User steve lack - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:43:54Z http://mathoverflow.net/feeds/user/10862 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85323/topos-associated-to-a-category/85379#85379 Answer by Steve Lack for Topos associated to a category Steve Lack 2012-01-11T00:49:45Z 2012-01-11T00:49:45Z <p>This is described in the paper </p> <blockquote> <p>Bunge and Carboni, The symmetric topos, Journal of Pure and Applied Algebra 105:233-249, 1995.</p> </blockquote> <p>which describes the sense in which the construction you call Spec is analogous to the symmetric algebra construction. </p> <p>Bunge and Carboni give a biadjunction between the bicategory R, which is the opposite of the bicategory of Grothendieck toposes, and the bicategory A of locally presentable categories and cocontinuous functors (equivalently, left adjoints). </p> http://mathoverflow.net/questions/41305/classification-of-quasi-topoi/67433#67433 Answer by Steve Lack for Classification of Quasi-topoi Steve Lack 2011-06-10T12:25:19Z 2011-06-10T12:25:19Z <p>This is really an answer to the question raised in Mike's reformulation of the question, but is too long for a comment and may be of interest.</p> <p>Richard Garner and I have considered when a reflective subcategory of a presheaf category has the form considered in condition 2 of his answer. It turns out that preservation of finite products and monomorphisms is not enough: to see this, consider the reflection of directed graphs into preorders, which preserves finite products and monomorphisms but is not of this form. </p> <p>In fact for a full reflective subcategory <em>E</em> of a presheaf category [<em>C</em>^{op},Set], the following conditions are equivalent:</p> <ol> <li>there is a topology <em>j</em> and a larger topology <em>k</em> for which <em>E</em> consists of the objects which are sheaves for <em>j</em> and separated for <em>k</em></li> <li>the reflection preserves finite products and monomorphisms and is semi-left-exact</li> <li>the reflection preserves monomorphisms and has stable units</li> </ol> <p>Here the notions of semi-left-exactness and stable units come from </p> <blockquote> <p>Cassidy, Hebert, Kelly, Reflective subcategories, localizations, and factorization systems, J. Austral. Math. Soc. Ser. A, 38:287-329, 1985.</p> </blockquote> <p>Let <em>R</em> be the reflection and <em>r</em> the unit of the reflection. Semi-left-exactness says that <em>R</em> preserves each pullback of a component <em>rX:X->RX</em> of the unit along a map <em>A->RX</em> with <em>A</em> in the subcategory.</p> <p>Stable units says the same thing, but without the requirement that <em>A</em> be in the subcategory. This turns out to be equivalent to <em>R</em> preserving all pullbacks over an object of the subcategory. </p> http://mathoverflow.net/questions/64382/comparing-discrete-fibrations-and-their-duals/64600#64600 Answer by Steve Lack for Comparing discrete fibrations and their duals Steve Lack 2011-05-11T12:12:19Z 2011-05-11T12:12:19Z <p>The fact that codiscrete cofibrations and discrete fibrations are equivalent is a very special exactness property of the 2-category Cat. I can't, off the top of my head, think of any other interesting examples. It's not true, for example in the 2-category 2-Cat of 2-categories, 2-functors, and 2-natural transformations you refer to, or in most (all?) other 2-categories of the form V-Cat. Consider, for example, the case where V=Cat and A=B=1. Then CodCofib(1,1) is just (the underlying ordinary category of) Cat. On the other hand, Fib(1,1) is 2-Cat, but discreteness of a fibration says that the 2-category A has an underlying ordinary category which is discrete. </p> <p>If you want to move from 2-Cat to Bicat, you'd first have to decide how you want to make Bicat into a 2-category (or decide what you mean by internal discrete fibrations or codiscrete cofibrations in a tricategory). But I wouldn't hold out too much hope ....