User karol szumiło - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:59:23Z http://mathoverflow.net/feeds/user/12547 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130616/reference-for-lax-monoidal-functors-monoids-under-day-convolution/130619#130619 Answer by Karol Szumiło for Reference for "lax monoidal functors" = "monoids under Day convolution" Karol Szumiło 2013-05-14T18:46:07Z 2013-05-14T18:46:07Z <p>This observation appears already in Day's thesis as Example 3.2.2. For some reason it is only stated for commutative monoids and symmetric (pro)monoidal functors and only as a correspondence of objects not an equivalence of categories, also no proof is given. (Day's thesis used to be available on <a href="http://maths.mq.edu.au/~street/Day.pub.html" rel="nofollow">Street's homepage</a> but the link is dead now.)</p> http://mathoverflow.net/questions/128976/can-one-make-the-category-of-pairs-of-topological-spaces-a-model-category/128982#128982 Answer by Karol Szumiło for Can one make the category of pairs of topological spaces a model category? Karol Szumiło 2013-04-28T07:18:00Z 2013-04-28T07:18:00Z <p>As Tyler Lawson points out you can use the category of all diagrams on $[1]$. Then the projective and injective model structures are both instances of Reedy model structures. This is discussed in Section 5.2 of Hovey's <em>Model Categories</em> and will work with both Quillen's and Strøm's model structures. (In fact it works with completely arbitrary model category.)</p> <p>I'm not sure what happens if you want to restrict to the category of pairs of spaces. However, if you insist that your pairs are cofibered, then the first problem you run into is that an uncountable product of (Hurewicz or Serre) cofibrations is not necessarily a cofibration again. So I suspect that his category might not be complete.</p> http://mathoverflow.net/questions/107188/does-the-monoid-axiom-hold-for-k-spaces/121646#121646 Answer by Karol Szumiło for Does the Monoid Axiom hold for k-spaces? Karol Szumiło 2013-02-12T20:28:40Z 2013-02-13T07:59:04Z <p>It seems to me that there is a relatively simple answer to this question but perhaps I am overlooking something.</p> <p>The category of K-spaces does satisfy the monoid axiom. (If I read the question correctly K-spaces are what is usually called "compactly generated spaces" and compactly generated spaces are what is usually called "weak Hausdorff compactly generated spaces.") We need to check that a (transfinite) sequential colimit of pushouts of maps of the form <code>$X \times i$</code> where <code>$X$</code> is an arbitrary K-space and <code>$i$</code> is an acyclic cofibration is a weak equivalence. First, observe that <code>$X \times i$</code> is a weak equivalence and while it is not necessarily a Serre cofibration it is a Hurewicz cofibration.</p> <p>Now, the conclusion will follow if we know the two following facts.</p> <ol> <li>Pushouts of Hurewicz cofibrations which are weak equivalences are again weak equivalences (and of course Hurewicz cofibrations.)</li> <li>(Transfinite) sequential colimits of Hurewicz cofibrations which are weak equivalences are again weak equivalences.</li> </ol> <p>The first one is proven by Boardman and Vogt in Proposition 4.8 (b) in the appendix of <em>Homotopy Invariant Algebraic Structures on Topological Spaces</em>. (The argument doesn't use any fancy point-set topology, in particular separation axioms play no role, so this holds in K-spaces.)</p> <p>For the second one let's consider a sequence of Hurewicz cofibrations between spaces <code>$(X_\beta \mid \beta &lt; \alpha)$</code>. We want to show that if they are all weak equivalences then so is <code>$X_0 \to \mathrm{colim}_{\beta &lt; \alpha} X_\beta$</code>. First observe that the canonical map from the telescope <code>$\mathrm{Tel}_{\beta &lt; \alpha} X_\beta \to \mathrm{colim}_{\beta &lt; \alpha} X_\beta$</code> is a homotopy equivalence so it suffices to show that <code>$X_0 \to \mathrm{Tel}_{\beta &lt; \alpha} X_\beta$</code> is a weak equivalence. By fattening the stages of the telescope slightly we can write it as a colimit of open subspaces homotopy equivalent to the original ones. The conclusion follows since compact spaces are small with respect to open embeddings.</p> http://mathoverflow.net/questions/120382/reedy-model-structures-on-oplax-limits/120388#120388 Answer by Karol Szumiło for Reedy model structures on oplax limits Karol Szumiło 2013-01-31T08:08:41Z 2013-01-31T08:08:41Z <p>You may be thinking of</p> <p>Johnson, Mark W. <em>On modified Reedy and modified projective model structures.</em> Theory Appl. Categ. 24 (2010), No. 8, 179–208.</p> <p>but his constructions (Definitions 3.3 and 5.2) have a fixed category at each object of <code>$R$</code> and only the model structures are allowed to vary.</p> <p>As a side note, a particular instance of this construction is used in these <a href="http://www.math.uni-bonn.de/people/schwede/global.pdf" rel="nofollow">new notes</a> about global homotopy theory, but here the indexing category is not even a Reedy category, but some sort of "enriched generalized Reedy category".</p> http://mathoverflow.net/questions/117736/is-a-conservative-finite-limit-preserving-functor-of-infinity-1-categories-homo Is a conservative finite limit preserving functor of (infinity,1)-categories homotopically faithful? Karol Szumiło 2012-12-31T15:27:28Z 2012-12-31T17:47:02Z <p>In classical category theory we have a following criterion. If <code>$\mathcal{C}$</code> and <code>$\mathcal{D}$</code> are finitely complete categories and <code>$F : \mathcal{C} \to \mathcal{D}$</code> is a functor which preserves finite limits and reflects isomorphisms, then <code>$F$</code> is faithful. It follows easily from an observation that an equalizer of two parallel morphisms is an isomorphism if and only if those morphisms are equal.</p> <p>My question is</p> <blockquote> <p>Given two finitely complete <code>$(\infty, 1)$</code>-categories <code>$\mathcal{C}$</code> and <code>$\mathcal{D}$</code> and a functor <code>$F : \mathcal{C} \to \mathcal{D}$</code> which preserves finite limits and reflects equivalences, does it follow that <code>$\mathrm{Ho} F : \mathrm{Ho} \mathcal{C} \to \mathrm{Ho} \mathcal{D}$</code> is faithful?</p> </blockquote> <p>The observation I mentioned previously doesn't work in higher category theory since for example an equalizer of the identity morphism of an object <code>$X$</code> with itself is a free loop object on <code>$X$</code>, but I am unable to decide whether this faithfulness criterion is valid.</p> http://mathoverflow.net/questions/116051/localizations-of-non-nilpotent-spaces Localizations of non-nilpotent spaces Karol Szumiło 2012-12-11T07:24:04Z 2012-12-11T19:00:17Z <p>For simplicity let's talk about <code>$p$</code>-localizations of spaces for a fixed prime <code>$p$</code>. Every space <code>$X$</code> has a well-defined <code>$p$</code>-localization which can be constructed by the small object argument and which becomes a fibrant replacement in the <code>$p$</code>-local model structure on the category of spaces. It is well-known that nilpotent spaces have nice enough Postnikov towers and we can localize such spaces by taking the Postnikov tower, localizing step by step and putting it back together by taking the limit of the resulting tower of fibrations. My question is:</p> <blockquote> <p>Is there an example of a non-nilpotent space <code>$X$</code> whose <code>$p$</code>-localization we can explicitly describe?