User omar antol&#237;n-camarena - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T12:59:14Z http://mathoverflow.net/feeds/user/644 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128854/what-are-some-examples-of-weak-categories What are some examples of weak ω-categories? Omar Antolín-Camarena 2013-04-26T20:10:26Z 2013-04-30T13:33:51Z <p>As is usual, let's say an (<em>n</em>, <em>k</em>)-category is something with objects, morphisms, 2-morphisms, ..., <em>n</em>-morphisms, such that all <em>j</em>-morphisms for <em>j</em> > <em>k</em> are invertible, everything meant in the weak sense. We can also take <em>n</em> = ∞ or <em>n</em> = <em>k</em> = ∞. In this terminology the weak ω-categories in the title question are (∞,∞)-categories.</p> <p>I think the only examples I know of weak ω-categories that are not (∞, <em>k</em>)-categories for some finite <em>k</em> are the ∞-category of all ∞-categories and the ∞-category <code>Cob</code> whose <em>n</em>-morphisms are <em>n</em>-dimensional manifolds (with corners) thought of as cobordisms between some specified (<em>n</em>-1)-dimensional manifolds (with corners). (I saw Dominic Verity give a very nice talk about his construction of a PL-version of this as a weak complicial set.) Of course, <code>Cob</code> has many variants, and we could also look at constructions such as functor categories, coproducts, products, etc., starting from these.</p> <p>I'd be very interested in hearing about other examples of (∞,∞)-categories, even if they haven't really been constructed in the literature yet. Specially examples like <code>Cob</code> which are not internal to the theory of (∞,∞)-categories.</p> <p>EDIT: I think that Sam Gunningham is right and I forgot (again) that the difference between having duals and having inverses is supposed to fall of the edge of the world when you go all the way out to ∞, so that <code>Cob</code> is an ∞-groupoid (specifically, it should be the well-known space classifying whatever kind of cobordism you used to build <code>Cob</code>). This means that I actually don't know any examples of genuinely (∞,∞)-categories that come from outside higher category theory.</p> <p>EDIT 2: I somehow missed <a href="http://mathoverflow.net/questions/73772/concrete-example-of-infty-categories" rel="nofollow">this earlier question</a>. Maybe my question should be closed as a duplicate.</p> <p>EDIT 3: Jeremy Hahn has convinced me that Sam's comment is true or false depending on how you define the equivalences of (∞,∞)-categories, and that it is not clear whether you really want every (∞,∞)-category with all adjoints to be an ∞-groupoid.</p> http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids/127787#127787 Answer by Omar Antolín-Camarena for Connected groupoids and action groupoids Omar Antolín-Camarena 2013-04-17T04:27:45Z 2013-04-18T00:28:39Z <p>Here's what I wrote on Math Stack Exchange:</p> <p>A connected groupoid <em>A</em> can be written as an action groupoid for many different groups <em>G</em>. All the groupoid determines is <em>H</em>, the group of automorphisms of any object in the groupoid, and the index of <em>H</em> in <em>G</em>, which is the cardinality of the set of objects of the groupoid. And any group <em>G</em> with subgroup <em>H</em> of the correct index, the action of <em>G</em> on the set of cosets of <em>H</em> has action groupoid isomorphic to <em>A</em>.</p> <p>This is to be expected, because if we think of <em>H</em> as a one object groupoid and of <em>X</em> as an indiscrete groupoid (the set of objects is <em>X</em>, and there is a unique morphism between any pair of objects) then the original groupoid <em>A</em> is isomorphic to the product <em>H</em> &times; <em>X</em>, so the isomorphism class of <em>A</em> depends only on the group <em>H</em> and the cardinality of <em>X</em>.</p> <p>As an extreme example of this, let <em>G</em> act on itself by translation and take the action groupoid. This has set of objects <em>G</em> and a unique morphism between every pair of elements. Notice all traces of the group structure of <em>G</em> are gone: the isomorphism class of this indiscrete groupoid only depends on the cardinality of <em>G</em>.</p> <p>UPDATE 2: Here is a sloganized answer to the question: the <strong>equivalence</strong> class of a connected groupoid <em>A</em> is determined by the isomorphism class of the group <em>H</em> = Aut(<em>x</em><sub>0</sub>); the <strong>isomorphism</strong> class of a category is given by the data of its equivalence class plus the number of isomorphic copies of each object in a skeleton.</p> <p>UPDATE: Here is a proof of the claims above "UPDATE 2".