User reid barton - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:53:45Z http://mathoverflow.net/feeds/user/350 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/984/algebraic-k-theory-and-tensor-products algebraic K-theory and tensor products Reid Barton 2009-10-18T02:22:27Z 2013-03-15T13:21:45Z <p>Algebraic K-theory defines a functor K taking commutative rings to E_\infty ring spectra. I'm interested in which pushouts (tensor/smash products) K preserves. For example, if R is a regular noetherian ring then (I believe) K(R[t, t^{-1}]) = K(R) \/ &Sigma;K(R) = K(R) /\<sub>K(Z)</sub> K(Z[t, t^{-1}]). On the other hand, K(Q) = K(Q &otimes; Q) is not the same as K(Q) /\<sub>K(Z)</sub> K(Q) as you can check by computing &pi;<sub>1</sub>.</p> <p>Are there useful conditions under which K-theory preserves pushouts?</p> <p>Edit: I'm equally interested in more general positive answers and more geometric counterexamples. For example, what is an example of smooth schemes X and Y over Spec k such that K(X) /\<sub>K(k)</sub> K(Y) -> K(X x<sub>k</sub> Y) is not an equivalence?</p> <p>Also, what if I only cared about K<sub>0</sub>? Is the product map more often an isomorphism then?</p> <p>More generally, is there a spectral sequence to compute the K-theory of a fiber product of schemes?</p> http://mathoverflow.net/questions/7561/naive-z-2-spectrum-structure-on-e-smash-e Naive Z/2-spectrum structure on E smash E? Reid Barton 2009-12-02T05:47:27Z 2013-02-12T14:40:57Z <p>Let $E$ be a spectrum. Then $E \wedge E$ is a $\mathbb{Z}/2$-spectrum in the naivest possible sense, i.e., an object with $\mathbb{Z}/2$-action in the (&infin;,1)-category of spectra. Can I make it a $\mathbb{Z}/2$-spectrum in the less naive, but still not genuine, sense? (That is, a $\mathbb{Z}/2$-spectrum indexed on the trivial universe.)</p> <p>I'm thinking of something like the following. I may represent $E$ as an (reduced &amp; continuous) excisive functor from pointed spaces to pointed spaces. Then define</p> <p>$$G(X) = \mathrm{colim}_{I \times I} \mathrm{Map}(S^{x_1} \wedge S^{x_2}, E(S^{x_1}) \wedge E(S^{x_2}) \wedge X)$$</p> <p>where $I$ is the category of finite sets and inclusions. Hopefully $G$ is a functor from spaces to $\mathbb{Z}/2$-spaces. If I forget about the $\mathbb{Z}/2$-fixed point set, I can think of it as $E \wedge E$ with its $\mathbb{Z}/2$ action. What spectrum does $G(X)^{\mathbb{Z}/2}$ correspond to? Is there a more familiar name for it? <strong>Edit:</strong> I seem to be getting $E \vee (E \wedge E)^{h\mathbb{Z}/2}$, but without much confidence.</p> <p>[Leftover part of the question: If so, by my question <a href="http://mathoverflow.net/questions/3154/1-categorical-description-of-equivariant-homotopy-theory" rel="nofollow">here</a> I can think of the resulting object as a functor from the opposite of the orbit category of $\mathbb{Z}/2$ to spectra. Unpacking this amounts to giving some spectrum $F$ together with a map $F \to (E \wedge E)^{h\mathbb{Z}/2}$. What is $F$?]</p> http://mathoverflow.net/questions/7762/are-bicategories-of-lax-functors-also-bicategories-of-of-pseudofunctors Are bicategories of lax functors also bicategories of of pseudofunctors? Reid Barton 2009-12-04T07:17:58Z 2012-08-15T13:19:01Z <p>Let A be a bicategory. Can I construct in some "natural" way a bicategory L(A) such that for every bicategory B, the bicategory of lax functors Lax(A, B) is (bi)equivalent to the bicategory of pseudofunctors Pseudo(L(A), B)? (Choose the morphisms in these functor categories as you wish. Ideally the same construction L would work for any choice.)</p> <p>For example, when A = &bull;, I believe we may take L(A) = B&Delta;<sub>+</sub>, the delooping of the augmented simplex category with monoidal structure given by ordinal sum. In general I imagine L(A) as being formed as something like the free category on the objects, 1-morphisms and 2-morphisms of A--denote the generator corresponding to a 1-morphism f of A by [f]--as well as 2-morphisms id &rarr; [id] and [f] o [g] &rarr; [f o g] for every composable f and g in A.</p> <p>(By Chris's question <a href="http://mathoverflow.net/questions/7666/lax-functors-and-equivalence-of-bicategories" rel="nofollow">here</a>, L(A) cannot be an equivalence invariant of A.)</p> http://mathoverflow.net/questions/9951/limits-in-category-theory-and-analysis/9962#9962 Answer by Reid Barton for Limits in category theory and analysis Reid Barton 2009-12-28T16:25:06Z 2012-02-23T22:51:15Z <p>In very special cases, the notions coincide. Let $R$ be the category (poset) whose objects are the real numbers and in which $Hom(x, y)$ has a single element if $x \leq y$ and is empty otherwise. Then for a nonincreasing sequence of real numbers, its limit in the classical sense (if not $-\infty$) is also its limit in the categorical sense (if it exists).</p> http://mathoverflow.net/questions/11178/homotopy-pushouts-via-model-structure-in-top/11180#11180 Answer by Reid Barton for Homotopy Pushouts via Model Structure in Top Reid Barton 2010-01-08T20:25:40Z 2012-02-16T21:53:29Z <p>Question 1: The model category $\mathcal{C}$ should be <em>left proper</em>, i.e. the pushout of a weak equivalence along a cofibration is again a weak equivalence. (Dually, there is a notion of right proper.) Top is left proper, as is any model category in which every object is cofibrant, such as SSet. There is some information on this notion of properness <a href="http://ncatlab.org/nlab/show/proper+model+category" rel="nofollow">at the nlab</a>, and I think it's also discussed more thoroughly in Hirschhorn's book <em>Model Categories and their Localizations</em> (and probably many other places).