</p> http://mathoverflow.net/questions/58010/on-a-corollary-in-mitchells-book/58029#58029 Answer by Steve Lack for On a corollary in Mitchell's book Steve Lack 2011-03-10T02:25:54Z 2011-03-10T09:38:33Z <p>I don't remember what an exact category is in this context, but in any case this seems to be true in any pointed category. Name the arrows $i:A\to B$, $p:B\to C$, $j:B'\to B$, and $q:B\to B''$, and suppose that $pj$ is epi. If $fqi=0$ then by the universal property of $p:B\to C$ we have $fq=gp$ for a unique $g$. Now $gpj=fqj=f0=0$ and $pj$ is epi so $g=0$. Thus $fq=gp=0$, but $q$ is epi so $f=0$. This proves that $qi$ is epi.</p> <p><strong>Edit:</strong> the comments below point out that this only shows that if $fqi=0$ then $f=0$. So this argument, as is, would need the category to be <strong>Ab</strong>-enriched. If instead it is exact, in the sense specified in the comments, then to finish I should prove:</p> <p><strong>Lemma:</strong> Let $h$ be an arrow with the property that if $fh=0$ then $f=0$. Then $h$ is epi. </p> <p><strong>Proof:</strong> The assumptions tell us that the cokernel of $h$ is $0$. Factorize $h$ as $me$, with $e$ epi and $m$ mono. Since $e$ is epi, we have $cok(m)=cok(me)=0$. But $m$ is a mono, so is the kernel of its cokernel, in this case 0, so $m$ must be invertible. This now proves that $h$ is epi.</p> http://mathoverflow.net/questions/57099/why-do-filtered-colimits-commute-with-finite-limits Why do filtered colimits commute with finite limits? Steve Lack 2011-03-02T10:34:02Z 2011-03-03T09:42:38Z <p>It's not hard to show that this is true in the category Set, and proofs have been written down in many places. But all the ones I know are a bit fiddly.</p> <p><strong>Question 1:</strong> is there a soft proof of this fact?</p> <p>For example, here's a soft proof of the fact that filtered colimits in Set commute with binary products. If $J$ is a filtered category, and $R,S:J\to$ Set are functors, then </p> <p>$$colim_{j\in J} R(j)\times colim_{k\in J} S(k) \cong colim_{j\in J} colim_{k\in J} R(j)\times S(k)$$ $$\cong colim_{(j,k)\in J\times J} R(j)\times S(k) \cong colim_{j\in J} R(j)\times S(j)$$</p> <p>where the first isomorphism uses the fact that Set is cartesian closed, so that the functors $X\times-$ and $-\times X$ are cocontinuous; the second isomorphism is the "Fubini theorem"; and the third isomorphism follows from the fact that the diagonal functor $\Delta:J\to J\times J$ is final. </p> <p>Is there some way to extend this to deal with equalizers and/or pullbacks? (The case of the terminal object is easy.)</p> <p>For the sort of person who'd rather just prove the fact directly (which after all is not that hard), it's worth pointing out that this proof works not just in Set but for any cartesian closed category with filtered colimits. It works without knowing how to construct colimits in Set.</p> <p>So another way to ask my question might be</p> <p><strong>Question 2:</strong> what is a class of categories in which you can prove that filtered colimits commute with finite limits (without first proving that this is true in Set)?</p> <p>So yes, I know that the commutativity holds in any locally finitely presentable category, but the only proofs of this I know depend on the fact that it is true in Set.</p> http://mathoverflow.net/questions/56577/historical-question-about-simplicial-sets/56589#56589 Answer by Steve Lack for Historical Question about Simplicial Sets Steve Lack 2011-02-25T02:28:42Z 2011-02-25T02:28:42Z <p>According to Mac Lane (see p19 of <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183537593" rel="nofollow">this paper</a>) they were introduced by Eilenberg-Zilber in 1950 under the name complete semisimplicial complexes.