</p> <p><strong>Claim 1</strong>: <em>A</em> is isomorphic to <em>H</em> &times; <em>X</em>.</p> <p><em>Proof.</em> Choose an object <em>x</em><sub>0</sub> of <em>A</em>, identify <em>H</em> with Aut(<em>x</em><sub><em>0</em></sub>) and choose arbitrary morphisms <em>a</em><sub><em>x</em></sub> : <em>x</em><sub><em>0</em></sub> &rarr; <em>x</em>. The isomorphism <em>H</em> &times; <em>X</em> &rarr; <em>A</em> is the identity on objects and sends a morphism (<em>h</em>, <em>u</em>) : <em>x</em> &rarr; <em>y</em> to the morphism <em>a</em><sub><em>y</em></sub> <em>h a</em><sub><em>x</em></sub><sup>-1</sup>. (Here <em>u</em> is the unique morphism in <em>X</em> from <em>x</em> to <em>y</em>.) The inverse <em>A</em> &rarr; <em>H</em> &times; <em>X</em> sends a morphism <em>a</em> : <em>x</em> &rarr; <em>y</em> to (<em>a</em><sub><em>y</em></sub><sup>-1</sup> <em>a a</em><sub><em>x</em></sub>, <em>u</em>) --same <em>u</em> as above.</p> <p><strong>Claim 2</strong>: For any group <em>G</em> with a subgroup <em>H</em> of index |<em>X</em>|, the action groupoid of <em>G</em> acting on the set <em>G/H</em> of cosets is isomorphic to <em>A</em>. </p> <p><em>Proof.</em> Both groupoids are isomorphic to <em>H</em> &times; <em>X</em>.</p> http://mathoverflow.net/questions/105878/how-should-one-understand-orbifold-fundamental-groups/126347#126347 Answer by Omar Antolín-Camarena for How should one understand orbifold fundamental groups? Omar Antolín-Camarena 2013-04-03T04:06:06Z 2013-04-03T10:55:00Z <p>The answer to question 1, which asks whether this definition recovers homotopy groups of a manifold, is yes. As in the question, we'll take $M$ to be manifold, $G_0 = \bigsqcup_i U_i$ the disjoint union of the sets of an open cover of $M$ and $G_1 = G_0 \times_M G_0 = \bigsqcup_{i,j} U_i \cap U_j$. It is easy to see that $G_2 = \bigsqcup_{i,j,k} U_i \cap U_j \cap U_k$, and generally, $G_n$ is the disjoint union of the $(n+1)$-fold intersections of the $U_i$. To prove that $B\mathcal{G}$ has the same homotopy groups as $M$, we'll show in fact that these two spaces are homotopy equivalent.</p> <p>First of all there is a clear map $B \mathcal{G} \to M$, that sends any point $(x,p) \in G_n \times \Delta^n$ to just $x$. This map is well defined since (1) $G_n$ is just a disjoint union of some open subsets of $M$, so any point of $G_n$ can be thought of as a point of $M$, and (2) all the face and degeneracy maps in $\mathcal{G}$ just perform bookkeeping: they change which multiple intersection of $U_i$'s a point is regarded as being in, but don't actually change the point.</p> <p>Now, we'll define a section $M \to B \mathcal{G}$. For this, take a locally finite partition of unity $\phi_i$ subordinate to the cover <code>$\{U_i\}_i$</code>. Given $x \in M$, it'll belong to the support of finitely many $\phi_i$, say $\phi_{i_0}, \ldots, \phi_{i_k}$ and we can send $x$ to the equivalence class of $(x, (\phi_{i_0}(x), \ldots, \phi_{i_k}(x))) \in G_k \times \Delta^k$. Here, $x$ is thought of as belonging to the $U_{i_0} \cap \cdots \cap U_{i_k}$ term of $G_k$, and we regard $\Delta^k$ as the set of points $(t_0,\ldots,t_k) \in \mathbb{R}^{k+1}$ such that $t_i \ge 0$ and $t_0 + \cdots + t_k = 1$. It's straightforward to check that the definition above does give a function $M \to B\mathcal{G}$ (for example, adding extra $\phi_{i_j}$ with $\phi_{i_j}(x)=0$, does not change the image of $x$ in $B\mathcal{G}$), and it is clear this map is a section of the map $B\mathcal{G}\to X$ above.</p> <p>Finally, the composite $B\mathcal{G} \to M \to B\mathcal{G}$ is homotopic to the identity by a straight line homotopy within the $\Delta^k$'s (that is, via homotopy that is the identity on the $G_k$ coordinate and moves in a straight line segment in the $\Delta^k$ coordinate).</p> <p>This argument has probably been discovered many times and is well-known, I fairly recently found out that it does appear in print at least in Segal's <a href="http://www.maths.ed.ac.uk/~aar/papers/segalclass.pdf" rel="nofollow"><em>Classifying Spaces and Spectral Sequences</em></a>, proposition 4.1. It doesn't use that $M$ is a manifold, just that the open cover has a continuous subordinate partition of unity, so for example it works for any cover of a paracompact space $M$. Even without subordinate partitions of unity, for any open cover $U_i$ of any any space $M$ the homotopy groups of $M$ and $B \mathcal{G}$ are the same: these two spaces might not be homotopy equivalent in this general setting, but they are still weakly homotopy equivalent. See Dugger and Isaksen’s <a href="http://hopf.math.purdue.edu/Dugger-Isaksen/hypercover.pdf" rel="nofollow"><em>Hypercovers in Topology</em></a>, theorem 2.1. This simplicial space $\mathcal{G}$ is called a <em>Čech cover</em>, and Dugger and Isaksen's paper also deals with the more general notion of a <em>hypercover</em>.