</p> </blockquote> <p>I leave the meaning of "explicitly" ambiguous. I would be interested in any construction not using the small object argument.</p> <p>Here's my stab at a possible example. For a group <code>$G$</code> we define its lower central series by setting <code>$G_0 = G$</code> and <code>$G_{n + 1} = [G_n, G]$</code> and we can also continue transfinitely by setting <code>$G_\beta = \bigcap_{\alpha &lt; \beta} G_\alpha$</code> for limit ordinals <code>$\beta$</code>. The group <code>$G$</code> is nilpotent if this construction terminates at the trivial subgroup at a finite stage. It is called <em>hypocentral</em> if it terminates at the trivial subgroup at some not necessarily finite stage. According to Wikipedia it is a result of Malcev that there are hypocentral groups with arbitrarily long lower central series.</p> <blockquote> <p>If we start with a hypocentral group <code>$G$</code> and convert its lower central series into a (transfinite) tower of fibrations (whose limit is a <code>$K(G, 1)$</code>), <code>$p$</code>-localize it step by step and take the limit of the resulting tower of fibrations, do we obtain the <code>$p$</code>-localization of <code>$K(G, 1)$</code>?</p> </blockquote> http://mathoverflow.net/questions/114779/local-finality-condition-for-re-indexing-parameterized-colimits/114799#114799 Answer by Karol Szumiło for Local finality condition (for re-indexing parameterized colimits) Karol Szumiło 2012-11-28T18:25:08Z 2012-11-28T18:25:08Z <p>I assume you meant <code>$\Sigma_G \Delta_F \to \Sigma_x$</code>, otherwise this doesn't parse.</p> <p>Here's my condition. It's not completely straightforward, but it is a generalization of the classical one you mentioned. I guess it's a matter of taste whether it's messy.</p> <p>Consider a triple <code>$(b, f, c)$</code> where <code>$b$</code> is an object of <code>$B$</code>, <code>$c$</code> is an object of <code>$C$</code> and <code>$f : x b \to c$</code> is a morphism in <code>$C$</code>. To each such triple one can associate a kind of "double slice", namely the "category of factorizations of <code>$f$</code> through objects of the form <code>$F a$</code>". Its objects are triples <code>$(g, a, h)$</code> where <code>$a$</code> is an object of <code>$A$</code>, <code>$g : b \to F a$</code> and <code>$h : x F a \to c$</code> such that <code>$h (x g) = f$</code>. The morphisms are morphisms of <code>$A$</code> making two evident triangles commute. My claim is that <code>$\Sigma_G \Delta_F \to \Sigma_x$</code> is a natural isomorphism precisely when all such "categories of factorizations" are connected (which I take to include non-empty).</p> <p>The argument is as follows. For any functor <code>$W : A^\mathrm{op} \to \mathrm{Set}$</code> there is a <code>$W$</code>-weighted colimit functor <code>$\mathrm{colim}^W : \mathrm{Set}^A \to \mathrm{Set}$</code> given by the coend formula <code>$\mathrm{colim}^W Y = \int^{a \in A} Y_a \times W_a$</code> for <code>$Y \in \mathrm{Set}^A$</code>. A transformation <code>$\phi : V \to W$</code> induces a transformation <code>$\phi_* : \mathrm{colim}^V \to \mathrm{colim}^W$</code>. It is easy to see that <code>$\phi_*$</code> is an isomorphism if and only if <code>$\phi$</code> is since we can recover <code>$\phi$</code> form <code>$\phi_*$</code> by evaluating it on representable functors.</p> <p>We have formulas for Kan extensions via coends which say that <code>$(\Sigma_x Y) c = \mathrm{colim}^{C(x -, c)} Y$</code> and similarly <code>$(\Sigma_G \Delta_F Y)_c = \mathrm{colim}^{\int^a B(-, F a) \times C(x F a, c)} Y$</code> and the natural transformation in question is induced by the transformation <code>$\int^a B(-, F a) \times C(x F a, c) \to C(x -, c)$</code> which takes a pair of morphisms <code>$g : b \to F a$</code> and <code>$h : x F a \to c$</code> and sends it to the composite <code>$h (x g)$</code>. We need to check that it is an isomorphism i.e. that the fiber over every point <code>$f : x b \to c$</code> is a singleton. There is an explicit description of this coend as the quotient set of an equivalence relation and it yields a description of the fiber over <code>$f$</code> as the quotient of the set of objects of the above "category of factorizations of <code>$f$</code>" by the relation which turns out to be the relation of being in the same component of this category.</p> http://mathoverflow.net/questions/114553/loop-space-of-a-category/114566#114566 Answer by Karol Szumiło for Loop space of a category Karol Szumiło 2012-11-26T18:30:17Z 2012-11-26T23:35:45Z <p>My French is not good enough to be sure about it, but it seems that this paper has the definition you are after.</p> <p>Evrard, Marcel <em>Fibrations de petites catégories.</em> Bull. Soc. Math. France 103 (1975), no. 3, 241–265. (<a href="http://www.numdam.org/item?id=BSMF_1975__103__241_0" rel="nofollow">Numdam</a>)</p> http://mathoverflow.net/questions/110309/injective-objects-in-morab/110319#110319 Answer by Karol Szumiło for Injective objects in Mor(Ab) Karol Szumiło 2012-10-22T11:44:45Z 2012-10-22T12:17:27Z <p>I will use notation <code>$A_0 \to A_1$</code> for objects of <code>$\mathrm{Mor}(\mathrm{Ab})$</code>.</p> <p>EDIT: previously I claimed something stronger (that I can produce lifting properties in the functor category without factorizations), but I am not so sure about it.</p> <p>The following is a lot more general than necessary, but I think this added generality is also useful. Let <code>$(\mathcal{L}, \mathcal{R})$</code> be a weak factorization system in a category <code>$\mathcal{C}$</code> with enough colimits and limits for the following to make sense. Let $J$ be a Reedy category. Then in the functor category <code>$\mathcal{C}^J$</code> the "Reedy <code>$\mathcal{L}$</code>-cofibrations" and "Reedy <code>$\mathcal{R}$</code>-fibrations" form a weak factorization system. By "Reedy <code>$\mathcal{L}$</code>-cofibrations" I mean morphisms of diagrams <code>$X \to Y$</code> such that for every <code>$j \in J$</code> the latching morphism <code>$X_j \sqcup_{L_j X} L_j Y \to Y_j$</code> is in <code>$\mathcal{L}$</code> and dually "Reedy <code>$\mathcal{R}$</code>-fibrations" are morphisms <code>$X \to Y$</code> such that for every <code>$j \in J$</code> the matching morphism <code>$X_j \to M_j X \times_{M_j Y} Y_j$</code> is in <code>$\mathcal{R}$</code>. The proof is exactly as in the construction of the Reedy model structures and can be found for example in Hovey's <em>Model Categories</em>.</p> <p>Now we take <code>$\mathcal{C} = \mathrm{Ab}$</code>, <code>$\mathcal{L} =$</code> monomorphisms and <code>$J = [1]$</code>. Then <code>$\mathcal{R}$</code> are split epimorphisms with injective kernel. The lifting properties are easily verified while the factorizations use the fact that there are enough injectives in <code>$\mathrm{Ab}$</code>. If <code>$f : A \to B$</code> is a map in <code>$\mathrm{Ab}$</code>, pick an injective hull <code>$i : A \to \hat A$</code>, then <code>$f$</code> factors as an injection <code>$[i, f] : A \to \hat A \oplus B$</code> followed by a split surjection with injective kernel <code>$\hat A \oplus B \to B$</code>. We consider <code>$J$</code> as a Reedy category where <code>$0$</code> has degree <code>$1$</code> and <code>$1$</code> has degree <code>$0$</code>. Then "Reedy <code>$\mathcal{L}$</code>-cofibrations" are monomorphisms again, so an object <code>$X$</code> is injective if and only if the map <code>$X \to 0$</code> is a "Reedy <code>$\mathcal{R}$</code>-fibration" i.e. when both <code>$X_1 \to 0$</code> and <code>$X_0 \to X_1$</code> are split epimorphisms with injective kernel i.e. when <code>$X_0 \to X_1$</code> is a split epimorphism with injective source.