</p> http://mathoverflow.net/questions/55182/what-is-known-about-the-category-of-monads-on-set/55356#55356 Answer by Steve Lack for What is known about the category of monads on Set? Steve Lack 2011-02-13T23:25:10Z 2011-02-13T23:25:10Z <p>The category of all monads on Set (often this is called $Mnd$, and $Mon$ means the category of monoids) is not just large, but has large hom-sets. </p> <p>But if you restrict it to the full subcategory $Mnd_f$ of all <em>finitary</em> monads, then you get a beautiful category: it is not just complete and cocomplete but locally finitely presentable. It is even monadic over the category $Set^N$ of families of sets indexed by the natural numbers. </p> <p>The finitary monads are the ones that correspond to finitary (one-sorted) algebraic theories, and so include the monads for monoids, commutative monoids, groups, rings, Lie algebras (over a given field), and so on. More formally, a monad is finitary when its endofunctor part $T:Set\to Set$ preserves filtered colimits, or equivalently when $T$ is the left Kan extension of its restriction to finite sets. </p> <p>An example of a monad which is not finitary is the ultrafilter monad, whose algberas are compact Hausdorff spaces. </p> <p>Instead of finitary monads, you can take monads of rank $\alpha$, for some regular cardinal $\alpha$ - these are the ones whose endofunctors preserve $\alpha$-filtered colimits, and which can be described in terms of $\alpha$-small operations and equations. The full subcategory $Mnd_\alpha$ of $Mnd$ consisting of the monads of rank $\alpha$ is also locally finitely presentable. For example $Mnd_f$ is just the case where $\alpha=\aleph_0$.</p> <p>The inclusion $Mnd_\alpha\to Mnd$ does preserve colimits (in fact it has a right adjoint), and so $Mnd$ does have colimits of diagrams of monads with bounded rank. </p> <p>An example of a monad which does not have a rank is the powerset monad, whose algebras are the complete semilattices. </p> http://mathoverflow.net/questions/54530/2-morphisms-in-structured-2-categories/54993#54993 Answer by Steve Lack for 2-morphisms in structured 2-categories Steve Lack 2011-02-10T06:21:08Z 2011-02-10T08:01:31Z <p>A. This is really just an aspect of Mike Shulman's answer, but could be of some use in particular cases.</p> <p>There's a 2-categorical limit called the power (or cotensor) of an object $B$ by the arrow-category $2$. This is an object $B^2$ with the property that morphisms from $A$ to $B^2$ are in bijection with pairs of morphism from A to B with a 2-cell between them. For example if B is a category then $B^2$ is the functor category $[2,B]$. If $B$ is a monoidal category then $B^2$ is $[2,B]$ with the evident (pointwise) monoidal structure.</p> <p>In each of your examples, and more generally in Mike's setting, this limit exists in the structured 2-category, and is preserved by the forgetful 2-functor into <strong>Cat</strong>. Normally you would <strong>prove</strong> this given the definition of 2-cell. But you can also turn this around. Given a structure on B, if you know how to make $B^2$ into a structured object, then you can use this to <strong>define</strong> the structured 2-cells. </p> <p>In examples where the structure is given by a 2-monad, and in particular in examples which involve structure described by operations $B^n\to B$, natural transformations between these, and equations, then you can always do this in a "pointwise way". (But if you choose a strange way to make $B^2$ into a structured object you will get a strange notion of 2-cell.)</p> <p>Suppose, for example, that $B$ is a monoidal category. Once you agree to make $[2,B]$ monoidal in the pointwise way, then you can define a monoidal transformation to be a monoidal functor with codomain $[2,B]$, and this will agree with the standard definition which you referred to.</p> <p>In the case of a cocomplete category $B$, you don't need to choose how to make $[2,B]$ cocomplete, it just is. And then you can consider cocontinuous functors with codomain $[2,B]$; once again this will give no extra condition to be satisfied by a natural transformation between cocontinuous functors </p> <p>The case of symmetric monoidal categories can be treated in the same way.