</p> http://mathoverflow.net/questions/126191/are-subfunctors-of-left-exact-functors-also-left-exact/126196#126196 Answer by Omar Antolín-Camarena for Are subfunctors of left exact functors also left exact? Omar Antolín-Camarena 2013-04-01T17:05:22Z 2013-04-02T15:28:36Z <p>Not necessarily. Consider these two functors <code>Set</code> &rarr; <code>Set</code>: <em>F</em> is the constant functor with value the empty set, and <em>G</em> is the identity functor. Then <em>F</em> is a subfunctor of <em>G</em>, <em>G</em> is obviously left exact, but <em>F</em> is not: it does not send the terminal object to the terminal object.</p> http://mathoverflow.net/questions/125663/are-luries-operads-special-smcs/125875#125875 Answer by Omar Antolín-Camarena for Are Lurie's operads special SMCs? Omar Antolín-Camarena 2013-03-29T04:38:44Z 2013-03-29T14:14:16Z <p>Given a symmetric monoidal category one can construct its <em>underlying operad</em> (well, symmetric colored operad, but I won't keep mentioning this). This operad has the same objects as the SMC. The operads arising this way can be characterized, so that we can regard a SMC as a special kind of operad.</p> <p>Conversely, given an operad, one can construct its <em>symmetric monoidal envelope</em>, a symmetric monoidal category whose objects are not just the objects of the operad but rather formal (commutative) products of them. The SMCs arising this way can be characterized, as David mentions in the question, so that we can also regard operads as special SMCs.</p> <p>Now, when operads and SMC are implemented by Lurie's definitions, the direction in which it is easy to go is that of regarding a SMC as an operad. If the SMC is given by a functor $p : C \to Fin_{\ast}$, then regarded as an operad it is the same functor, and whether we view it as a SMC or as an operad its objects are the objects of the category $p^{-1}\langle 1 \rangle$.</p> <p>To go in the other direction, we need to change the functor: given an operad $p : C \to Fin_{\ast}$, its symmetric monoidal envelope will be a different functor $q : D \to Fin_{\ast}$. The objects of the envelope, which are the objects of $q^{-1}\langle 1 \rangle$, will be all the objects of $C$, not just the objects of the subcategory $p^{-1}\langle 1 \rangle$.</p> <p>Lurie constructs the symmetric monoidal envelope in section 2.2.4 of <em>Higher Algebra</em>.</p> http://mathoverflow.net/questions/122115/problems-from-the-scottish-book/123003#123003 Answer by Omar Antolín-Camarena for problems from the scottish book Omar Antolín-Camarena 2013-02-26T17:14:13Z 2013-02-26T17:14:13Z <p>Luis Montejano solved problem 68 in 1990 (he also solved the limiting 0 density case of problem 19, but I think this is already mentioned in Mauldin's book). The paper is called <a href="https://eudml.org/doc/140207" rel="nofollow"><em>About a problem of Ulam concerning flat sections of manifolds</em></a> and appeared in Commentarii Mathematici Helvetici.</p> http://mathoverflow.net/questions/122643/realization-of-maps-between-classifying-spaces-of-categories/122693#122693 Answer by Omar Antolín-Camarena for realization of maps between classifying spaces of categories Omar Antolín-Camarena 2013-02-23T04:17:47Z 2013-02-23T04:17:47Z <p>Here's one instance where the answer is affirmative: when the target category is a groupoid (a little more general than my remark on John Klein's answer).</p> <p>Let $\mathcal{C}$ be any category and $\mathcal{G}$ be a groupoid. Then $B\mathcal{G}$ is a 1-type (i.e., all $\pi_n$ with $n\ge 2$ vanish, with any base points), and therefore, for any space $X$, $\mathrm{maps}(X, B\mathcal{G})$ is a 1-type too, and in fact is just weakly homotopy equivalent to $B \mathrm{Fun}(\pi_{\le 1}(X), \mathcal{G})$ where $\pi_{\le1}(X)$ is the fundamental groupoid of $X$ (notice that since $\mathcal{G}$ is a groupoid, so is any functor category $\mathrm{Fun}(\mathcal{A}, \mathcal{G})$). Applying this to $X=B\mathcal{C}$, we get a weak homotopy equivalence $\mathrm{maps}(B \mathcal{C}, B \mathcal{G}) \simeq B \mathrm{Fun}(\pi_{\le 1}(B\mathcal{C}), B\mathcal{G})$. Now, $\pi_{\le 1}B \mathcal{C}$ is just $\mathcal{C}[\mathcal{C}^{-1}]$, the localization of $\mathcal{C}$ obtained by adding formal inverses for all morphisms of $\mathcal{C}$; and since $\mathcal{G}$ is a groupoid, any functor $\mathcal{C} \to \mathcal{G}$ automatically factors through $\mathcal{C}[\mathcal{C}^{-1}]$, so composition with the canonical functor $\mathcal{C} \to \mathcal{C}[\mathcal{C}^{-1}]$ induces an equivalence of categories $\mathrm{Fun}(\mathcal{C}, \mathcal{G}) \cong \mathrm{Fun}(\mathcal{C}[\mathcal{C}^{-1}], \mathcal{G})$. Equivalences of categories induce homotopy equivalences of classifying spaces, so we get a weak equivalence $\mathrm{maps}(B \mathcal{C}, B \mathcal{G}) \simeq B\mathrm{Fun}(\mathcal{C}, \mathcal{G})$, which is a little better than just saying that every function between the classifying spaces is homotopic to one coming from a functor.