</p> http://mathoverflow.net/questions/108605/does-the-following-categorial-sum-preserve-weak-equivalences/108609#108609 Answer by Karol Szumiło for Does the following categorial sum preserve weak equivalences? Karol Szumiło 2012-10-02T09:10:14Z 2012-10-02T09:10:14Z <p>The general case can be concluded from the finite one as follows. Let <code>$(A_i \to B_i \mid i \in I)$</code> be a family of weak equivalences between simplicial $O$-categories. I'm going to assume that $I = \mathbb{N}$, the general case can be handled similarly, but the notation would be a bit more tedious (you can well-order $I$ or consider the directed poset of finite subsets of $I$).</p> <p>The coproduct $\coprod_{i \in \mathbb{N}} A_i$ can be written as a colimit of the sequence</p> <p>$$A_0 \to A_0 \sqcup A_1 \to A_0 \sqcup A_1 \sqcup A_2 \to \ldots$$</p> <p>which is computed hom-set-wise (since morphisms in the big coproduct can be written as finite composites of morphisms of $A_i$s so each occurs at some stage in the sequence). Moreover, all the maps induced on hom-sets are injective i.e. cofibrations of simplicial sets. The transformation between this sequence and the corresponding one for $B_i$s is a natural weak equivalence by the result you mentioned. Putting this all together you get that the induced map on each hom-set in the colimit is also a weak equivalence.</p> http://mathoverflow.net/questions/11456/unpointed-brown-representability-theorem/105156#105156 Answer by Karol Szumiło for unpointed brown representability theorem Karol Szumiło 2012-08-21T13:12:54Z 2012-08-21T13:21:14Z <p>This is a copy of my answer to <a href="http://mathoverflow.net/questions/104866/brown-representability-for-non-connected-spaces/" rel="nofollow">http://mathoverflow.net/questions/104866/brown-representability-for-non-connected-spaces/</a> which I repost here per request in the comment.</p> <p>A negative answer to the question can be concluded from this paper:</p> <p>Freyd, Peter; Heller, Alex <em>Splitting homotopy idempotents. II.</em> J. Pure Appl. Algebra 89 (1993), no. 1-2, 93–106.</p> <p>This paper introduces a notion of conjugacy idempotent. It is a triple <code>$(G, g, b)$</code> consisting of a group <code>$G$</code>, an endomorphism <code>$g \colon G \to G$</code> and an element <code>$b \in G$</code> such that for all <code>$x \in G$</code> we have <code>$g^2(x) = b^{-1} g(x) b$</code>. The theory of conjugacy idempotents can be axiomatized by equations, so there is an initial conjugacy idempotent <code>$(F, f, a)$</code>. The Main Theorem of the paper says (among other things) that <code>$f$</code> does not split in the quotient of the category of groups by the conjugacy congruence.</p> <p>Now <code>$f$</code> induces an endomorphism <code>$B f \colon B F \to B F$</code> which is an idempotent in <code>$\mathrm{Ho} \mathrm{Top}$</code> and it follows (by the Main Lemma of the paper) that it doesn't split. It is then easily concluded that <code>$(B f)_+ \colon (B F)_+ \to (B F)_+$</code> doesn't split in <code>$\mathrm{Ho} \mathrm{Top}_*$</code>.</p> <p>The map <code>$(B f)_+$</code> induces an idempotent of the representable functor <code>$[-, (B F)_+]_*$</code> which does split since this is a <code>$\mathrm{Set}$</code> valued functor. Let <code>$H \colon \mathrm{Ho} \mathrm{Top}_*^\mathrm{op} \to \mathrm{Set}$</code> be the resulting retract of <code>$[-, (B F)_+]_*$</code>. It is half-exact (i.e. satisfies the hypotheses of Brown's Representability) as a retract of a half-exact functor. However, it is not representable since a representation would provide a splitting for <code>$(B f)_+$</code>.</p> <p>The same argument with <code>$B f$</code> in place of <code>$(B f)_+$</code> shows the failure of Brown's Representability in the unbased case.</p> http://mathoverflow.net/questions/104866/brown-representability-for-non-connected-spaces/105049#105049 Answer by Karol Szumiło for Brown representability for non-connected spaces Karol Szumiło 2012-08-19T18:18:54Z 2012-08-19T20:44:50Z <p>I was thinking more about this question and found another paper by Heller which offers an answer (unfortunately a negative one). The paper is</p> <p>Freyd, Peter; Heller, Alex <em>Splitting homotopy idempotents. II.</em> J. Pure Appl. Algebra 89 (1993), no. 1-2, 93–106.</p> <p>This paper introduces a notion of conjugacy idempotent. It is a triple <code>$(G, g, b)$</code> consisting of a group <code>$G$</code>, an endomorphism <code>$g \colon G \to G$</code> and an element <code>$b \in G$</code> such that for all <code>$x \in G$</code> we have <code>$g^2(x) = b^{-1} g(x) b$</code>. The theory of conjugacy idempotents can be axiomatized by equations, so there is an initial conjugacy idempotent <code>$(F, f, a)$</code>. The Main Theorem of the paper says (among other things) that <code>$f$</code> does not split in the quotient of the category of groups by the conjugacy congruence.</p> <p>Now <code>$f$</code> induces an endomorphism <code>$B f \colon B F \to B F$</code> which is an idempotent in <code>$\mathrm{Ho} \mathrm{Top}$</code> and it follows (by the Main Lemma of the paper) that it doesn't split. It is then easily concluded that <code>$(B f)_+ \colon (B F)_+ \to (B F)_+$</code> doesn't split in <code>$\mathrm{Ho} \mathrm{Top}_*$</code>.</p> <p>The map <code>$(B f)_+$</code> induces an idempotent of the representable functor <code>$[-, (B F)_+]_*$</code> which does split since this is a <code>$\mathrm{Set}$</code> valued functor. Let <code>$H \colon \mathrm{Ho} \mathrm{Top}_*^\mathrm{op} \to \mathrm{Set}$</code> be the resulting retract of <code>$[-, (B F)_+]_*$</code>. It is half-exact (i.e. satisfies the hypotheses of Brown's Representability) as a retract of a half-exact functor. However, it is not representable since a representation would provide a splitting for <code>$(B f)_+$</code>.</p> <p>The same argument with <code>$B f$</code> in place of <code>$(B f)_+$</code> shows the failure of Brown's Representability in the unbased case.</p> http://mathoverflow.net/questions/86016/alternative-characterization-of-homotopy-equivalence/86021#86021 Answer by Karol Szumiło for Alternative characterization of homotopy equivalence Karol Szumiło 2012-01-18T18:35:04Z 2012-06-11T12:51:59Z <p>EDIT: This is an old question, but I have stumbled upon it by accident and realized that my answer is wrong. It turns out that genuine homotopy equivalences cannot be characterized in terms of their homotopy fibers. Here's a counterexample. Let <code>$X = \{ 1, 2, 3, \ldots, \infty \}$</code> with discrete topology and <code>$Y = \{1, \frac{1}{2}, \frac{1}{3}, \ldots, 0 \}$</code> with the topology inherited from <code>$\mathbb{R}$</code>. Let <code>$f \colon X \to Y$</code> be a map given by <code>$f(m) = \frac{1}{m}$</code> (and $f(\infty) = 0$). Then <code>$f$</code> has contractible homotopy fibers (in fact they are all one-point spaces), but it is not a homotopy equivalence since the only candidate for a homotopy inverse is not continuous.</p> <p>It is well-known that a map with all homotopy fibers weakly contractible is a weak equivalence and my mistake was to assume that there is a similar result for homotopy equivalences.