</p> <p>Question 2: Yes. People often say that a square in a model category is a homotopy pushout square if the induced map from the (strict) pushout of a cofibrant replacement (meaning cofibrant objects and maps) of the "initial" three objects to the last object is a weak equivalence, and that is the case here.</p> http://mathoverflow.net/questions/10246/model-category-structure-on-set-without-axiom-of-choice Model category structure on Set without axiom of choice Reid Barton 2009-12-31T07:07:40Z 2012-02-16T01:14:49Z <p>There is a model category structure on Set in which the cofibrations are the monomorphisms, the fibrations are maps which are either epimorphisms or have empty domain, and the weak equivalences are the maps $f : X \rightarrow Y$ such that $X$ and $Y$ are both empty or both nonempty.</p> <p>In order for the lifting axioms to hold we need the axiom of choice. Suppose we want to avoid the axiom of choice. One option seems to be to replace "epimorphism" with "map which has a section" everywhere. Can we instead leave the definition of fibration unchanged and change the definition of cofibration?</p> <p>Note that if $A$ is a cofibrant set in this hypothetical model structure then any surjection $X \rightarrow A$ has a section. So it would be necessary that every set admits a surjection from a set $A$ of this type, which seems rather implausible to me. Perhaps the notion of model category needs to be modified in a setting without the axiom of choice.</p> <p>(Apologies if this question turns out to be meaningless or trivial; I have not thought about it much nor do I often try to avoid using the axiom of choice.)</p> <hr> <p>Let me explain the motivation behind this question. I am trying to get a better picture of what category theory without the axiom of choice looks like.</p> <p>My rough understanding is that in the absence of the axiom of choice we should use <a href="http://ncatlab.org/nlab/show/anafunctor" rel="nofollow">anafunctors</a> in place of functors in some if not all contexts. An anafunctor from $C$ to $D$ is a span of functors $C \leftarrow E \rightarrow D$ in which the left leg $E \leftarrow C$ is a surjection on objects and fully faithful. The point is that even under these conditions $E \rightarrow C$ may not have a section. For instance, suppose $C$ is a category with all binary products. There may not be a product functor $- \times -: C \times C \rightarrow C$, because we cannot simply choose a distinguished product of each pair of objects. However, if we define $Prod(C)$ to be the category whose objects are diagrams of the shape $\bullet \leftarrow \bullet \rightarrow \bullet$ in $C$ which express the center object as the product of the outer two, there is a forgetful functor $Prod(C) \rightarrow C \times C$ remembering only the outer objects, and it is surjective on objects (because $C$ has binary products) and fully faithful. Furthermore there is another forgetful functor $Prod(C) \rightarrow C$ which remembers only the center object. Together these define a product anafunctor $C \times C \leftarrow Prod(C) \rightarrow C$.</p> <p>"Classically" (:= under $AC$) the following paragraph holds: There is a model category structure on $Cat$ in which the cofibrations are functors which are injective on objects, fibrations are the functors with the right lifting property w.r.t. the inclusion of an object into the contractible groupoid on two objects, and weak equivalences are equivalences of categories. In particular the acyclic fibrations are the functors which are surjective on objects and fully faithful, exactly the functors we allow as left legs of anafunctors. Since every category is fibrant in this model structure, we can view an anafunctor from $C$ to $D$ as a representative of an element of $RHom(C, D)$, i.e., a functor from a cofibrant replacement for $C$ to $D$. Of course, every category is cofibrant too so for our cofibrant "replacement" we can just take $C$, and we learn that anafunctors from $C$ to $D$ are the same as functors when we consider both up to natural isomorphism (homotopy).</p> <p>I would like to understand anafunctors as a kind of $RHom$ also without the axiom of choice. But I cannot use the same definition of the model category structure, because the lifting axioms require $AC$. I would like to keep the same acyclic fibrations, since they appear in the definition of anafunctor, and I would like every object to be fibrant. I cannot really imagine what a cofibrant replacement could look like in this model structure, but then I am not accustomed to working without $AC$.</p> <p>My original question is related to this one via the functor which assigns to a set $S$ the codiscrete category on $S$, and it seems to contain the same kinds of difficulties.</p> http://mathoverflow.net/questions/47533/uncountable-preimage-of-every-point/47537#47537 Answer by Reid Barton for Uncountable preimage of every point Reid Barton 2010-11-27T20:43:03Z 2010-11-27T20:43:03Z <p>No. For example, let $g : [0, 1] \to [0, 1] \times [0, 1]$ be a continuous surjective map (space-filling curve) and let $p : [0, 1] \times [0, 1] \to [0, 1]$ be the projection onto the first coordinate. Then $p \circ g$ is continuous and the preimage of every point is uncountable.