</p> http://mathoverflow.net/questions/122412/simplicial-sets-from-bisimplicial-sets-and-their-realisations/122482#122482 Answer by Omar Antolín-Camarena for Simplicial sets from bisimplicial sets, and their realisations. Omar Antolín-Camarena 2013-02-20T23:44:32Z 2013-02-20T23:44:32Z <p>In the term "canonical presentation", "presentation" just means that the diagram you wrote is a coequalizer diagram <em>presenting</em> $T$ as being obtained from "simpler" bisimplicial spaces (so, yes, "cokernel of a pair" means "coequalizer"). This use of the term presentation comes from algebra, where presenting, say, a group, can be interpreted to mean writing it as a coequalizer of maps between free groups (and similarly for other algebraic structures).</p> <p>The "canonical" part just means the whole diagram is a functor of $T$, or more concretely, that both $F(T):=\coprod_{(r,s) \to (r',s')} h^{r',s'} \times T_{rs}$ and $G(T) := \coprod_{(r,s)} h^{r,s} \times T_{rs}$ define (the object part) of functors and all three arrows in $F(T) \rightrightarrows G(T) \to T$ are components of natural transformations $F \rightrightarrows G \to \mathrm{Id}$.</p> <p>Finally, the way the argument goes, once you have this functorial diagram, is that if $R$ denotes any of the three realization functors (from bisimplicial spaces all the way to spaces) you mention, then $R$ commutes with colimits (called "inductive limits" in the paper). This means that $R(F(T)) = \coprod_{(r,s) \to (r',s')} R(h^{r',s'} \times T_{rs})$, that $R(G(T)) = \coprod_{(r,s)} R(h^{r,s} \times T_{rs})$, and that the diagram</p> <p>$$\coprod_{(r,s) \to (r',s')} R(h^{r',s'} \times T_{rs}) \rightrightarrows \coprod_{(r,s)} R(h^{r,s} \times T_{rs}) \to R(T)$$</p> <p>is still a coequalizer diagram (but now in spaces). Since Quillen first proved that the spaces $R(h^{r,s} \times T_{pq})$ do not depend on which of the three $R$'s you take, the result, $R(T)$ does not depend on which $R$ you take either.</p> http://mathoverflow.net/questions/122221/is-there-a-discrete-monoid-m-injecting-into-its-group-completion-g-for-which-bm Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG? Omar Antolín-Camarena 2013-02-18T19:34:34Z 2013-02-18T22:22:31Z <p>For a (discrete) monoid $M$, the classifying space $BM$ is the geometric realization of the nerve of the one object category whose hom-set is $M$. (This definition gives the usual classfiying space when $M$ is a group.) The group completion of $M$ can be constructed as the fundamental group of $BM$, and is characterized by the universal property that any monoid homomorphism from $M$ to a group factors uniquely through the group completion.</p> <p>My question is whether there is an example of a monoid for which the canonical map to its group completion is injective, but for which this canonical map does not induce a homotopy equivalence of the classifying spaces.</p> <p>As background here are some facts:</p> <ol> <li><p>Classifying spaces of monoids produce all connected homotopy types! This is proved in Dusa McDuff's 1979 paper <em>On the classifying spaces of discrete monoids</em>. For a neat concrete example, see Zbigniew Fiedorowicz's <a href="http://arxiv.org/abs/math/0202260" rel="nofollow"><em>A counterexample to a group completion conjecture of JC Moore</em></a>; it shows a specific 5 element monoid whose classifying space is homotopy equivalent to $S^2$.</p></li> <li><p>If $G$ is the group completion of a <em>commutative</em> monoid $M$, the canonical map $BM \to BG$ is a homotopy equivalence; even if $M \to G$ is <em>not</em> injective. (This is easy to prove: think of $M \to G$ as a functor between one object categories and apply Quillen's Theorem A to it. There is only one slice category to check and using commutativity it is easy to see this category is filtered and thus contractible.)