</p> <p>B. Regarding the case of symmetric monoidal categories, there is a general phenomenon here. As you add structure to your objects in the form of operations $B^n\to B$ (like a tensor product) you generally introduce preservation conditions on both morphisms and 2-cells (although there are special cases, as in your Example 2, where the 2-cell part is automatic). But if you introduce structure in the form of natural transformations between the operations $B^n\to B$ (such as a symmetry), this results in new preservation conditions for the morphisms but not for the 2-cells. </p> <p>C. Despite all this, there can be more than one choice for the 2-cells. The general principles described by Mike (and by me) would suggest that if our structure is categories with pullback, so that our morphisms are pullback-preserving functors, the 2-cells should be all natural transformations between these. But sometimes it's good to consider only those natural transformations for which the naturality squares are pullbacks. (These are sometimes called cartesian natural transformations.) See <a href="http://arxiv.org/abs/0904.2486" rel="nofollow">this paper</a> for example.</p> http://mathoverflow.net/questions/54277/do-plain-functors-out-of-monoidal-categories-factor-into-a-nontrivial-monoidal-pa/54285#54285 Answer by Steve Lack for Do plain functors out of monoidal categories factor into a nontrivial monoidal part and a plain part? Steve Lack 2011-02-04T03:44:57Z 2011-02-04T03:44:57Z <p>Regarding the specific example, the construction of $C'$ can be tightened up as follows. The first functor, from $C$ to $C'$ is not just hom-preserving (closed) but bijective on objects. The second functor, from $C'$ to $D$, has the property that it is fully faithful on maps out of the monoidal unit. </p> <p>(This is enough to determine $C'$ uniquely. As is implicit in Scott's comment, you need more than just the fact that the first leg is closed, otherwise you could take the first map to be the identity.)</p> <p>So you could consider the (2-)category of symmetric monoidal closed categories, and (strong) symmetric monoidal functors. Any such morphism $F:C\to D$ comes with a canonical comparison $F[c,d]\to[Fc,Fd]$, and we can call it (strongly) closed if these comparisons are invertible. There's a class E of morphisms consisting of those which are bijective on objects and closed, and a class M of morphisms consisting of those for which $F$ induces a bijection between maps $i\to c$ and maps $Fi\to Fc$, for all $c$, where $i$ is the monoidal unit. (You might call these maps pre-fully-faithful, or something like that.) These classes E and M are closed under composition. I haven't checked in detail, but it looks like they have a reasonable chance of being orthogonal to each other, and so you would have a factorization system.</p> <p>I don't see a way of doing something similar with mere functors between monoidal categories. But perhaps this is too much to expect. In the other example, although the internal hom is not preserved, there is a canonical comparison. So perhaps rather than looking at plain functors you should look at the lax monoidal ones (often just called monoidal, with the word strong being added to mean preservation up to isomorphism). Now it is true that every lax monoidal functor $C\to D$ factorizes as a lax monoidal $C\to C'$ followed by a strict monoidal $C'\to D$, moreover in a universal (initial) way. For any monoidal category $C$ there is a lax monoidal $p:C\to C'$ with the property that composition with $p$ induces a bijection between lax monoidal $C\to D$ and strict monoidal $C'\to D$, for any $D$. (Notice that the order is opposite to that in your example: the strict map comes second.)</p> <p>This situation is quite common, and holds for many different types of structure. It is much less common, but does sometimes happen that there's a universal factorization $C\to D'\to D$ with the map $C\to D'$ strict and the map $D'\to D$ non-strict. </p> <p>Your example involves an extra ingredient involving the bijective-on-objects and pre-fully-faithful conditions.