</p> http://mathoverflow.net/questions/3114/whats-an-example-of-an-adjunction-up-to-adjunction What's an example of an "adjunction up to adjunction"? Reid Barton 2009-10-28T18:58:18Z 2010-11-03T20:19:16Z <p>(There are several dialects of 2-categorical language. Mine is the one where everything is weak by default. You might need to insert the word "weak" or "pseudo" (or even "strong", if your default is lax) here and there to translate this to your own dialect.)</p> <p>The usual notion of an adjunction between two 2-categories C and D is a pair of 2-functors F : C &rarr; D and G : D &rarr; C together with a suitably natural equivalence between the categories Hom<sub>D</sub>(FX, Y) and Hom<sub>C</sub>(X, GY). One could ask for something weaker&mdash;rather than an equivalence, just a natural functor &phi;<sub>X,Y</sub> : Hom<sub>D</sub>(FX, Y) &rarr; Hom<sub>C</sub>(X, GY) which itself has a right adjoint. (We could instead ask for &phi;<sub>X,Y</sub> to have a left adjoint; this gives a different notion for any particular C and D, but we can interchange the two notions by reversing all the 2-morphisms in C and D, so we'll just pick this notion arbitrarily.) This is called a "lax 2-adjunction" at the <a href="http://ncatlab.org/nlab/show/lax+2-adjunction" rel="nofollow">nlab</a>.</p> <p>A boring example: If D = &bull; is the final 2-category, then an adjunction C &rarr; D is a final object of C, while an adjunction up to adjunction is merely an object Z of C such that Hom<sub>C</sub>(X, Z) has an initial object for every X and these initial objects are preserved by precomposition by f : X' &rarr; X.</p> <p>Does anyone know of a more interesting, natural example?</p> http://mathoverflow.net/questions/636/quotient-of-a-category-by-a-free-group-action Quotient of a category by a free group action Reid Barton 2009-10-15T19:19:13Z 2010-05-02T20:06:03Z <p>Let Cat denote the 1-category of small categories. The functor Mor : Cat -> Set which assigns to a category its set of morphisms (aka Hom([&bull; -> &bull;], -)) does not commute with most colimits. Does it commute with quotients by free group actions? In other words, if C is a small category and G is a group acting on C such that the action of G on the objects of C is free, does Mor(C/G) = (Mor C)/G?</p> http://mathoverflow.net/questions/22941/is-an-infinite-compositions-of-arrows-meaningful/22947#22947 Answer by Reid Barton for Is an "infinite compositions of arrows" meaningful? Reid Barton 2010-04-29T06:08:34Z 2010-04-29T06:08:34Z <p>This is probably not what the OP is looking for, but there is a notion of "infinite composition of arrows" which often appears for example in categorical homotopy theory:</p> <p>If $f_0 : X_0 \to X_1$, $f_1 : X_1 \to X_2$, $\ldots$ are morphisms in a category $C$ and the colimit of the diagram $X_0 \to X_1 \to X_2 \to \cdots$ exists (call it $X$) then $X$ is equipped in particular with a canonical map $X_0 \to X$ which is called the <em>transfinite composition</em> of the maps $f_i$. Of course, technically this morphism of $C$ is specified only up to canonical isomorphism because $X$ may be replaced by a (uniquely isomorphic) different colimit of the $X_i$. More generally given an ordinal $\alpha$ which we may view as a category (poset) and a colimit-preserving functor $X_\cdot : \alpha \to C$ (so that for each limit ordinal $\beta \in \alpha$, we have <code>$X_\beta = \operatorname{colim}_{\gamma &lt; \beta} X_\gamma$</code>), we may form the transfinite composition <code>$X_0 \to \operatorname{colim}_\alpha X_\alpha$</code>. One is often concerned with questions such as whether a given class of maps is closed under transfinite compositions (for example, the class of cofibrations in a model category has this property).</p> http://mathoverflow.net/questions/22624/example-of-a-good-zero-knowledge-proof/22631#22631 Answer by Reid Barton for Example of a good Zero Knowledge Proof. Reid Barton 2010-04-26T18:38:36Z 2010-04-26T18:38:36Z <p>Demonstrating an attack on a cryptosystem is very similar to the colored balls example in Ryan's answer. Suppose Alice and Bob have a means of communicating messages and Eve wants to prove that it is insecure, without revealing the method used to exploit the system. Alice and Eve can simply agree that Alice will send a sequence of random messages to Bob. If Eve can tell Alice the contents of the messages, then with high probability Eve must have an attack on the cryptosystem.</p> http://mathoverflow.net/questions/22359/why-havent-certain-well-researched-classes-of-mathematical-object-been-framed-by/22380#22380 Answer by Reid Barton for why haven't certain well-researched classes of mathematical object been framed by category theory? Reid Barton 2010-04-23T18:04:37Z 2010-04-23T18:04:37Z <p>Fundamentally I agree with Mike Shulman's comment and I do not really want to claim the following fancy language is at all necessary to answer this question, but you may (or may not) find it illuminating.</p> <p>From the standpoint of higher category theory, categories (i.e., 1-categories) are just one level among many in a family of mathematical structures. Typically a mathematical object will "naturally" exist as an n-category for some particular n. For example, <strong>Set</strong> is naturally a 1-category, while <strong>Cat</strong> is naturally a 2-category. Your examples <strong>Series</strong> and so on seem to just be 0-categories, i.e., sets, since as Pete explained in his answer, there is no obvious natural notion of morphism between infinite series. Asking why <strong>Series</strong> is not a 1-category is like asking why <strong>Set</strong> is not a 2-category; these are just not the natural categorical levels that these objects live at.