</p></li> <li><p>If $M$ is a free monoid and the free group $G$ is its completion, the map $BM \to BG$ is a homotopy equivalence. It fact, more generally, if $C$ is the free category on some directed graph $X$, the nerve of $C$ is homotopy equivalent to the geometric realization of $X$. This is proved in Dwyer and Kan's <a href="http://www.nd.edu/~wgd/Dvi/SimplicialLocalizations.pdf" rel="nofollow"><em>Simplical Localization of Categories</em></a>, proposition 2.9, but the proof is simple enough to sketch here: for each $k$, the inclusion of the $k$-skeletion of $NC$ into the $(k+1)$-skeleton is a weak homotopy equivalence (since you get the $(k+1)$-skeleton by filling in some horns); so the $1$-skeleton, $X$, is weakly equivalent to $NC$. (The claim for free monoids is the case where $X$ consists of a single vertex with some loops.)</p></li> <li><p>Even if a monoid has left and right cancellation the canonical map to its group completion might not be injective. Here's an example from Malcev's <em>On the Immersion of an Algebraic Ring into a Field</em>: let $M$ be the <em>monoid</em> presented by $(a,b,c,d,x,y,u,v : ax=by, cx=dy, au=bv)$. Malcev shows that $M$ is cancellative, but that in $M$, $cu \neq dv$; in any group the relations listed for $M$ would imply that $cu=dv$.</p></li> </ol> http://mathoverflow.net/questions/98609/algorithmically-unsolvable-problems-in-topology/121624#121624 Answer by Omar Antolín-Camarena for Algorithmically unsolvable problems in topology Omar Antolín-Camarena 2013-02-12T17:20:18Z 2013-02-12T17:20:18Z <p>I'm not sure whether these problems count as being topological: they're about topologically defined classes of groups:</p> <p>Determining whether a presentation of a certain kind of knot groups in fact presents a more restrictive class of knots groups tends to be algorithmically unsolvable. I'm talking about the results in the paper <a href="http://arxiv.org/abs/0908.4009" rel="nofollow">Unsolvable problems about higher-dimensional knots and related groups</a> by Francisco González Acuña, Cameron Gordon and Jonathan Simon:</p> <p>Consider the classes of groups $\mathcal K_0 \subset \mathcal K_1 \subset \mathcal K_2 \subset \mathcal K_3 \subset \mathcal S \subset \mathcal M \subset \mathcal G$, where $\mathcal K_n$ is the class of fundamental groups of complements of $n$-spheres in $S^{n+2}$ (it is known that $\mathcal K_n = \mathcal K_3$ for $n\ge 3$); $\mathcal S$ (resp. $\mathcal M$, $\mathcal G$) is the class of fundamental groups of complements of orientable, closed surfaces in $S^4$ (resp. in a 1-connected 4-manifold, in a 4-manifold). (It is known that $\mathcal G$ is in fact the class of all finitely presented groups and that all the inclusions are strict.) </p> <p>Their main theorem says that if $\mathcal B \subsetneq \mathcal A$ are two of the above classes of groups <em>and</em> $\mathcal K_3 \subseteq \mathcal A$, then there does not exist an algorithm that can decide, given a finite presentation of a group in $\mathcal A$ whether or not the group is in $\mathcal B$. And they conjecture this is true assuming only $\mathcal K_2 \subset \mathcal A$.</p> http://mathoverflow.net/questions/93262/which-colimits-commute-with-which-limits-in-the-category-of-sets Which colimits commute with which limits in the category of sets? Omar Antolín-Camarena 2012-04-05T23:06:53Z 2012-09-10T12:55:46Z <p>Given two categories $I$ and $J$ we say that <em>colimits of shape $I$ commute with limits of shape $J$</em> in the category of sets, if for any functor $F : I \times J \to \text{Set}$ the canonical map <code>$$\textrm{colim}_{i\in I} \text{lim}_{j\in J} F(i,j) \to \textrm{lim}_{j\in J} \text{colim}_{i\in I} F(i,j)$$</code> is an isomorphism.</p> <p>The standard examples are a) filtered colimits commute with finite limits and b) sifted colimits commute with finite products. (Those statements can be regarded as definitions of which categories $I$ are filtered or sifted respectively, but both terms have independent definitions for which these commutation results are propositions.) A third, less known example is to take $I$ a finite group and $J$ a cofiltered category, in other words, if $G$ is a finite group and $X_j$ is an inverse system of $G$-sets, then the canonical map <code>$$(\varprojlim_{j\in J} X_j)/G \to \varprojlim_{j \in J}(X_j/G)$$</code> is an isomorphism.</p> <p>Now, all of these examples are easy to prove separately (<a href="http://math.harvard.edu/~oantolin/notes/fingrpcomm.html" rel="nofollow">here</a>'s a proof of the $G$-set result, for example) but I see no unifying pattern. Is there a simple criterion for when $I$-colimits and $J$-colimits commute in the category of sets?