</p> http://mathoverflow.net/questions/54128/singly-generated-monoids-in-mathematics/54157#54157 Answer by Steve Lack for singly-generated monoids in mathematics Steve Lack 2011-02-03T00:37:51Z 2011-02-03T00:37:51Z <p>Think about the sheaves on some site as a full subcategory of presheaves: $Sh(C)\to PSh(C)$. This has a left adjoint, called sheafification. There are various ways to construct the sheafification, but one of them uses something called the plus-construction. For any presheaf $F$ it gives an associated presheaf $F^+$, which is always separated but may not be a sheaf. (Being separated means that there is at most one way to glue local bits of data, as opposed to exactly one way for a sheaf.)</p> <p>If you do this a second time, to get $F^{++}$, you do get a sheaf, and if you do it a third time it has no further effect. Thus $sss=ss$ if $s$ is this plus-construction. </p> <p>There is a higher-dimensional version involving Cat-enriched presheaves and stacks, where you have to do the analogous construction three times. This would give an example of $ssss=sss$. </p> <p>I don't know if this continues in still higher dimensions. </p> http://mathoverflow.net/questions/53470/enriched-locally-presentable-categories/53474#53474 Answer by Steve Lack for Enriched locally presentable categories Steve Lack 2011-01-27T09:51:24Z 2011-01-27T09:51:24Z <p>The standard reference is <a href="http://archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1982__23_1/CTGDC_1982__23_1_3_0/CTGDC_1982__23_1_3_0.pdf" rel="nofollow">this paper by Max Kelly</a>.</p> <p>Perhaps the most unexpected thing is how well the theory works!</p> http://mathoverflow.net/questions/52989/reference-request-2-monads-and-2-adjunctions/53007#53007 Answer by Steve Lack for Reference request: 2-Monads and 2-Adjunctions Steve Lack 2011-01-23T23:12:05Z 2011-01-23T23:12:05Z <p>As Todd says, there are several flavours of 2-monad. </p> <p>If you are interested in strict 2-monads, strict algebras for these, and strict morphisms, then yes you have an adjunction (even an enriched adjunction) as usual. </p> <p>If you mean something weaker, then you will have something weaker than an adjunction. In particular, if you are considering pseudomorphisms of algebras - those which preserve the structure only up to (suitably coherent) isomorphism - then you'll have an <em>equivalence</em> between the category of algebra morphisms from a free algebra $TX$ to an algebra $B$ and the category of morphisms in the base 2-category from $X$ to (the underlying object of) $B$. So rather than an adjunction you'll get some sort of biadjunction. See Corollary 5.6 of the Blackwell-Kelly-Power paper for the case of strict monads, strict algebras, and pseudomorphisms, which is in fact the most important case for many purposes.</p> <p>An important aspect of this is that free algebras $TX$ are <em>flexible</em>, which means among other things that any pseudomorphism from $TX$ to $B$ is isomorphic to a strict one. This is false for a general algebra $A$ in place of $TX$.</p> http://mathoverflow.net/questions/51783/enriched-monoidal-categories/52016#52016 Answer by Steve Lack for Enriched monoidal categories Steve Lack 2011-01-13T22:56:28Z 2011-01-13T22:56:28Z <p>There is a theorem in category theory, generally regarded as folklore, which says that for a symmetric monoidal closed category $V$, the following structures are equivalent:</p> <ol> <li>a category $C$ with an action $V\times C\to C$ of the monoidal category $V$ on $C$, which we may write as $(v,c)\mapsto v*c$, for which $-*c:V\to C$ has a right adjoint for each $c\in C$ (here the action amounts to a strong monoidal functor $V\to [C,C]$.