</p> http://mathoverflow.net/questions/21591/when-does-adding-inverses-of-morphisms-preserve-commutativity-of-a-diagram/21596#21596 Answer by Reid Barton for When does adding inverses of morphisms preserve commutativity of a diagram? Reid Barton 2010-04-16T18:01:43Z 2010-04-16T18:26:58Z <p>There is an easy conceptual proof using the fact that the category obtained by formally inverting all the arrows in a category C is equivalent to the fundamental groupoid of the nerve NC of C, and that the nerve of a category with a final object is contractible. Without the assumption of a final object your assertion is false in general, e.g., reverse the arrows from c in your example.</p> <p>But it should also be easy to prove by induction: for any zigzag of arrows between a and b, the corresponding map in the category with all arrows inverted, when composed with the map from b to the original terminal object, is equal to the map from a to the original terminal object (this is by induction); and so any two maps from a to b in the category with all arrows inverted are equal. In symbols: let me write $t_x$ for the unique morphism in C from $x$ to the terminal object and $[f]$ for the image of $f$ in the category with all arrows inverted. Suppose $[f_1]^{\pm 1} \cdots [f_n]^{\pm 1}$ is a typical map in the category with all arrows inverted with domain $a$ and target $b$. Then the inductive claim is that $[t_b] [f_1]^{\pm 1} \cdots [f_n]^{\pm 1} = [t_a]$, and so $[f_1]^{\pm 1} \cdots [f_n]^{\pm 1} = [t_b]^{-1} [t_a]$.</p> http://mathoverflow.net/questions/21373/what-functor-is-adjoint-to-the-tensor-product-of-2-vector-spaces/21390#21390 Answer by Reid Barton for What functor is adjoint to the tensor product of 2-vector spaces? Reid Barton 2010-04-14T21:20:30Z 2010-04-14T21:20:30Z <p>I'll denote your category of 2-vector spaces by 2Vect. By your preliminary remarks, 2Vect is actually the category of Vect-valued presheaves on &Delta;<sub>&le;1</sub> where &Delta;<sub>&le;1</sub> denotes the full subcategory of &Delta; on the objects [0] and [1]. Therefore, colimits in 2Vect are computed objectwise under this identification. So the functor &ndash; &otimes; X certainly has a right adjoint <strong>Hom</strong>(X, &ndash;) (by the adjoint functor theorem for locally presentable categories). This adjunction also respects the Vect enrichment.</p> <p>To compute this adjoint, we can use the "Vect-enriched Yoneda lemma": writing &Delta;<sup>i</sup> for the 2-vector space (&Delta;<sup>i</sup>)<sub>j</sub> = Hom<sub>&Delta;<sub>&le;1</sub></sub>([j], [i]) &bull; &#8477;, we have hom<sub>2Vect</sub>(&Delta;<sup>i</sup>, X) = X<sub>i</sub> as vector spaces, where hom denotes the Vect-enriched Hom. So</p> <p><strong>Hom</strong>(X, Y)<sub>0</sub> = hom(&Delta;<sup>0</sup>, <strong>Hom</strong>(X, Y)) = hom(&Delta;<sup>0</sup> &otimes; X, Y) = hom(X, Y)</p> <p>since &Delta;<sup>0</sup> happens to be the unit for this &otimes;, but</p> <p><strong>Hom</strong>(X, Y)<sub>1</sub> = hom(&Delta;<sup>1</sup> &otimes; X, Y)</p> <p>will have a more complicated formula which you'll have to work out (my guess is it will look similarly complicated to your expression for the other tensor product).</p> http://mathoverflow.net/questions/21358/does-every-cocontinuous-functor-between-categories-of-presheaves-on-small-categor/21363#21363 Answer by Reid Barton for Does every cocontinuous functor between categories of presheaves on small categories have a right adjoint? Reid Barton 2010-04-14T17:39:39Z 2010-04-14T17:51:44Z <p>The part "<em>F</em> will always have a right adjoint when <em>C</em>, <em>E</em> are small" is definitely right. Using some mildly overkill machinery: in this case <em>C&#x0302;</em> and <em>E&#x0302;</em> are locally presentable categories, and the result is then the <strong>adjoint functor theorem for locally presentable categories</strong>:</p> <p>Theorem: Let C and D be locally presentable categories and F : C &rarr; D a functor. Then</p> <ol> <li><p>F has a right adjoint iff F preserves small colimits.</p></li> <li><p>F has a left adjoint iff F is accessible (preserves &kappa;-filtered colimits for some &kappa;) and preserves small limits.</p></li> </ol> <p>(Reference: <em>Higher Topos Theory</em> Corollary 5.5.2.9 for the (&infin;,1)-categorical version)</p> http://mathoverflow.net/questions/14402/semiadditivity-and-dualizability-of-2 Semiadditivity and dualizability of 2 Reid Barton 2010-02-06T17:25:25Z 2010-04-14T05:15:56Z <p><strong>Short version</strong>: Let (C, &otimes;, 1) be a locally presentable closed symmetric monoidal category with a zero object, and write 2 = 1 ∐ 1. Suppose the object 2 has a dual. Does it follow that C is a category with biproducts?</p> <p><strong>Longer version, with motivation</strong>: Let (C, &otimes;, 1) be a locally presentable closed symmetric monoidal category. If you don't know what "locally presentable" means, you can replace these conditions with "complete and cocomplete symmetric monoidal category in which &otimes; commutes with colimits in each variable". Familiar examples include (Set, &times;, &bull;), (Set<sub>*</sub>, &and;, S<sup>0</sup>) (the category of pointed sets with the smash product), and (Ab, &otimes;, &#8484;). Any such category C has a unique "unit" functor F<sub>C</sub> : Set &rarr; C preserving colimits and the unit object: the set S is sent to the coproduct in C of S copies of 1. For a nonnegative integer n, let me also write n for the image under this functor of the n-element set. For instance, 0 represents the initial object of C.