</p> <hr> <p>I have to say I don't understand the motivation behind your question. Why exactly do you want to characterize homotopy equivalences without using homotopies? Moreover I'm not sure the question is well-posed, the answers would probably refer to homotopies implicitly and it would be a matter of taste whether they count as "not mentioning homotopies". To illustrate my point, here's an attempt at an answer.</p> <p>First, define the space $X$ to be contractible if it is a retract of its cone (more precisely if the canonical inclusion $X \to C X$ admits a retraction). Then define a map $X \to Y$ to be a homotopy equivalence if its homotopy fiber at every point $y \in Y$ is contractible. The homotopy fiber can be defined as a mapping cocylinder i.e. the pullback $Y^I \times_{Y \times Y} (X \times *)$.</p> <p>Now, I didn't use the word "homotopy" (except in "homotopy fiber", but I explained how to "go around it"). However, for example the retraction $C X \to X$ is nothing else but a homotopy from $\mathrm{id}_X$ to some constant map. I suspect that you won't be satisfied with this kind of hiding the homotopies backstage. If this is the case you should explain precisely what counts as "not mentioning homotopies".</p> http://mathoverflow.net/questions/96071/general-gluing-theorem-for-adjunction-spaces/96116#96116 Answer by Karol Szumiło for General gluing theorem for adjunction spaces Karol Szumiło 2012-05-06T05:59:22Z 2012-05-06T05:59:22Z <p>Yes, this is true even for not necessarily closed cofibrations. If you want a single source that gives a complete proof, then the only one that comes to my mind is this <a href="http://arxiv.org/abs/math/0610009v4" rel="nofollow">preprint</a>.</p> <p>Definition 1.1.1 introduces cofibration categories and then Lemma 1.4.1 says that the desired result holds in any cofibration category. Section 3.1 contains a detailed proof that the category of topological spaces equipped with Hurewicz cofibrations and homotopy equivalences is a cofibration category and thus the lemma applies. The crux of the matter is that acyclic cofibrations are closed under pushouts and this follows from a classical result of Dold (Lemma 3.1.9) that acyclic cofibrations admit deformation retractions, which doesn't depend on closedness.</p> http://mathoverflow.net/questions/93281/is-it-possible-to-define-the-notion-of-a-localization-of-a-category-without-refer/93284#93284 Answer by Karol Szumiło for Is it possible to define the notion of a localization of a category without reference to a set of morphisms, $S$? Karol Szumiło 2012-04-06T06:45:14Z 2012-04-06T06:45:14Z <p>I don't know a general result, but since you also asked for special cases here's one. If $F$ admits a right adjoint, then it is a localization with respect to some class of morphisms if and only if its right adjoint is fully faithful (which is also equivalent to the counit being an isomorphism). The proof is not difficult, it can be found in Proposition I.1.3 of Gabriel, Zisman <em>Calculus of Fractions and Homotopy Theory</em>.</p> http://mathoverflow.net/questions/92492/evaluation-functors-and-injective-model-structure-on-diagram-categories/92545#92545 Answer by Karol Szumiło for Evaluation functors and injective model structure on diagram categories Karol Szumiło 2012-03-29T09:25:48Z 2012-03-29T09:25:48Z <p>As already observed, this is not always true, but I will give a more general sufficient condition, which I believe may also be necessary. The condition is that <code>$\alpha$</code> is an epimorphism in <code>$C$</code>.</p> <p>I will denote the one arrow category by <code>$[1]$</code>. The left adjoint to <code>$\mathrm{ev}_\alpha : \mathcal{M}^C \to \mathcal{M}^{[1]}$</code> is the left Kan extension <code>$\mathrm{Lan}_\alpha : \mathcal{M}^{[1]} \to \mathcal{M}^C$</code>. It can be computed explicitly. If $X \in \mathcal{M}^{[1]}$ and $c \in C$, then $(\mathrm{Lan}_\alpha X)_c$ is the pushout of $C(\alpha_1, c) \times X_0 \to C(\alpha_1, c) \times X_1$ along $C(\alpha_1, c) \times X_0 \to C(\alpha_0, c) \times X_0$. If $\alpha$ is an epimorphism, then $C(\alpha_1, c) \to C(\alpha_0, c)$ is injective for all $c$ and the pushout in question is a pushout along a cofibration. It follows from the Gluing Lemma that $\mathrm{Lan}_\alpha$ preserves levelwise (acyclic) cofibration, so it is a left Quillen functor.</p> http://mathoverflow.net/questions/92206/what-properties-make-0-1-a-good-candidate-for-defining-fundamental-groups/92279#92279 Answer by Karol Szumiło for What properties make $[0,1]$ a good candidate for defining fundamental groups? Karol Szumiło 2012-03-26T14:50:19Z 2012-03-26T15:04:18Z <p>The above answers explain what is needed for the definition of a fundamental group to make sense. Let me try to answer the question from a different angle and explain what properties of the interval are needed for this notion to be reasonably well-behaved.</p> <p>I believe that the key property is that, intuitively speaking, "small pieces of $I$ look just like the whole thing". More precisely, the interval can be subdivided arbitrarily finely into smaller intervals, i.e. given any open cover $\mathcal{U}$ of $I$ there is a sequence $0 = t_0 &lt; \ldots &lt; t_m = 1$ such that for every $i$ we have $[t_{i - 1}, t_i] \subseteq U$ for some $U \in \mathcal{U}$. In some sense this is a stronger version of the observation that gluing two intervals yields a space that is again homeomorphic to the interval. (It is interesting that the universal property of the theorem mentioned by Tom Leinster already implies the "strong version" of the subdivision property even though it is stated purely in terms of the "weak version".) This is easily proven using the Lebesgue's Lemma and is the starting point of standard techniques for calculating with fundamental groups like the path lifting property for coverings or the Van Kampen Theorem. A similar property of cubes leads to similar techniques for higher homotopy groups.</p> <p>I cannot think of any other space with a property of this kind we could use in place of $I$. However, it would be interesting to see if there is some analogy between this standard approach to the fundamental group and approaches to something like Čech fundamental group (which I am unfamiliar with).</p> http://mathoverflow.net/questions/91115/sheaf-cohomology-invariant-of-weak-homotopy-type/91123#91123 Answer by Karol Szumiło for Sheaf cohomology invariant of weak homotopy type? Karol Szumiło 2012-03-13T21:24:50Z 2012-03-13T21:24:50Z <p>No. For paracompact spaces sheaf cohomology coincides with Čech cohomology. In particular it applies to the closed topologist's sine curve $C$. There is a map $C \to S^1$ inducing an isomorphism on Čech cohomology, but $C$ is weakly contractible.