</p> <p>[Note: It's true that $I$ is filtered (resp. sifted) if and only if for all finite (resp. finite discrete) $J$ the diagonal functor $I \to I^J$ is final; but I don't think that for arbitrary $I$ and $J$, if the diagonal $I \to I^J$ is final then $I$-colimits commute with $J$-limits. If I'm wrong and that condition on the diagonal actually is sufficient for commutation: why? and is it also necessary?]</p> http://mathoverflow.net/questions/17539/is-there-a-description-of-sheaf-cohomology-in-algebraic-topological-terms Is there a description of sheaf cohomology in algebraic-topological terms? Omar Antolín-Camarena 2010-03-08T22:17:46Z 2012-08-11T13:21:36Z <p>Is there a description of of sheaf cohomology for the sheaf of sections of a continuous function in terms of common constructions in Algebraic Topology?</p> <p>In more detail: Any sheaf on a space X can be described as the sheaf of sections of some continuous map from the étale space Y to X. In fact, the category of sheaves (of sets) on X is equivalent to the category of maps to X which are local homeomorphisms. A sheaf of Abelian groups is the same as an Abelian group object in the category of sheaves of sets, so instead of talking about cohomology of sheaves, we could talk about cohomology of an Abelian group object in the category of local homeomorphisms to X, that is, a local homeomorphism from some space Y to X such that, roughly, every fibre has an Abelian group structure where all the multiplications (of the fibres) put together form a continuous map from Y &times; <sub>X</sub> Y to Y.</p> <p>It seems like there should be a simple description of cohomology of X with coefficient in a sheaf of Abelian groups in terms of the corresponding map Y &rarr; X that uses only usual constructions in Algebraic Topology and the (fibrewise) group structure of Y. Is there one?</p> http://mathoverflow.net/questions/76505/in-which-situations-can-one-see-that-topological-spaces-are-ill-behaved-from-the/76594#76594 Answer by Omar Antolín-Camarena for In which situations can one see that topological spaces are ill-behaved from the homotopical viewpoint? Omar Antolín-Camarena 2011-09-28T02:59:32Z 2011-09-28T02:59:32Z <p>The smash product of pointed topological spaces is not associative, i.e., $(X \wedge Y)\wedge Z$ need not be homeomorphic to $X \wedge (Y \wedge Z)$. (It fails, for example, for $X = Y = \mathbb{Q}$ and $Z = \mathbb{N}$.)</p> http://mathoverflow.net/questions/74386/how-should-i-think-of-the-infty-category-of-spectra/74465#74465 Answer by Omar Antolín-Camarena for How should I think of the $\infty$-category of spectra? Omar Antolín-Camarena 2011-09-03T22:26:40Z 2011-09-03T22:26:40Z <p>Basically you want to know what the space of maps between two spectra $X$ and $Y$ is. Well, each map from $X$ to $Y$ is a sequence of maps from $X_n$ to $Y_n$ and thus $\mathrm{map}(X,Y)$ is a subspace of the product of the $Y_n^{X_n}$. And you can build an entire spectrum of maps from $X$ to $Y$, whose nth space is just maps from $X$ to the nth suspension of $Y$.</p> <p>For modern constructions (S-modules, orthogonal spectra, symmetric spectra) of spectra which give symmetric monoidal categories this internal hom is adjoint to the smash product. These modern categories have model structures and between cofibrant and fibrant objects the mapping space described above is homotopically correct. (And, in Boardman's construction --which is not monoidal before passing to the homotopy category--, you still get that this mapping spectrum is the right one for maps between a CW-spectrum and an $\Omega$-spectrum). So one version the stable $(\infty,1)$-category of spectra is the topologically enriched category of fibrant-cofibrant spectra in any of these modern categories, with mapping spaces as above.</p> <p>Of course, for these model categories of spectra, any of the other ways of getting at the $(\infty,1)$-category they represent, such as Dwyer-Kan localization, will give an equivalent stable $(\infty,1)$-category of spectra.</p> http://mathoverflow.net/questions/61761/topological-sort-of-partial-order-into-sorted-sets/61762#61762 Answer by Omar Antolín-Camarena for Topological sort of partial order into sorted sets Omar Antolín-Camarena 2011-04-14T23:00:12Z 2011-04-15T23:03:50Z <p>EDIT: This answer was for a previous version of the question.</p> <p>There is usually no such list: consider the case where some element is incomparable to everything else.</p> http://mathoverflow.net/questions/35097/nerves-of-braided-or-symmetric-monoidal-categories/35172#35172 Answer by Omar Antolín-Camarena for Nerves of (braided or symmetric) monoidal categories Omar Antolín-Camarena 2010-08-10T21:32:58Z 2010-08-10T21:43:13Z <p>For plain old monoidal categories, you could regard them as a bicategory with a single object and use the <a href="http://www.tac.mta.ca/tac/volumes/9/n10/9-10abs.html" rel="nofollow">Duskin nerve</a>. For braided or symmetric categories there might be higher nerves you can take, but I'm not sure how well those work. You might be better off using the k-fold simplicial sets that Scott suggested.