</p> <p>A <em>dual</em> for an object X of C is another object X<sup>*</sup> together with maps 1 &rarr; X &otimes; X<sup>*</sup> and X<sup>*</sup> &otimes; X &rarr; 1 which satisfy the triangular identities; see <a href="http://en.wikipedia.org/wiki/Dual_object" rel="nofollow">wikipedia</a> for more details. The data of X<sup>*</sup> together with these maps is unique up to unique isomorphism if it exists, so it makes sense to ask whether an object has a dual or not.</p> <p>I'm interested in the relationship between which objects in the image of F<sub>C</sub> have duals and the existence of more familiar structures on C. In our examples,</p> <ul> <li>C = Set: Only 1 has a dual.</li> <li>C = Set<sub>*</sub>: Only 1 and 0 = &bull; have duals.</li> <li>C = Ab: n has a dual for any nonnegative integer n.</li> </ul> <p>It's easy to show that 1 is always its own dual, and slightly less trivially, that 0 has a dual iff 0 is also a final object, i.e., C has a zero object, or equivalently C is enriched in Set<sub>*</sub>. Moreover, if C is <a href="http://mathoverflow.net/questions/12410/terminology-is-there-a-name-for-a-category-with-biproducts/12419#12419" rel="nofollow">semiadditive</A>, i.e., enriched in commutative monoids, or equivalently has biproducts, then n has a dual (in fact, n is its own dual) for every nonnegative integer n. Conversely, if 0 has a dual, so that C is pointed, and 2 also has a dual, then there is a canonical map 2 = 1 ∐ 1 &rarr; 1 &times; 1 = 2<sup>*</sup>. My question, then, is: is this map is always an isomorphism? Or, could it happen that 2<sup>*</sup> exists but is not isomorphic to 2 via this map?</p> http://mathoverflow.net/questions/20604/are-rings-really-more-fundamental-objects-than-semi-rings/20654#20654 Answer by Reid Barton for Are rings really more fundamental objects than semi-rings? Reid Barton 2010-04-07T18:32:06Z 2010-04-07T18:32:06Z <p>Of course the real question is whether abelian groups are really more fundamental objects than commutative monoids. In a sense, the answer is obviously no: the definition of commutative monoid is simpler and admits alternative descriptions such as the one I give <a href="http://mathoverflow.net/questions/2551/why-do-groups-and-abelian-groups-feel-so-different/9302#9302" rel="nofollow">here</a>. The latter description can be adapted to other settings, such as to the 2-category of locally presentable categories, which shares many formal properties with the category of commutative monoids (such as being closed symmetric monoidal, having a zero object, having biproducts). As such I would claim that any locally presentable closed symmetric monoidal category is itself a categorified version of a semiring, not in the sense you describe, but in that it is an algebra object in a closed symmetric monoidal category, so we may talk of modules over it, etc.</p> <p>However, it is undeniable that there is a large qualitative difference between the theories of abelian groups and commutative monoids. Observe that an abelian group is just a commutative monoid which is a module over $\mathbb{Z}$ (more precisely a commutative monoid has either a unique structure of $\mathbb{Z}$-module, if it has additive inverses, and no structure of $\mathbb{Z}$-module otherwise). The situation is analogous to the (smaller) difference between abelian groups and $\mathbb{Q}$-vector spaces. I do not know of a characterization of $\mathbb{Z}$ as a commutative monoid that can be transported to other settings. It seems that there is something deep about the fact that $\mathbb{Z}$-modules are so much nicer than commutative monoids, which often is taken for granted.</p> http://mathoverflow.net/questions/20580/coend-computation-continued/20586#20586 Answer by Reid Barton for Coend computation continued Reid Barton 2010-04-07T02:51:14Z 2010-04-07T03:52:57Z <p>The step $\int_{a \in A} \mathrm{Set}(\mathrm{hom}_A(a, a), S) = \mathrm{Nat}(\mathrm{hom}_A(-,-),S)$ does not really make sense, because $a \mapsto \mathrm{hom}_A(a,a)$ is not a functor. And $\int^{a \in A} \mathrm{hom}_A(a,a)$ is not equal to $\mathrm{Ob}(A)$ in general. For instance, let $G$ be a group and let $A = BG$ be the groupoid with a single object with automorphism group $G$. Then $\int^{a \in A} \mathrm{hom}_A(a,a)$ can be identified with the <strike>abelianization</strike> set of conjugacy classes of $G$. In general, $\int^{a \in A} \mathrm{hom}_A(a,a)$ is the "Hochschild homology" of the category $A$.</p> <hr> <p>Edit: I was originally thinking of the case of $G$ abelian, and generalized wrongly to the nonabelian case. As atonement, let me write out the computation directly from the definition of coend:</p> <p><code>$$\int^{a \in A} \mathrm{hom}_A(a, a) = \operatorname{colim} \left[ \coprod_{f:a \to b} \mathrm{hom}_A(b,a)\rightrightarrows \coprod_{a\in A} \mathrm{hom}_A(a,a)\right] = \operatorname{colim} [G \times G \rightrightarrows G]$$</code></p> <p>where the two maps send $(g, h)$ to $gh$ and $hg$ respectively. So the coend is the quotient of $G$ in the category of sets by the relation $gh = hg$, or $g = hgh^{-1}$, i.e., it is the set of conjugacy classes of elements of $G$.</p> http://mathoverflow.net/questions/20558/what-is-the-minimum-n-for-which-there-exist-n-points-in-the-plane-that-cannot-be/20566#20566 Answer by Reid Barton for What is the minimum N for which there exist N points in the plane that cannot be covered by any number of non-overlapping closed unit discs? Reid Barton 2010-04-06T23:52:11Z 2010-04-06T23:52:11Z <p>The trick for N = 10 (which I heard from a friend earlier today) is to check that the density of the triangular packing of unit diameter circles is high enough that some translate of this packing must cover all the points.