</p> http://mathoverflow.net/questions/90291/does-the-bordism-homology-theory-satisfy-the-weak-equivalence-axiom Does the bordism homology theory satisfy the weak equivalence axiom? Karol Szumiło 2012-03-05T16:22:42Z 2012-03-08T09:38:17Z <p>There is an interesting and important homology theory called <em>bordism</em>. Briefly speaking, a <em>singular manifold</em> in a space <code>$X$</code> is a pair <code>$(M, f)$</code> where <code>$M$</code> is a closed smooth manifold and <code>$f : M \to X$</code> is a map. Two singular manifolds of the same dimension <code>$(M, f)$</code> and <code>$(N, g)$</code> in <code>$X$</code> are <em>bordant</em> if there is a pair <code>$(W, H)$</code> where <code>$W$</code> is a bordism from <code>$M$</code> to <code>$N$</code> and <code>$H : W \to X$</code> is a map restricting to <code>$f$</code> and <code>$g$</code> on <code>$M$</code> and <code>$N$</code>. We define the <em>bordism group</em> <code>$MO_n X$</code> to be the set of bordism classes of all $n$-dimensional singular manifolds in <code>$X$</code> (this is indeed an abelian group under the addition induced by the coproduct of manifolds). It turns out that this gives rise to a homology theory, i.e. the functors <code>$MO_*$</code> satisfy the Eilenberg--Steenrod axioms. Strictly speaking, one needs to define the relative bordism groups which are defined using bordism classes of singular compact manifolds with boundary, i.e. maps <code>$(M, \partial M) \to (X, A)$</code> and there are some issues with manifolds with corners to be resolved. The complete construction is explained very nicely in <em>Differentiable Periodic Maps</em> by Conner and Floyd. (Beware that there are a book and a paper by the same authors and with the same title, I mean the book: <em>Differentiable Periodic Maps</em>, LNM 738, 1979.) There are also (even more interesting) variants of the bordism theory like <code>$MU$</code> and <code>$M \mathrm{Spin}$</code>, which arise by considering additional structure on singular manifolds. I hope that a good answer to my question will handle the general case.</p> <p>A homology theory <code>$h_*$</code> satisfies the <em>weak equivalence axiom</em> if for every weak equivalence of pairs of spaces <code>$f : (X, A) \to (Y, B)$</code> (i.e. a map such that both <code>$f : X \to Y$</code> and <code>$f | A : A \to B$</code> are weak equivalences) the induced map <code>$h_*(X, A) \to h_*(Y, B)$</code> is an isomorphism. My question is exactly as in the title.</p> <blockquote> <p>Does the bordism homology theory satisfy the weak equivalence axiom?</p> </blockquote> <p>An example of a homology theory that satisfies the weak equivalence axiom is singular homology. The way to prove this is as follows. Fix a pair of spaces <code>$(X, A)$</code> and a natural number <code>$n$</code>. Consider the <em>Eilenberg subcomplex</em> <code>$\mathrm{Sing}^{(n, A)} X$</code> of the singular complex <code>$\mathrm{Sing} X$</code> which consist in degree <code>$k$</code> of maps of pairs <code>$(\Delta^k, (\Delta^k)^{(n)}) \to (X, A)$</code> where <code>$(\Delta^k)^{(n)}$</code> is the <code>$n$</code>-skeleton of <code>$\Delta^k$</code>. One can prove that if <code>$(X, A)$</code> is <code>$n$</code>-connected, then the induced inclusion of singular chain complexes <code>$S_\bullet^{(n, A)} X \to S_\bullet X$</code> is a chain homotopy equivalence and thus the relative groups <code>$H_*(X, A)$</code> are zero up to degree <code>$n$</code>.</p> <p>This gave me an idea that maybe in case of bordism the weak equivalence axiom could be verified by choosing a triangulation on a manifold <code>$M$</code> and considering maps <code>$(M, M^{(n)}) \to (X, A)$</code> where <code>$M^{(n)}$</code> is the <code>$n$</code>-skeleton of <code>$M$</code> with respect to this triangulation. I was unable to find a proof along these lines. However, if this approach has any merit at all, then it means that the question has something to do with existence of triangulations of manifolds and thus the following may be a subtler variant of the question.</p> <blockquote> <p>Does the topological bordism homology theory (i.e. the one constructed using compact topological manifolds, which do not necessarily admit triangulations) satisfy the weak equivalence axiom?</p> </blockquote> <p>Some people may feel that the issue is somewhat immaterial since even if a homology theory doesn't satisfy the weak equivalence axiom, then we simply prolong it from CW-complexes to all spaces by means of CW-replacement. However, I feel that if some geometrically defined (co)homology theory satisfies the weak equivalence axiom for some class of spaces larger than CW-complexes, then it is good to know how large exactly this class is. For example for singular homology this class consists of all spaces which makes the theory unexpectedly well-behaved. An example where it fails very badly is topological K-theory. There the geometric definition is wrong even for non-finite CW-complexes and one needs to use the representing spectrum to prolong the theory to all spaces.</p> http://mathoverflow.net/questions/89822/exact-sequences-in-homotopy-categories/89884#89884 Answer by Karol Szumiło for Exact sequences in homotopy categories Karol Szumiło 2012-02-29T17:21:55Z 2012-02-29T17:21:55Z <p>As it was already pointed out, those statements will not follow from such general principles. What you would like to conclude is that homotopy pushouts are (or to certain degree behave like) homotopy pullbacks. Such a phenomenon is called <em>excision</em>. In the language of $\infty$-categories you can define a functor $\mathcal{C} \to \mathcal{D}$ to be <em>excisive</em> if it sends homotopy pushouts to homotopy pullbacks. However, being excisive is rather strong condition and proving that some functor satisfies it usually takes some work. In particular I don't think there is some abstract way around the usual point-set arguments used to prove Seifert--van Kampen theorem or various excision theorems (i.e. in homotopy or singular homology).</p> <p>There are lots of generalizations of this notion that consider higher connectivity, functors of many variables etc. Those ideas give rise to Goodwillie calculus. Jacob Lurie has recently updated his <em>Higher Algebra</em> which now containes a full chapter developing Goodwillie calculus in the language of $\infty$-categories. Some more basic properties of excisive functors are discussed in section 1.4.</p> http://mathoverflow.net/questions/89213/when-is-a-cube-of-cofibrations-are-lattice/89215#89215 Answer by Karol Szumiło for When is a cube of cofibrations are "lattice"? Karol Szumiło 2012-02-22T17:59:52Z 2012-02-23T17:46:34Z <p>The notion you are looking for is well-known in homotopy theory under the name <em>Reedy cofibration</em>, but for some reason this name doesn't show up in papers about Waldhausen categories, even though the concept is used all the time.</p> <p>To keep things close to your question let's say that $J$ is a finite poset (in general it can be any <em>direct category</em>). For a diagram $X : J \to \mathcal{C}$ and $j \in J$ we define a <em>latching object</em> $L_j X$ as the colimit of the restriction of $X$ to the subposet $\lbrace i \in J \mid i &lt; j \rbrace$. If $L_j X$ exists, then it comes with a canonical map $L_j X \to X_j$. We say that a diagram $X$ is <em>Reedy cofibrant</em> if all $L_j X$s exist and the canonical maps $L_j X \to X_j$ are cofibrations. More generally, a map of Reedy cofibrant diagrams $X \to Y$ is a <em>Reedy cofibration</em> if all the induced maps $X_j \sqcup_{L_j X} L_j Y \to Y_j$ are cofibrations.</p> <p>You can easily verify that if $\mathcal{C}$ is a category with cofibrations, then so is the category of Reedy cofibrant diagrams $J \to \mathcal{C}$ (and Reedy cofibrations as cofibrations). The same holds for Waldhausen categories.</p> <p>Now, $F_n \mathcal{C}$ is nothing else but the category of Reedy cofibrant diagrams $[n] \to \mathcal{C}$ and you can verify that both categories $F_n F_m \mathcal{C}$ and $F_m F_n \mathcal{C}$ can be identified with the category of Reedy cofibrant diagrams $[m] \times [n] \to \mathcal{C}$ (as categories with cofibrations), which fills in the gap in the proof.