</p> http://mathoverflow.net/questions/30511/ebook-readers-for-mathematics/30630#30630 Answer by Omar Antolín-Camarena for eBook readers for mathematics Omar Antolín-Camarena 2010-07-05T15:26:55Z 2010-07-05T15:26:55Z <p>I have a Sony PRS-505. It works reasonably well for reading straight through, but is too slow for jumping around comfortably. It can read PDFs directly, but cannot zoom (increasing font-size will reflow the text to fit the margins which actually works pretty well for text only PDFs but messes up formulas and diagrams). You can rotate the view 90 degrees which leaves the text a decent size provided you chop off all the margins before putting the PDF on the device. Adobe Acrobat (full Acrobat, not the Reader) can do this, or a variety of commandline tools: on Linux you can use pdfcrop, pdftk or pdfmanipulate (that last one comes with calibre), a windows program called soPDF I found on some ebook forum, etc. Removing the whitespace in the margins really improves the experience, and should probably be done with any of the current eBook readers: even the ones with zoom will probably zoom very slowly.</p> http://mathoverflow.net/questions/26537/connectedness-and-the-real-line/26594#26594 Answer by Omar Antolín-Camarena for Connectedness and the real line Omar Antolín-Camarena 2010-05-31T15:32:29Z 2010-05-31T15:45:14Z <p>If you've already developed basic facts about compactness you can prove it this way:</p> <p>Let [0,1] = <em>A</em> &cup; <em>B</em> with <em>A</em> and <em>B</em> closed and disjoint. Then since <em>A</em> &times; <em>B</em> is compact and the distance function is continuous, there is a pair (<em>a</em>, <em>b</em>) &isin; <em>A</em> &times; <em>B</em> at minimum distance. If that distance is zero, <em>A</em> and <em>B</em> intersect. If not, you get a contradiction by taking any point in the interval from <em>a</em> to <em>b</em>: it can't be in either <em>A</em> or <em>B</em> because its distance from <em>b</em> or <em>a</em> is smaller than the minimum.</p> <p>That shows a compact interval in &#8477; is connected. If &#8477; = <em>A</em> &cup; B with <em>A</em> and <em>B</em> closed and disjoint, then for any closed interval <em>I</em> with one endpoint in <em>A</em> and one in <em>B</em>, <em>I</em> = (<em>A</em> &cap; <em>I</em>) &cup; (<em>B</em> &cap; <em>I</em>) is disconnection of <em>I</em>. Alternatively, you could write &#8477; as a union of closed intervals with a common point.</p> http://mathoverflow.net/questions/3003/in-what-sense-are-fields-an-algebraic-theory In what sense are fields an algebraic theory? Omar Antolín-Camarena 2009-10-28T05:34:44Z 2010-05-27T20:14:07Z <p>Since there is no "free field generated by a set", it would seem that</p> <p>1) there is no monad on Set whose algebras are exactly the fields</p> <p>and</p> <p>2) there is no Lawvere theory whose models in Set are exactly the fields</p> <p>(Are 1) and 2) correct?)</p> <p>Fields don't form a variety of algebras in the sense of universal algebra since the field axioms can´t be written as identities (since the axiom for multiplicative inverses has the restriction that the element be non-zero).</p> <p>I guess fields are an algebraic theory in a more general universal algebra sense of being defined by operations on a single set with a set of first order axioms.</p> <p>Is there any better sense in which they are algebraic or are fields just not really algebraic in nature?</p> http://mathoverflow.net/questions/17786/why-are-local-systems-and-representations-of-the-fundamental-group-equivalent/17810#17810 Answer by Omar Antolín-Camarena for Why are local systems and representations of the fundamental group equivalent Omar Antolín-Camarena 2010-03-11T03:14:56Z 2010-03-11T03:14:56Z <p><a href="http://www.math.lsu.edu/~pramod/teaching/07s-7280/notes3.pdf" rel="nofollow">Pramod Achar's notes</a> (from a lecture in an course he taught on perverse sheaves) are two pages. </p> http://mathoverflow.net/questions/129265/for-a-monoid-with-zero-m-how-many-additive-operations-on-m-can-there-be-maki/129272#129272 Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-05-01T15:27:46Z 2013-05-01T15:27:46Z @Noah: Whoops! You're right, I got confused. http://mathoverflow.net/questions/129265/for-a-monoid-with-zero-m-how-many-additive-operations-on-m-can-there-be-maki/129272#129272 Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-05-01T02:25:42Z 