</p> http://mathoverflow.net/questions/20332/what-is-a-reference-for-an-explicit-logic-based-statement-of-duality-in-categor/20349#20349 Answer by Reid Barton for What is a reference for an explicit, logic-based, statement of duality in category theory (in ''complicated'' situations)? And what are the prerequisites for a beginner in logic? Reid Barton 2010-04-05T00:57:15Z 2010-04-05T00:57:15Z <p>I would go even farther than the comments above, at least in the specific case you mention about computing (co)limits objectwise in a functor category. Once you know the statement for limits, deducing the statement for colimits is not even a syntactic transformation to the proof, and needs no arguments from formal logic at all. Instead, to prove that colimits are computed objectwise in the functor category [I, C], simply <em>use the fact</em> that limits are computed objectwise in the functor category [I<sup>op</sup>, C<sup>op</sup>] = [I, C]<sup>op</sup>, and the fact that colimits in a category are the same as limits in the opposite category.</p> <p>I think this is the common case in this kind of argument, but perhaps someone can come up with an example where the argument really needs to be repeated in the dual situation.</p> http://mathoverflow.net/questions/19786/when-do-the-reedy-and-injective-model-category-structures-agree When do the Reedy and injective model category structures agree? Reid Barton 2010-03-30T02:25:09Z 2010-03-30T21:04:01Z <p>Let $R$ be a <a href="http://ncatlab.org/nlab/show/Reedy+category" rel="nofollow">Reedy category</a> and consider the category $\mathcal{P}(R) = \mathbf{sSet}^{R^{\mathrm{op}}}$ of simplicial presheaves on $R$. When are the <a href="http://ncatlab.org/nlab/show/Reedy+model+structure" rel="nofollow">Reedy</a> and <a href="http://ncatlab.org/nlab/show/global+model+structure+on+functors" rel="nofollow">injective</a> model structures on $\mathcal{P}(R)$ the same? Are there useful sufficient conditions on $R$?</p> <p>I know this is the case when all the morphisms of $R$ raise degree (I hope that's the right direction), and also when $R = \Delta$. I am specifically interested about the case where $R$ is <a href="http://ncatlab.org/nlab/show/Theta+category" rel="nofollow">$\Theta_n$</a>, or a product of several copies of $\Theta_n$.</p> http://mathoverflow.net/questions/19732/ackermann-related-function/19742#19742 Answer by Reid Barton for Ackermann-related function Reid Barton 2010-03-29T17:26:59Z 2010-03-29T17:26:59Z <p>Here's a sketch of an argument which I expect could be made into a proof.</p> <p>The key fact is that the Ackermann function fails to be primitive recursive only because it grows so quickly. More formally:</p> <p><strong>Claim</strong>. There exists a Turing machine T and a primitive recursive function f(a, b, c) (which is an increasing function of c) such that on input (a, b), T computes A(a, b) in at most f(a, b, A(a, b)) steps.</p> <p>"Proof". Starting with the expression "A(a, b)", repeatedly expand terms of the form A(x, y) with the recursive definition, but do not simplify any of the resulting additions. The length of the string increases at every step, if we agree that the symbol "+" is "longer" than "A". The resulting string is a formal sum of positive integers, so its length is bounded by (a multiple of) A(a, b); hence the number of steps is also bounded above in terms of A(a, b), and we may perform each step in time polynomial in A(a, b).</p> <p>Now, we can simulate a given Turing machine for a fixed number of steps using a primitive recursive function. We may therefore compute the graph of the Ackermann function with a primitive recursive function as follows: Given a, b, c,</p> <ul> <li>Compute f(a, b, c).</li> <li>Simulate the Turing machine T on input (a, b) for f(a, b, c) steps.</li> <li>If T has halted, then return whether c is equal to the output of T. If T has not halted, then c &lt; A(a, b) so return false.</li> </ul> http://mathoverflow.net/questions/19644/what-is-the-definition-of-canonical/19645#19645 Answer by Reid Barton for What is the definition of "canonical" ? Reid Barton 2010-03-28T18:36:26Z 2010-03-28T20:07:32Z <ol> <li><p>Not a <em>definition</em>, exactly; I would say the situation is similar to that of <a href="http://mathoverflow.net/questions/19405/definition-of-forgetful-functor" rel="nofollow">forgetful functor</a>. If I say there is a canonical isomorphism between X and Y, then what I mean is that if asked, pretty much everyone would choose the same isomorphism. A canonical isomorphism is very often a natural isomorphism in the sense of category theory, but the converse need not hold. A canonical isomorphism does not need to be the unique isomorphism between X and Y, though sometimes it is when X and Y are considered as equipped with some additional structure.</p></li> <li><p>"There is a canonical isomorphism between the set of elements of a ring R and the set of ring maps $\mathbb{Z}[x] \to R$." Obviously, I mean for $r \in R$ to correspond to the ring map sending $x$ to $r$, although I could just as well send $x$ to $-r$.</p></li> <li><p>"There is a canonical isomorphism between a finite-dimensional vector space V and its dual." No explanation needed, I suppose.</p></li> </ol> <p>Maybe more interesting would be an example where the word "canonical" is arguably correct or incorrect; I can't think of one off-hand.