</li> <li>a $V$-category $C$ for which the $V$-functor $C(c,-):C\to V$ has a left adjoint for each $c\in C$. (such a $V$-category is said to be "tensored'' or "copowered'')</li> </ol> <p>You can see this, for example, in the appendix to <a href="http://www.tac.mta.ca/tac/volumes/9/n4/n4.pdf" rel="nofollow">this paper</a>.</p> <p>In your case, unless I've misunderstood, the centre $Z(C)$ plays little role. The point is that your functor $z:V\to C$ induces an action via $v*c=z(v)\otimes c$, and $-*c$ has a right adjoint by assumption, so you get the $V$-enrichment.</p> <p>(There is an analogous characterization of $V$-categories $C$ which are cotensored/powered: this means that each $C(-,d):C^{op}\to V$ has a left adjoint.)</p> http://mathoverflow.net/questions/51163/reference-for-factorization-of-left-adjoints/51271#51271 Answer by Steve Lack for Reference for factorization of left adjoints? Steve Lack 2011-01-06T01:34:51Z 2011-01-06T01:34:51Z <p>This should probably be a comment but I don't know how to include links in comments. </p> <p>Results of this type are often called <em>adjoint triangle theorems</em>. There are many such: see John Power's paper <a href="http://archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1988__29_1/CTGDC_1988__29_1_67_0/CTGDC_1988__29_1_67_0.pdf" rel="nofollow">A unified approach to the lifting of adjoints</a> for a summary (and a unified approach). </p> <p>A sufficient condition for $H$ to admit a left adjoint is that $C$ has coequalizers and that $G$ is of descent type''. (This means that if we form the monad $S$ on $A$ induced by $G$ and its left adjoint, then the induced functor $B\to A^S$ is fully faithful.) This includes the important examples of the type considered in Todd's answer. </p> <p>for a summary. </p> http://mathoverflow.net/questions/7090/what-should-the-definition-of-yoneda-property-be/51076#51076 Answer by Steve Lack for What should the definition of "Yoneda property" be? Steve Lack 2011-01-03T23:53:12Z 2011-01-03T23:53:12Z <p>Hi Jim, not sure if this is the sort of thing you are after, but here is one possibility. </p> <p>Let $K$ be a category and $F=(f_i:A_i\to B_i)_{i\in I}$ a family of morphisms in $K$. Say that an object $X$ is injective to $F$ if for each $i\in I$ and each $a:A_i\to X$ there exists a morphism $b:B_i\to X$ whose restriction along $f_i$ is $a$. The collection of all such objects $X$ (for given $F$) is called an <em>injectivity class</em> in $K$. </p> <p>Any injectivity class in $[C^{op},Set]$ defines a property of objects of $C$: those objects $c$ for which the representable functor $C(-,c)$ lies in the injectivity class. Your various examples arise in this way:</p> <ul> <li>For injectivity, just take all maps of the form $C(-,i):C(-,a)\to C(-,b)$ with $i:a\to b$ mono. </li> <li>For terminal object, take all the maps $0\to C(-,a)$ and $\nabla:C(-,a+a)\to C(-,a)$ (where $\nabla$ is the codiagonal)</li> <li>and similarly being formally smooth or formally unramified. </li> </ul> <p>Your properties "of uniqueness type'' can be seen as a special case of those "of existence type'', by using codiagonal maps, as in the case of terminal object. </p> <p>Of course in the first example, of injectivity, you could just as well work in the original category $C$ itself, rather than $[C^{op},Set]$. More generally, you would always be able to do this if $C$ had colimits. </p> <p>You will get better properties if the family of morphisms defining the injectivity class is, or can be taken to be, small. Not sure if this is important for your purposes. </p> http://mathoverflow.net/questions/48361/when-does-the-2-category-v-cat-have-pseudo-pullbacks/48422#48422 Answer by Steve Lack for When does the 2-category V-Cat have pseudo-pullbacks? Steve Lack 2010-12-06T06:52:34Z 2010-12-06T06:52:34Z <p>As I mentioned in the comment above, "weak limit" is normally defined as for limit, but with the universal property modified to ask only for existence not uniqueness. The 2-dimensional limit notion in which all equations between 1-cells are replaced by suitably coherent invertible 2-cells is usually given the prefix "bi". </p> <p>Any 2-category with finite limits (in the strict 2-categorical sense) also has isocomma objects (defined like comma objects but with an invertible 2-cell), pseudopullbacks, and bipullbacks. </p> <p>If V has finite limits (in the ordinary sense) then V-Cat has finite limits (in the strict 2-categorical sense). </p> <p>(The parts of this that relate to the original version of your question are dealt with in Finn's answer.)