2013-05-01T02:25:42Z given a Boolean algebra structure, the poset is recovered by setting $a\le b$ iff $a\wedge b=a$. http://mathoverflow.net/questions/128854/what-are-some-examples-of-weak-categories Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-27T12:51:44Z 2013-04-27T12:51:44Z Did you mean $\infty$-groupoid instead of contractible, @JeremyHahn? I mean an ordinary groupoid has all adjoints, right? I certainly don't want it to be contractible as an $(\infty,\infty)$-category. (Or do I?) If this second homotopy theory of $(\infty,\infty)$-categories makes either (1) a all homotopy types contractible, or (2) homotopy inverses not count as adjoints, it doesn't sound like such a great idea. http://mathoverflow.net/questions/128854/what-are-some-examples-of-weak-categories Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-27T02:30:23Z 2013-04-27T02:30:23Z Yes, I think that's right @SamGunningham. I've added a remark about this to the question. http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids/127787#127787 Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-18T04:41:54Z 2013-04-18T04:41:54Z I hope that is clear enough, @MikhailBorovoi. If not let me know. As you can see, the proof is pretty much just what Sam Cunningham wrote (only slightly generalized). If you want an explicit isomorphism between A and G/H, you can just compose the isomorphisms A -&gt; H x X -&gt; G/H given in the proof of Claim 1. http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids/127787#127787 Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-17T11:20:43Z 2013-04-17T11:20:43Z @Mikhail: Pick a set of representatives for the cosets of <i>H</i>, and pick a bijection between those representatives and the objects of <i>A</i>. I can add more detail later when I'm at a computer with a real keyboard. http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids/127747#127747 Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-17T04:30:29Z 2013-04-17T04:30:29Z These are some of the groups you can use, but in general you don't need $G$ to be a product, just to have a subgroup isomorphic to $G_0$ with index $|X|$. http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-17T04:23:00Z 2013-04-17T04:23:00Z This is easier than everyone is making it sound. http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-17T04:21:57Z 2013-04-17T04:21:57Z I answered on math.stack exchange. http://mathoverflow.net/questions/105878/how-should-one-understand-orbifold-fundamental-groups/126347#126347 Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-03T10:56:47Z 2013-04-03T10:56:47Z I've reworded the last paragraph to make it clearer, since Dan Petersen's question made me see I wasn't being very clear. http://mathoverflow.net/questions/105878/how-should-one-understand-orbifold-fundamental-groups/126347#126347 Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-03T10:45:18Z 2013-04-03T10:45:18Z No Dan, the argument I gave works whenever you have a continuous partition of unity subordinate to the cover (I.e.,for paracompact spaces M even if M is not a manifold) and gives you a homotopy equivalence. What I'm saying in the last paragraph is that without a subordinate partition of unity you don't get a homotopy equivalence anymore, but you still get a weak equivalence. http://mathoverflow.net/questions/126191/are-subfunctors-of-left-exact-functors-also-left-exact/126196#126196 Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-04-02T15:30:52Z 2013-04-02T15:30:52Z Everyone can stop down-voting my answer: I've corrected the spelling of 'necessarily'. :P (Seriously now, I agree that Eric's example is much nicer and closer to the context Samuel Mf is interested in.) http://mathoverflow.net/questions/125663/are-luries-operads-special-smcs/125875#125875 Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-03-31T17:10:29Z 2013-03-31T17:10:29Z Yes, that's right. http://mathoverflow.net/questions/125663/are-luries-operads-special-smcs Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-03-27T17:28:48Z 2013-03-27T17:28:48Z I think the construction you are looking for, that given an operad produces a symmetric monoidal category, is the monoidal envelope described in <i>Higher Algebra</i>, section 2.2.4. http://mathoverflow.net/questions/125663/are-luries-operads-special-smcs Comment by Omar Antolín-Camarena Omar Antolín-Camarena 2013-03-27T01:53:30Z 2013-03-27T01:53:30Z For what you want, given an operad $p : C \to Fin_{\ast}$, you would need to construct a $q : D \to Fin_{\ast}$ where the objects of $q^{-1}\langle 1 \rangle$ come from all of $C$, not just from $p^{-1} \langle 1 \rangle$.