</p> <p><strong>Edit:</strong> I realized that to answer all your concerns raised in the comments, I would have to go through the basic theory of Reedy cofibrations and an MO answer is not a good place for this. Instead I will try to indicate what needs to be done and point you to a reference (unfortunately, I don't know a reference that discusses exactly what you need).</p> <p>Initially, I thought that I would simplify things by restricting to finite posets, but this only resulted in obscuring an important part of the story. So let's go to general direct categories (by the way, "direct" is very different from "directed"). A small category $J$ is <em>direct</em> if there is a functor $\mathrm{deg} : J \to \mathbb{N}$ such that for any non-identity morphism $i \to j$ in $J$ we have $\mathrm{deg} i &lt; \mathrm{deg} j$. Everything I said above goes through, you just have to modify the definition of latching objects, $L_j X$ is defined to be the colimit over $\partial J \downarrow j$ i.e. the full subcategory of the slice $J \downarrow j$ spanned by non-identity morphisms.</p> <p>The key fact is that the colimit of a Reedy cofibrant diagram always exists, the proof can be found in <a href="http://arxiv.org/abs/math/0610009v4" rel="nofollow">http://arxiv.org/abs/math/0610009v4</a> (Theorem 9.3.5). If you go through the proof you will see that this means in particular that if you know that some diagram is Reedy cofibrant below certain degree $m$, then the latching objects in degree $m$ exist. This should help you with seeing why $L_j (X_{i,\bullet})$ from your comment exists.</p> <p>If $I$ and $J$ are direct, then so is $I \times J$ (just take the sum of degrees). Using the remarks above you should be able to verify that a diagram $I \to \mathcal{C}^J$ is Reedy cofibrant if and only if the corresponding diagram $I \times J \to \mathcal{C}$ is Reedy cofibrant. Moreover a morphism $X \to Y$ in $\mathcal{C}^J$ is a Reedy cofibration between Reedy cofibrant diagrams if and only if the corresponding diagram in $\mathcal{C}^{J \times [1]}$ is Reedy cofibrant. Those things are hopefully sufficient to answer your questions.</p> http://mathoverflow.net/questions/88448/mapping-spaces-of-diagrams/88473#88473 Answer by Karol Szumiło for mapping spaces of diagrams Karol Szumiło 2012-02-15T00:10:31Z 2012-02-15T00:10:31Z <p>Let me mention yet another approach using the simplicial localization due to Dwyer and Kan.</p> <p>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.</p> <p>The precise reference is Proposition 4.4 of <em>Function Complexes in Homotopical Algebra</em> by Dwyer and Kan.</p> http://mathoverflow.net/questions/87271/when-do-functors-induce-monadic-adjunctions-of-presheaf-categories/87365#87365 Answer by Karol Szumiło for When do functors induce monadic adjunctions of presheaf categories Karol Szumiło 2012-02-02T19:59:21Z 2012-02-02T19:59:21Z <p>I've been wondering about the same thing recently and here's my best guess. Let $\widetilde C$ denote the Cauchy completion of $C$. if the associated functor $\widetilde F : \widetilde C \to \widetilde D$ induces a bijection of the sets isomorphism classes of objects in $\widetilde C$ and $\widetilde D$, then $\Delta_F$ is monadic.</p> <p>Here's a rough idea why this is true. The monad $M$ is given by a coend formula $M X = \int^c X c \times D(F c, F-)$. For simplicity let's assume that $C$ and $D$ are Cauchy complete (we can do this since $C$ and $\widetilde C$ are Morita equivalent) and $F$ is actually bijective on objects (replace $C$ and $D$ by their skeleta). Now a map $M X \to X$ is a transformation $X c \times D(F c, F c') \to X c'$ dinatural in $c$ and natural in $c'$ i.e. a transformation $D(F c, F c') \to \mathrm{Set}(X c, X c')$ natural in both in $c$ and $c'$. Since $F$ is bijective on objects we can rewrite it as $D(d, d') \to \mathrm{Set}(X d, X d')$ where $X (F c) = X c$. At this point this is just a family of maps and I do not claim any compatibility with composition in $D$. However, if $M X \to X$ is an $M$-algebra structure on $X$, then this family actually obtains a structure of a functor $D \to \mathrm{Set}$. It shouldn't be to difficult to verify (I admit I haven't done it in full detail myself) that this is an inverse to the canonical functor $\mathrm{Set}^D \to M$-$\mathrm{Alg}$.</p> <p>I also have reasons to believe that the essential surjectivity is necessary. Here's a <em>very rough</em> idea why I suspect this. If $F$ is not essentially surjective, then we seem not to be able to control the behavior of functors $D \to \mathrm{Set}$ by means of $M$-algebra structure. If I wanted to try to actually verify this, I would try to show that in that case $\Delta_F$ fails to create limits. (This is what fails when $C$ is the terminal category and $D$ is a two-point discrete category.)</p> <p>On the other hand I don't know whether the injectivity on the isomorphism classes is necessary.</p> <p>I also don't know any answer to your second question and in fact I don't see any non-trivial example of $F$ such that $\Pi_F$ is monadic (if you have such an example, please share it).</p> http://mathoverflow.net/questions/84971/suspension-of-an-excisive-pair/84996#84996 Answer by Karol Szumiło for Suspension of an excisive pair Karol Szumiło 2012-01-05T19:32:44Z 2012-01-06T08:01:38Z <p>The answer to your first question is negative. Before I give a counterexample, let me rephrase the problem in terms I consider more natural.</p> <p>First, I believe it is more convenient to consider excisive triads instead of excisive triples, i.e. I will replace a triple $(X, A, U)$ by a triad $(X; A, B)$ where $B = X \setminus U$.</p> <p>Second, the excision (either in homotopy or homology) is not really a statement about excisive triples or triads, but about homotopy pushouts. Excisive triad is just a model for homotopy pushout with some specific point-set properties, which make topological arguments possible. By this I mean that $X$ is a homotopy pushout of $A$ and $B$ along $A \cap B$. (I don't think it is literally true that every excisive triad is a homotopy pushout, but those that aren't should be considered pathological anyway. However, every homotopy pushout is homotopy equivalent to an excisive triad.)</p> <p>Thus your question could be rephrased as follows: given an excisive triple $(X, A, U)$ is the triad $(S X; S A, S X \setminus S U)$ excisive or at least a homotopy pushout? As you observed this triad is not excisive, which doesn't really tell us much since it still could be a homotopy pushout. However, this also doesn't have to be true. Let $X = S^1$ (as a subspace of $\mathbb{C}$ to fix the notation), <code>$U = \{-1, 1\}$</code> and <code>$A = X \setminus \{-i, i\}$</code>. You can write down the suspended triad and observe that the homotopy pushout of $S A$ and $S X \setminus S U$ along $S A \setminus S U$ has the homotopy type of the wedge of three circles, so it cannot be $S X$.</p> <p>On the other hand, it is easy to see that given an excisive triad $(X; A, B)$, the triad $(S X; S A, S B)$ is again excisive, which seems like a more natural thing to expect.</p> <p>To answer your second question, I don't know the book you mention, but I assume that the proof of the Homotopy Excision Theorem is more or less the same as in tom Dieck's <em>Algebraic Topology</em>. In this proof the only moment when the point-set properties of $A$ and $B$ are used is when we map a cube into $X$ and use the Lebesgue Lemma to subdivide it into cubes mapping into $A$ or $B$. To do this we only need to assume that interiors of $A$ and $B$ cover $X$. This is equivalent to saying that closure of $U$ is contained in the interior of $A$ in the corresponding triple.