holmstrom pointed me to Voevodsky's preprint "Homotopy theory of simplicial presheaves in completely decomposable topologies", <a href="http://front.math.ucdavis.edu/0805.4578" rel="nofollow">http://front.math.ucdavis.edu/0805.4578</a> , where I discovered lemma 4.1 , which I think aswers your question.</p> <p>Although it is not proved, Voevodsky says that it is straightforward. I could manage myself to write a proof of it, at least under the hypothesis of Brown-Gernstern's "Algebraic K-theory as generalized sheaf cohomology", in LNM 341/1973, that is, with sheaves defined on a Noetherian space of finite Krull dimension.</p> <p>In this situation at least, fibrations are <em>global fibrations</em>. That is, a morphism of sheaves $p: E \longrightarrow B$ is a global fibration if and only if if for every inclusion of open sets $U\subset V$ the natural map $E(V) \longrightarrow B(V) \times_ {B(U)} E(U)$ is a (Kan) fibration of simplicial sets.</p> <p>As a corollary, if you take $B = *$, this condition tells you that fibrant objects are those for which each restriction map $E(V) \longrightarrow E(U)$ is a Kan fibration. In particular, put $U=\emptyset$ and this implies that each $E(V)$ must be a Kan complex.</p> <p>All this with Brown-Gersten's hypothesis, but Voevodsky doesn't seem to need them, so maybe it is also true in your situation.</p> http://mathoverflow.net/questions/9267/what-is-an-explicit-example-of-a-sequence-converging-to-two-different-points/9287#9287 Answer by Agusti Roig for What is an explicit example of a sequence converging to two different points? Agusti Roig 2009-12-18T16:55:48Z 2009-12-18T16:55:48Z <p>An easy example, in the same vein as Greg's one. Take the real line $\mathbb{R}$ with the <strong>finite complement topology</strong>, <a href="http://en.wikipedia.org/wiki/Finite_complement_topology" rel="nofollow">http://en.wikipedia.org/wiki/Finite_complement_topology</a> .</p> <p>That is, a subset $U \subset \mathbb{R}$ is open if and only if it is the empty set or its complement $\mathbb{R}\backslash U$ is a finite set. Then every sequence $(x_n)$ of points of $\mathbb{R}$ converges to every point $x \in \mathbb{R}$.</p> <p>To see this, take any open set $U$ containing $x$. Because $\mathbb{R} \backslash U$ has only a finite number of points, an infinite number of points of the sequence $(x_n)$ must be in $U$; i.e., there exists $n_0 \in \mathbb{N}$ such that, for every $n \geq n_0$, $x_n \in U$. Thus, $(x_n) \longrightarrow x$.</p> http://mathoverflow.net/questions/7625/hypercohomology-of-a-dg-algebra/7705#7705 Answer by Agusti Roig for Hypercohomology of a dg-algebra Agusti Roig 2009-12-03T19:23:25Z 2009-12-03T19:23:25Z <p>Since you want references, not a proof, maybe you can look at the paper by Hinich that Leonid already mentioned, but also at the original book by Godement, "Topologie Algébrique et Théorie des Faisceaux". There he treats also the problem with multiplicative structures in sheaf cohomology (and you can adapt it to the derived functor of direct images). As for "Thom-Whitney functors", mentioned by Minhyong, you should look at the original paper by V. Navarro-Aznar, "Sur la thérorie de Hodge-Deligne", Inv. Math. 90 (1987), 11-76. But be aware that if you are not working with <em>commutative</em> dg algebras you don't need in fact the essential tool of this paper (the "Thom-Whitney simple/total functor"): the usual total functor of double complexes (together with the Alexander-Whitney map) will work as well and looks easier. Or you can also ask me for a preprint about that subject that I'm going to finish one of these days. Sorry for this self-advertising. :-) (One last hint: you don't need to restrict yourself to the H^0 cohomology: you have a dg multiplicative structure already in R\pi_*C : the multiplicative structure in the H^0 is inherited from that one.)</p> http://mathoverflow.net/questions/4474/assumptions-on-the-category-c-for-sheafification-of-c-valued-presheaves/4496#4496 Answer by Agusti Roig for Assumptions on the category C for sheafification of C-valued presheaves Agusti Roig 2009-11-07T05:36:25Z 2009-11-07T05:36:25Z <p>There is also this paper by Gray: Category-valued sheaves, BAMS 68, (1962), </p> <p><a href="http://projecteuclid.org/DPubS?verb=Display&amp;version=1.0&amp;service=UI&amp;handle=euclid.bams/1183524852&amp;page=record" rel="nofollow">http://projecteuclid.org/DPubS?verb=Display&amp;version=1.0&amp;service=UI&amp;handle=euclid.bams/1183524852&amp;page=record</a> </p> <p>who addresses the question.