</p> <hr> <p>Addendum, after reading some of the other answers: I would emphasize that for me there is a difference between "natural" in the formal category-theoretic sense and "canonical". For one thing there is a linguistic distinction: if I am considering an isomorphism F between X and Y then "Theorem: F is a natural isomorphism" is perfectly acceptable but "Theorem: F is a canonical isomorphism" is very strange to me. There should be only one canonical isomorphism between two things, though what that isomorphism is could depend on context, e.g., "the canonical isomorphism $A \otimes B \to B \otimes A$" where $A$ and $B$ are graded abelian groups might mean different things to an algebraic geometer and an algebraic topologist.</p> <hr> <p>Finally, this is hardly a definition, more of a rule of thumb: there is a canonical isomorphism between X and Y if and only if you would feel comfortable writing "X = Y".</p> http://mathoverflow.net/questions/19116/colimits-in-the-category-of-smooth-manifolds/19123#19123 Answer by Reid Barton for Colimits in the category of smooth manifolds Reid Barton 2010-03-23T16:46:14Z 2010-03-23T16:46:14Z <p>I'll show that the pushout that glues two copies of $\mathbb{R}$ at the origin does not exist in <strong>Man</strong>. Suppose for the sake of contradiction that it did; call the resulting manifold $M$, and the common image of the origins $x \in M$. The real line $\mathbb{R}$ is a ring object in <strong>Man</strong>, and it represents the functor $X \mapsto C^\infty(X)$. So, we learn that as a ring, $C^\infty(M)$ consists of pairs of functions on the real line with the same value at the origin. Similarly we can identify the ideal $I$ in $C^\infty(M)$ of functions that vanish at $x$ with pairs of functions on the real line that both vanish at the origin. Now we may compute that $\dim_{\mathbb{R}} I^n/I^{n+1} = 2$ for all $n \ge 1$, which cannot happen for a point $x$ of a smooth manifold $M$.</p> http://mathoverflow.net/questions/19084/equivalences-in-model-categories/19089#19089 Answer by Reid Barton for Equivalences in Model Categories Reid Barton 2010-03-23T04:47:28Z 2010-03-23T04:47:28Z <p>Yes. The isomorphism in $\mathrm{Ho}(\mathcal{M})$ is represented by a morphism in $\mathcal{M}$ from a cofibrant replacement for $A$ to a fibrant replacement for $B$. The "converse to the Whitehead lemma" states that a map in a model category is a weak equivalence iff its image in the homotopy category is an isomorphism. Combining this with the definition of (co)fibrant replacement, we see that $A$ and $B$ are connected by a 3-step zig-zag of weak equivalences.</p> http://mathoverflow.net/questions/18698/homotopy-type-of-set-of-self-homotopy-equivalences-of-a-surface/18702#18702 Answer by Reid Barton for Homotopy type of set of self homotopy-equivalences of a surface Reid Barton 2010-03-19T01:41:43Z 2010-03-19T01:41:43Z <p>The connected oriented surface $\Sigma_g$ of genus $g \ge 1$ is a $K(G_g, 1)$ where $G_g$ has a well-known presentation. For a general group $G$, the mapping space $Map(K(G, 1), K(G, 1))$ has homotopy type $$\coprod_{f:G \to G} B(Z(f(G)))$$ where $f$ ranges over group endomorphisms of $G$ and $Z(f(G))$ denotes the centralizer of the image of $f$. In your case, you are interested in the case where $f$ is surjective, so the question reduces to whether $G_g$ has trivial center for $g \ge 2$, which I assume it does.</p> http://mathoverflow.net/questions/18513/k-theory-as-a-generalized-cohomology-theory/18515#18515 Answer by Reid Barton for K-theory as a generalized cohomology theory Reid Barton 2010-03-17T18:59:56Z 2010-03-17T18:59:56Z <p>1 is doubly wrong. First, you need to distinguished generalized cohomology theories and <em>reduced</em> generalized cohomology theories. If you want to work with the latter, you should replace "a point" in 1 by "$S^0$", and then the corrected version of 3 no longer holds. But even this new version 1' is false; a generalized cohomology theory is not determined by its coefficients, unless they are concentrated in a single degree (example: complex K-theory vs. integer cohomology made even periodic).</p> http://mathoverflow.net/questions/18392/signed-minimum/18448#18448 Answer by Reid Barton for Signed minimum? Reid Barton 2010-03-17T00:38:06Z 2010-03-17T00:38:06Z <p>$smin(x,y)$ can also be described as the number of smallest absolute value in the closed interval between $x$ and $y$.</p> <p>When $x \le y$ are integers, this is also the value of the game $\{x-1 \mid y+1\}$ in the sense of combinatorial game theory (Conway, <em>On Numbers And Games</em>).</p> http://mathoverflow.net/questions/17523/are-there-any-important-mathematical-concepts-without-discrete-analog/17779#17779 Answer by Reid Barton for Are there any important mathematical concepts without discrete analog? Reid Barton 2010-03-10T23:48:38Z 2010-03-10T23:48:38Z <p>Is there a discrete analogue of the notion of discreteness?</p> http://mathoverflow.net/questions/16930/terminology-lax-vs-oplax-colimits Terminology: lax vs. oplax colimits Reid Barton 2010-03-02T22:12:55Z 2010-03-03T06:00:11Z <p>I would like to know the standard usage of "lax colimit" and "oplax colimit" in the 2-categorical literature. The nLab does not give an explicit definition of "lax colimit", as far as I can see, and I don't know what the most reliable source is. I think I have seen at least one paper using each convention, but I have not encountered the notion often enough to have a good sense of whether this one of those places where the terminology is not really standardized, or there is general agreement with a few exceptions.