</p> http://mathoverflow.net/questions/83808/what-is-the-relation-between-a-homotopy-fiber-bundle-and-a-serre-fibration/83815#83815 Answer by Karol Szumiło for What is the relation between a ''homotopy fiber bundle'' and a Serre fibration? Karol Szumiło 2011-12-18T19:12:28Z 2011-12-18T19:12:28Z <p>There are many "homotopy fiber bundles" which are not fibrations. For example take any homotopy equivalence $E \to B$ that is not a fibration, then it is a "homotopy fiber bundle" with a one-point fiber.</p> <p>On the other hand the other implication is (almost) true. The following works for Hurewicz fibrations. I don't whether it is true that a fiber of a Serre fibration between CW-complexes has a homotopy type of a CW-complex. If this is the case, then the proof works also for Serre fibrations.</p> <p>Let $\pi : E \to B$ be a Hurewicz fibration between CW-complexes and $x \in B$. CW-complexes are locally contractible, so there is a contractible neighborhood $U$ of $x$. Let $\pi_U : E_U \to U$ be the restriction of $\pi$ to $U$. If $E_x$ is the fiber of $\pi$ at $x$, then the inclusion $E_x \to E_U$ is a pullback of the inclusion <code>$\{x\} \to U$</code> along a Hurewicz fibration, so it is a homotopy equivalence and admits a homotopy inverse $f : E_U \to E_x$. Thus $(\pi_U, f) : E_U \to U \times E_x$ is a homotopy equivalence over $U$.</p> http://mathoverflow.net/questions/83801/the-boundary-map-of-kan-simplicial-sets/83811#83811 Answer by Karol Szumiło for The boundary map of Kan simplicial sets Karol Szumiło 2011-12-18T18:33:30Z 2011-12-18T18:33:30Z <p>The surjectivity of your "Kan maps" is equivalent to the lifting property against all horn inclusions. Similarly the surjectivity of "boundary maps" is equivalent to the lifting property against all boundary inclusions $\partial \Delta^n \to \Delta^n$. The horn inclusions generate acyclic cofibartions and similarly boundary inclusions generate cofibrations. This means that "boundary maps" are surjective if and only if $S$ is a <em>contractible</em> Kan complex.</p> http://mathoverflow.net/questions/79825/two-kinds-of-equivalence-conjugate-vs-isomorphic-objects/79863#79863 Answer by Karol Szumiło for Two kinds of equivalence: conjugate vs. isomorphic objects Karol Szumiło 2011-11-02T20:44:27Z 2011-11-02T20:44:27Z <p>Here's a counterexample to Question 2. Let $\mathbb{Z}$ be the totally ordered set of integers regarded as a category. Then</p> <ol> <li>Distinct objects of $\mathbb{Z}$ are non-isomorphic.</li> <li>There is a morphism between every two objects of $\mathbb{Z}$.</li> <li>All objects of $\mathbb{Z}$ are conjugate (just apply the shift that maps one object to the other).</li> </ol> <p>I suspect that a finite example may be more difficult to construct (or even not exist at all).</p> <p>I also believe that notions of isomorphism and conjugacy are conceptually quite apart. The first one emphasizes "categories as universes" point of view, where we are interested in properties of objects in a category. On the other hand the notion of conjugacy seems to concern symmetries of the category itself, which emphasizes "categories as structures" point of view.</p> http://mathoverflow.net/questions/75094/semi-simplicial-versus-simplicial-sets-and-simplicial-categories/75101#75101 Answer by Karol Szumiło for Semi-simplicial versus simplicial sets (and simplicial categories) Karol Szumiło 2011-09-10T13:58:55Z 2011-09-10T13:58:55Z <p>I haven't thought about your second question, but the counit <code>$\varepsilon_K:j_!j^*K\to K$</code> is always a weak equivalence. It follows from the fact that the unit is a weak equivalence. First, observe that a map of simplicial sets <code>$f:X\to Y$</code> is a weak equivalence if and only if <code>$j^*f$</code> is a weak equivalence, because <code>$|j^*X|$</code> is the "fat geometric realization" of <code>$X$</code>, which is naturally homotopy equivalent to the usual geometric realization. Since the unit is always a weak equivalence, it follows from one of the triangular identities that <code>$j^*\varepsilon_K$</code> is a weak equivalence and by the above observation so is <code>$\varepsilon_K$</code>.</p> http://mathoverflow.net/questions/74955/strict-categorical-localization-is-automatically-a-2-localization/74992#74992 Answer by Karol Szumiło for Strict categorical localization is automatically a "2-localization"? Karol Szumiło 2011-09-09T12:08:03Z 2011-09-09T12:08:03Z <p>Let $\widetilde{\mathcal{C}}$ be the category with same objects as $\mathcal{C}$ and exactly one morphism between each pair of objects. Then there exists a unique functor $F:\mathcal{C} \to \widetilde{\mathcal{C}}$ that is the identity on objects. It sends all morphisms in $W$ to isomorphisms (since all morphisms of $\widetilde{\mathcal{C}}$ are isomorphisms). Thus it factors as $F = G q$ for a unique $G : \mathcal{C}[W^{-1}]\to\widetilde{\mathcal{C}}$. Since $F$ is the identity on objects, it follows that $q$ is injective on objects.</p> http://mathoverflow.net/questions/72992/localizing-at-homotopy-equivalences/72996#72996 Answer by Karol Szumiło for Localizing at homotopy equivalences Karol Szumiło 2011-08-16T16:23:16Z 2011-08-16T16:23:16Z <p>I'm not sure if I'm reading your question correctly. Are you asking for a proof of the following statement?</p> <p>If $f, g \colon X \to Y$ are homotopic map of topological spaces, then they are identified in the localization of the category of spaces with respect to homotopy equivalences.</p> <p>The proof is completely elementary. Let $i_0, i_1 \colon X \to X \times I$ be the bottom and top inclusions of the cylinder on $X$. They admit a common homotopy inverse, namely the projection $X \times I \to X$ (it is in fact strong deformation retraction). Thus $i_0$ and $i_1$ are identified in the localization. If $H \colon X \times I \to Y$ is a homotopy from $f$ to $g$, then $f = H i_0$ and $g = H i_1$, so $f$ and $g$ are also identified.</p> <p>In a general model category $\mathcal{C}$, if you assume that $X$ is cofibrant and $Y$ is fibrant, essentially the same argument tells you that $\mathrm{Ho}\mathcal{C}(X, Y)$ is $\mathcal{C}(X, Y)$ divided by the relation of left homotopy (or equivalently right homotopy). Of course you need to know that under (co)fibrancy assumptions every morphism in $\mathrm{Ho}\mathcal{C}(X, Y)$ is representable by a single arrow in $\mathcal{C}$ (as opposed to a longer zig-zag). In case of topological spaces and homotopy equivalences you don't need to do it, since every space is both cofibrant and fibrant in the Strøm model structure.