</p> http://mathoverflow.net/questions/761/undergraduate-level-math-books/3463#3463 Answer by Agusti Roig for Undergraduate Level Math Books Agusti Roig 2009-10-30T18:40:02Z 2009-10-30T18:40:02Z <p>Applied Linear Algebra, by B. Noble and J.W. Daniel, <a href="http://www.amazon.co.uk/Applied-Linear-Algebra-Ben-Noble/dp/0130412600/ref=sr_1_14?ie=UTF8&amp;s=books&amp;qid=1256927879&amp;sr=1-14" rel="nofollow">http://www.amazon.co.uk/Applied-Linear-Algebra-Ben-Noble/dp/0130412600/ref=sr_1_14?ie=UTF8&amp;s=books&amp;qid=1256927879&amp;sr=1-14</a> .</p> http://mathoverflow.net/questions/3247/internal-homs-and-adjunctions/3260#3260 Answer by Agusti Roig for internal homs and adjunctions? Agusti Roig 2009-10-29T13:46:39Z 2009-10-29T13:46:39Z <p>As Dimitri ans bhargav said, this is automatic for categories with internal homs. And you don't need x to be the product: you have the same result for any closed monoidal category (in your case it is cartesian closed). You can find it in "Basic concepts of enriched category theory", by G.M. Kelly, page 14, which you can get for free here: <a href="http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html" rel="nofollow">http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html</a> .</p> http://mathoverflow.net/questions/3079/most-helpful-heuristic/3135#3135 Answer by Agusti Roig for Most helpful heuristic? Agusti Roig 2009-10-28T20:24:08Z 2009-10-28T20:24:08Z <p>Three of the most useful heuristic rules I've learnt:</p> <p>(1) When trying to define a morphism that has to make a diagram commutative, first forget about signs: in the end, they will agree.</p> <p>(2) If you want to prove something using induction, first check if it works (besides for n=1) for n=2.</p> <p>(3) Trying to define a morphism that has to fulfil some universal property? -The first one (diferent from zero or constant) that you can imagine / find is probably: (a) the only non trivial one that exists, (b) that miraculously fulfils also your requirements.</p> http://mathoverflow.net/questions/77094/a-model-structure-on-the-category-of-dualizing-maps/77126#77126 Comment by Agusti Roig Agusti Roig 2011-10-04T18:43:31Z 2011-10-04T18:43:31Z Well, I thought it could give you inspiration. I conjecture that an analogous model structure could work for your category and the only problem should be to define an analogous of the push-out which is needed in order to define cofibrations. But maybe I'm too daring. http://mathoverflow.net/questions/56166/do-homotopy-groups-always-commute-with-filtered-colimits/56181#56181 Comment by Agusti Roig Agusti Roig 2011-02-22T15:51:57Z 2011-02-22T15:51:57Z [sequel] Since my index category $I$ is just filtered (finitely filtered) I would need my objects (my cofibrant objects, some of them at least) to be (finitely) small, not &quot;eventually&quot; small for some big enough cardinal. Thus, as far as I understand, &quot;combinatorial&quot; may be not the word in this case, is it? http://mathoverflow.net/questions/56166/do-homotopy-groups-always-commute-with-filtered-colimits/56181#56181 Comment by Agusti Roig Agusti Roig 2011-02-22T15:49:55Z 2011-02-22T15:49:55Z Harry: please, correct me if I'm wrong (which is almost sure the case), but if I understand all this story about cardinals (which is unlikely), the objects of a combinatorial model category are $\lambda$-small... for some big enough cardinal $\lambda$. Right? But being just (finitely) small is stronger than being $\lambda$-small for any other bigger (infinite) cardinal $\lambda$. [to be continued] http://mathoverflow.net/questions/56166/do-homotopy-groups-always-commute-with-filtered-colimits/56181#56181 Comment by Agusti Roig Agusti Roig 2011-02-22T02:15:00Z 2011-02-22T02:15:00Z @Harry. For the moment, I've only read what I could find at the nLab. I'll check Lurie's reference. Thank you very much. http://mathoverflow.net/questions/56166/do-homotopy-groups-always-commute-with-filtered-colimits/56181#56181 Comment by Agusti Roig Agusti Roig 2011-02-22T01:09:46Z 2011-02-22T01:09:46Z Thanks Harry for pointing me to combinatoriality. Nevertheless I think the problem I talked about &quot;filtered&quot; vs &quot;totally ordered&quot; remains, isn't it? I mean: combinatoriality, again, is a condition about colimits of $\lambda$-sequences, not filtered ones. Or am I missing something? http://mathoverflow.net/questions/28972/colimits-of-spectral-sequences/56202#56202 Comment by Agusti Roig Agusti Roig 2011-02-22T01:02:53Z 2011-02-22T01:02:53Z Thanks, Dylan. Later on, I've found the answer I needed: is the &quot;colimit lemma&quot; of Mitchell in &quot;Hypercohomology spectra and Thomason's descent theorem.