</p> <p>Concretely, given a diagram X : I &rarr; C in a 2-category C (for my purposes indexed by a 1-category I), suppose I have a cone (Y, {g<sub>i</sub>}<sub>i&isin;Ob I</sub>, {&alpha;<sub>f</sub>}<sub>f&isin;Mor I</sub>), with g<sub>i</sub> : X<sub>i</sub> &rarr; Y, and for f : i &rarr; j in I, &alpha;<sub>f</sub> : g<sub>j</sub>f &rarr; g<sub>i</sub> (such that various diagrams commute). Is this a lax colimit cone or a oplax colimit cone?</p> http://mathoverflow.net/questions/30049/profinite-spaces-coming-from-profinite-groups Comment by Reid Barton Reid Barton 2010-06-30T14:21:07Z 2010-06-30T14:21:07Z The answer to the first question is no: at a minimum, the automorphism group of the space must act transitively, ruling out a space like {1, 1/2, 1/3, ..., 0}. http://mathoverflow.net/questions/30032/equivalence-versus-isomorphism-of-categories Comment by Reid Barton Reid Barton 2010-06-30T11:00:52Z 2010-06-30T11:00:52Z This seems to be more or less a duplicate of <a href="http://mathoverflow.net/questions/11674/can-skeleta-simplify-category-theory" rel="nofollow" title="can skeleta simplify category theory">mathoverflow.net/questions/11674/&hellip;</a> http://mathoverflow.net/questions/24936/why-the-choice-of-the-simplex-for-defining-homology Comment by Reid Barton Reid Barton 2010-05-18T03:31:52Z 2010-05-18T03:31:52Z This question is closely related: <a href="http://mathoverflow.net/questions/691/simplicial-objects" rel="nofollow" title="simplicial objects">mathoverflow.net/questions/691/simplicial-objects</a> http://mathoverflow.net/questions/24693/neutral-tic-tac-toe Comment by Reid Barton Reid Barton 2010-05-15T14:17:26Z 2010-05-15T14:17:26Z If I understand the terminology correctly, Kevin and Patrik have analysed the normal version of the game, where the objective is to form a line of Xs, whereas in the misere version, the first player to form a line of Xs <i>loses</i>. http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23640#23640 Comment by Reid Barton Reid Barton 2010-05-11T02:12:55Z 2010-05-11T02:12:55Z Your fifth example reminds me of an even more plausible false belief I once held: if $A \otimes A = 0$, then $A = 0$. http://mathoverflow.net/questions/23828/can-someone-construct-in-zfc-bas-with-csr-equal-to-the-following-relation Comment by Reid Barton Reid Barton 2010-05-08T13:18:10Z 2010-05-08T13:18:10Z @Fran&#231;ois: please go ahead and expand the abbreviations yourself if you are confident about what the asker means. http://mathoverflow.net/questions/22941/is-an-infinite-compositions-of-arrows-meaningful/22947#22947 Comment by Reid Barton Reid Barton 2010-04-29T06:49:51Z 2010-04-29T06:49:51Z Yes. If $\alpha = n$ is a finite ordinal, then for the colimit we may take the last object $X_{n-1}$ and the structural map of the colimit cone is the composition of the maps $X_0 \to X_1 \to \cdots \to X_{n-1}$. http://mathoverflow.net/questions/22777/how-to-find-a-closest-integer-point-to-intersection-of-two-lines/22779#22779 Comment by Reid Barton Reid Barton 2010-04-27T21:05:24Z 2010-04-27T21:05:24Z It seems that you didn't notice the phrase &quot;in the same quarter [as (x, y)]&quot; in the problem statement... http://mathoverflow.net/questions/22678/disjointlize-an-arbitrary-sequence-in-a-ring Comment by Reid Barton Reid Barton 2010-04-27T21:04:24Z 2010-04-27T21:04:24Z Right, I assumed that the Fs were intended to be elements of <b>R</b>, since otherwise <b>R</b> plays no role... http://mathoverflow.net/questions/22737/how-can-i-calculate-the-expected-ranking-of-a-competitor-from-the-probabilities-o Comment by Reid Barton Reid Barton 2010-04-27T19:18:24Z 2010-04-27T19:18:24Z @Qiaochu: of course, there are many other possible models that can be chosen (resulting in different answers) and the question of which is most appropriate is not a mathematical one. http://mathoverflow.net/questions/22659/does-the-forgetful-functor-hopf-algebrasalgebras-have-a-right-adjoint/22698#22698 Comment by Reid Barton Reid Barton 2010-04-27T15:06:44Z 2010-04-27T15:06:44Z Are Hopf algebras really the category of co-whatsits in whatevers? Can you fill in the details a little more? http://mathoverflow.net/questions/22686/quotient-of-a-category-by-a-group-action Comment by Reid Barton Reid Barton 2010-04-27T06:21:52Z 2010-04-27T06:21:52Z The appropriate 2-categorical construction is to replace $C$ by $C \times EG$ where $EG$ is the codiscrete category ($\mathrm{Hom}(x, y) = *$ for all $x$ and $y$) with object set $G$ and $G$ acting by left translation, and forming the quotient $(C \times EG)/G$ as described there. But, I guess your question is about the 1-categorical colimit? http://mathoverflow.net/questions/636/quotient-of-a-category-by-a-free-group-action/22682#22682 Comment by Reid Barton Reid Barton 2010-04-27T04:52:22Z 2010-04-27T04:52:22Z What if I take C = BH, where H is a group with an action of G? I think the quotient category will be B(H with the relations x = gx for all x in H imposed), whose set of morphisms will be different from the set of orbits for the action of G on H as a set. http://mathoverflow.net/questions/22678/disjointlize-an-arbitrary-sequence-in-a-ring Comment by Reid Barton Reid Barton 2010-04-27T04:44:13Z 2010-04-27T04:44:13Z @Mariano: It does? Don't I need to form some infinite unions, which might take me outside of <b>R</b>? http://mathoverflow.net/questions/22131/name-for-an-inequality-of-isoperimetric-type Comment by Reid Barton Reid Barton 2010-04-27T03:06:58Z 2010-04-27T03:06:58Z Hmm, I wonder what happened there...?