</p> http://mathoverflow.net/questions/131255/objects-which-cant-be-defined-without-making-choices-but-which-end-up-independen Comment by Karol Szumiło Karol Szumiło 2013-05-21T06:07:11Z 2013-05-21T06:07:11Z For your choiceless definition of the trace you need to know that the canonical map <code>$V^&#42; \otimes V \to \mathrm{End}(V)$</code> is an isomorphism. How do you prove it without making choices? http://mathoverflow.net/questions/128976/can-one-make-the-category-of-pairs-of-topological-spaces-a-model-category/128982#128982 Comment by Karol Szumiło Karol Szumiło 2013-04-28T08:47:41Z 2013-04-28T08:47:41Z It's also worth pointing out that in the projective model structure on the category of arrows the cofibrant objects are exactly cofibrations between cofibrant objects of the original model category. http://mathoverflow.net/questions/84971/suspension-of-an-excisive-pair/84996#84996 Comment by Karol Szumiło Karol Szumiło 2013-04-20T12:42:22Z 2013-04-20T12:42:22Z @Fernando: I don't remember what exactly I meant at the time of writing that answer and I was somewhat confused then. By now I have clarified it. All excisive triads are homotopy pushouts with respect to weak homotopy equivalences (see tom Dieck <i>Algebraic Topology</i> Theorem 6.7.9) and all <i>numerable</i> excisive triads are homotopy pushouts with respect to genuine homotopy equivalences (loc. cit. Proposition 4.2.3) http://mathoverflow.net/questions/125834/homotopy-colimits-over-a-certain-subset-category Comment by Karol Szumiło Karol Szumiło 2013-03-28T22:43:29Z 2013-03-28T22:43:29Z The category <code>$I$</code> is not directed (unless <code>$S$</code> is finite, but that's a boring case). However, it is <i>direct</i> and for such categories there is a very explicit procedure for constructing homotopy colimits (of nice enough i.e. Reedy cofibrant diagrams). It proceeds by inductively attaching objects lying over <code>$I&#95;n$</code> and the taking the mapping telescope of the resulting sequence. It is described in detail in Theorem 9.3.5 of <a href="http://arxiv.org/abs/math/0610009v4" rel="nofollow">arxiv.org/abs/math/0610009v4</a>. http://mathoverflow.net/questions/124930/finitely-cocomplete-categories-of-compact-hausdorff-spaces Comment by Karol Szumiło Karol Szumiło 2013-03-19T12:32:17Z 2013-03-19T12:32:17Z I just wanted to point out that perhaps you are asking for more than you really need. If you only care about homotopy theory then the example that Chris described shouldn't worry you because the colimit of his diagram is not its homotopy colimit. http://mathoverflow.net/questions/124930/finitely-cocomplete-categories-of-compact-hausdorff-spaces Comment by Karol Szumiło Karol Szumiło 2013-03-19T07:59:53Z 2013-03-19T07:59:53Z (Except that I am not sure whether <code>$\mathrm{CHTop} \into \mathrm{Top}$</code> really preserves arbitrary finite colimits but I think it preserves pushouts along cofibrations which is good enough from the perspective of homotopy theory.) http://mathoverflow.net/questions/124930/finitely-cocomplete-categories-of-compact-hausdorff-spaces Comment by Karol Szumiło Karol Szumiło 2013-03-19T07:59:18Z 2013-03-19T07:59:18Z I don't really understand your motivations here. In particular, I don't understand what you dislike about simplicial sets. The category of finite simplicial sets has the structure of a cofibration category and this gives you tools to work with homotopy theory of finite simplicial sets including finite homotopy colimits. The category of finite simplicial sets and the geometric realization functor satisfy all the requirements of your Question 2. http://mathoverflow.net/questions/123739/is-there-a-conceptual-reason-why-topological-spaces-have-quotient-structures-whil Comment by Karol Szumiło Karol Szumiło 2013-03-06T12:47:32Z 2013-03-06T12:47:32Z I'm not sure whether this really answers your question, but the reason why limits and colimits of topological spaces exist is that on a given set there are always the coarsest and the finest topology satisfying some condition. On the other hand it doesn't seem sensible to talk about &quot;coarsest&quot; and &quot;finest&quot; metrics. http://mathoverflow.net/questions/123731/when-is-the-projective-model-structure-cartesian-when-is-the-internal-hom-invari Comment by Karol Szumiło Karol Szumiło 2013-03-06T11:01:46Z 2013-03-06T11:01:46Z One sufficient condition is that <code>$D$</code> has coproducts (and of course that <code>$\mathcal{M}$</code> is cartesian itself). Then the pushout product of generating projective cofibrations of <code>$\mathcal{M}^D$</code> is a cofibration in <code>$\mathcal{M}$</code> tensored with a representable copresheaf on <code>$D$</code> so it is again a projective cofibration. This handles Segal's <code>$\Gamma$</code> and the site of smooth manifolds (if we allow disconnected manifolds with components of varying dimensions). http://mathoverflow.net/questions/107188/does-the-monoid-axiom-hold-for-k-spaces/121646#121646 Comment by Karol Szumiło Karol Szumiło 2013-02-19T06:54:23Z 2013-02-19T06:54:23Z On the other hand it seems to me that the generating (acyclic) cofibrations in $\mathcal{K}$ are closed embeddings and that closed embeddings <i>are</i> preserved under pushouts and sequential colimits in $\mathcal{K}$. Hovey states this for $\mathcal{T}$ but implies that this is problematic for $\mathcal{K}$ but I don't see why it should fail. http://mathoverflow.net/questions/107188/does-the-monoid-axiom-hold-for-k-spaces/121646#121646 Comment by Karol Szumiło Karol Szumiło 2013-02-19T06:53:39Z 2013-02-19T06:53:39Z But again this doesn't seem useful since the mixed model structure is (as far as I understand) not known to be cofibrantly generated. In fact Hovey's remark on page 7 that you mention seems to say something similar about $\mathcal{K}$, namely that we don't have enough smallness properties to construct a model structure on the category of monoids. http://mathoverflow.net/questions/107188/does-the-monoid-axiom-hold-for-k-spaces/121646#121646 Comment by Karol Szumiło Karol Szumiło 2013-02-19T06:47:59Z 2013-02-19T06:47:59Z It seems to me that for the mixed model structure it is also rather simple. The fibrations of the mixed model structure are the same as those of the Str&#248;m model structure. Thus the acyclic cofibrations of the mixed model structure are the same as those of the Str&#248;m model structure i.e. acyclic Hurewicz cofibrations. Therefore the monoid axiom holds for the mixed model structure since it holds for the Str&#248;m model structure. http://mathoverflow.net/questions/107188/does-the-monoid-axiom-hold-for-k-spaces/121646#121646 Comment by Karol Szumiło Karol Szumiło 2013-02-18T08:50:28Z 2013-02-18T08:50:28Z By the way, for the Str&#248;m model structure this doesn't seem to be a useful observation since the construction of a model structure on the category of monoids (or modules over a monoid) uses the assumption that the original model category was cofibrantly generated. http://mathoverflow.net/questions/107188/does-the-monoid-axiom-hold-for-k-spaces/121646#121646 Comment by Karol Szumiło Karol Szumiło 2013-02-16T13:31:35Z 2013-02-16T13:31:35Z This of course applies to the Quillen model structure. For the Str&#248;m model structure there is nothing to check since every object is cofibrant so crossing with any space preserves all acyclic cofibrations. http://mathoverflow.net/questions/107188/does-the-monoid-axiom-hold-for-k-spaces/121646#121646 Comment by Karol Szumiło Karol Szumiło 2013-02-16T13:29:40Z 2013-02-16T13:29:40Z I can't seem to locate a version of this preprint which contains the lemma you are talking about so I can't comment on its proof. (In the version I've found 1.5 is the definition of smallness.) When I say compact I mean &quot;compact Hausdorff&quot; in the most classical topological sense. Homming out of compact Hausdorff spaces commutes with sequential colimits of open embeddings. In the last paragraph I reduce my colimit to a colimit of open embeddings and in the last line I apply this to spheres in order to see that <code>$X&#95;0 \to \mathrm{Tel}&#95;{\beta &lt; \alpha} X&#95;\beta$</code> is a <code>$\pi&#95;&#42;$</code>-isomorphism.