&quot; http://mathoverflow.net/questions/42755/non-standard-model-category-structure-on-cochain-complexes/42761#42761 Comment by Agusti Roig Agusti Roig 2010-10-20T19:01:43Z 2010-10-20T19:01:43Z Merci encore, Denis. Je crois qu'on s'est rencontr&#233; &#224; Barcelone il y a peut-&#234;tre deux ans, n'est-ce pas? Amiti&#233;s. http://mathoverflow.net/questions/42755/non-standard-model-category-structure-on-cochain-complexes/42761#42761 Comment by Agusti Roig Agusti Roig 2010-10-19T12:05:28Z 2010-10-19T12:05:28Z Thank you, Denis-Charles Cisinski: any reference for this result? http://mathoverflow.net/questions/39914/group-of-yoneda-extensions-and-the-ext-groups-defined-via-derived-category/39921#39921 Comment by Agusti Roig Agusti Roig 2010-09-26T04:44:32Z 2010-09-26T04:44:32Z @John. For the category of modules, assuming you already know the bijection between elements of $Ext^n(A,B)$ and $n$-extensions of $A$ by $B$ (see Hilton-Stammbach, &quot;A course in Homological Algebra&quot; if not), you can find the rest of the bijection in the first chapter of Borel's &quot;Algebraic D-modules&quot; for instance. http://mathoverflow.net/questions/32173/mitchells-embedding-theorem/39878#39878 Comment by Agusti Roig Agusti Roig 2010-09-25T08:26:20Z 2010-09-25T08:26:20Z [sequel] But both are far away from being small categories: the second has more objects than the category of Abelian groups. The first ones always have at least as many objects as the category in which sheaves take values (you always have constant sheaves). Anyway, I will be glad to rule out this enhancement of Mitchell's embedding theorem with a counter-example like the one you are proposing. http://mathoverflow.net/questions/32173/mitchells-embedding-theorem/39878#39878 Comment by Agusti Roig Agusti Roig 2010-09-25T08:21:42Z 2010-09-25T08:21:42Z Yes, but the only categories I know with &quot;few&quot; projectives are: (a) categories of sheaves, and (b) Freyd's category of $\mathbb{Z}[I]$-modules, where the set of indexes for your polynomials $I$ is a proper class (you can find it in Freyd's &quot;Abelian categories&quot;, or in Casacuberta-Neeman &quot;Brown representability does not come for free&quot;, Math. Res. Lett. 16 (2009), no. 1, 1–5). I don't know if the latter has no projectives at all, what is sure is it has no free modules. [to be continued] http://mathoverflow.net/questions/32173/mitchells-embedding-theorem/39878#39878 Comment by Agusti Roig Agusti Roig 2010-09-25T02:30:48Z 2010-09-25T02:30:48Z [sequel] Notice that this is <i>not</i> the same as being <i>small-complete</i>, as in the hypothesis of Freyd's adjoint functor theorem: unfortunately, &quot;small and complete&quot; is not the same as &quot;small-complete&quot;. Otherwise you would have proved that the solution set condition of this theorem is unnecessary, since <i>every</i> functor from a small category verifies it. http://mathoverflow.net/questions/32173/mitchells-embedding-theorem/39878#39878 Comment by Agusti Roig Agusti Roig 2010-09-25T02:29:52Z 2010-09-25T02:29:52Z Thank you for your interest. It would be nice to have a counter-example at least. But I'm not sure to understand the one you are proposing. As an aside question: $A$ is the category of sheaves over $[0,1]$ with values in which category? But the main point I would like to ask you is the following: are you sure that $A$ is <i>small</i>? That is, its objects form a set? [to be continued] http://mathoverflow.net/questions/18898/grothendiecks-tohoku-paper-and-combinatorial-topology/30015#30015 Comment by Agusti Roig Agusti Roig 2010-09-17T08:27:43Z 2010-09-17T08:27:43Z @Dam. It's been a while since you posted this answer, but I read it right now and I'm interested: which result is this in the Tohoku paper? http://mathoverflow.net/questions/36893/looking-for-applications-of-a-nice-result-in-linear-algebra Comment by Agusti Roig Agusti Roig 2010-08-27T17:49:47Z 2010-08-27T17:49:47Z @unkown(google). Despite its appearance, exp(A) <i>is</i> a polynomial on A. That is, for every A, you can find a polynomial r(t) such that exp(A)= r(A). It is the Lagrange-Sylvester interpolation polynomial: see the book of Gantmacher, The theory of matrices, <a href="http://books.google.es/books?id=ZQ4VPOG4St8C&amp;pg=PA103&amp;lpg=PA103&amp;dq=%22lagrange+-+sylvester+polynomial%22&amp;source=bl&amp;ots=-TcPXtTNZz&amp;sig=XXfsmXzYxAvECsmgjmAl4D1l_ig&amp;hl=ca&amp;ei=aPl3TOOcIqqJ4gaIjo2DBg&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=3&amp;ved=0CCEQ6AEwAg#v=onepage&amp;q=%22lagrange%20-%20sylvester%20polynomial%22&amp;f=false" rel="nofollow">books.google.es/&hellip;</a> .