</p> http://mathoverflow.net/questions/47079/line-bundles-in-abelian-otimes-categories/47160#47160 Answer by Steve Lack for Line bundles in abelian $\otimes$-categories Steve Lack 2010-11-23T23:14:56Z 2010-11-23T23:14:56Z <p>Question 2: this does not really depend on line bundles. If $\phi:\mathcal {L\to L}$ is a non-invertible epimorphism, then $\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is epi, since $-\otimes\mathcal L'$ is exact, and non-invertible, since otherwise $\phi\otimes 1\otimes 1:\mathcal{L\otimes L'\otimes L\to L\otimes L'\otimes L}$ would be invertible. Thus there is a non-invertible epimorphism $\mathcal {O\to O}$.</p> <p>Question 3: this has the same answer as Question 2. If the answer to Q2 is yes, then we can take $\psi$ to be the identity and get a positive answer to Q3. If the answer to Q2 is no, then (as above) if $\phi:\mathcal{ L\to L}$ is a non-invertible epimorphism, also $\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is a non-invertible epimorphism. But now if $\psi:\mathcal {L'\to L'}$ is any map, then $\phi\otimes\psi=(1\otimes\psi)\circ(\phi\otimes 1)$ and if this is invertible then $\phi\otimes 1$ is split monic and so invertible (since it is already known to be epi).</p> http://mathoverflow.net/questions/46877/natural-examples-of-sequences-of-adjoint-functors/46879#46879 Answer by Steve Lack for Natural examples of sequences of adjoint functors Steve Lack 2010-11-22T00:03:07Z 2010-11-22T00:03:07Z <p>For any category $B$ with small hom-sets one can form the yoneda embedding $y:B\to[B^{op},Set]$ (although if $B$ is not small, $[B^{op},Set]$ may not itself have small hom-sets).</p> <p>Rosebrugh and Wood showed <a href="http://www.jstor.org/stable/2161031" rel="nofollow">link text</a> here that if $B$ is itself the category of of sets then there is an adjoint string $u\dashv v\dashv w\dashv x\dashv y$, and that this characterizes $Set$.</p> <p>The adjunctions $\pi_0\dashv Dis\dashv U\dashv CoDis$s also work if we replace Top by any of the categories SSet of simplicial sets, Cat of categories, Gpd of groupoids, or Preord of preorders.</p> http://mathoverflow.net/questions/46700/for-which-categories-does-one-have-a-goursat-lemma/46709#46709 Answer by Steve Lack for For which categories does one have a Goursat Lemma? Steve Lack 2010-11-20T03:43:30Z 2010-11-20T03:43:30Z <p>To make life simple, suppose that finite limits and finite colimits exist. If we work with regular epimorphisms rather than epimorphisms then your condition is equivalent to saying that for any two (regular) epimorphisms $G\to L$ and $G\to R$, if you form the pushout $F$ then the canonical comparison from $G$ to the pullback $L\times_F R$ is a regular epimorphism.</p> <p>This is true in any exact Mal'cev category: see Theorem 5.7 of</p> <blockquote> <p>Carboni, Kelly and Pedicchio, Some remarks on Maltsev and Goursat categories, Applied Categorical Structures 1:385-421, 1993.</p> </blockquote> <p>Here exact means that the category (i) has finite limits (ii) has regular epi-mono factorizations (iii) the pullback of a regular epi is a regular epi (iv) any equivalence relation is the kernel pair of some map (one can choose the map to be the coequalizer) and Mal'cev can be characterized in many ways. For example, it says that if R and S are equivalence relations on some object A, then RS=SR.</p> <p>In fact if the category is regular, in the sense that (i)-(iii) hold, then your condition is equivalent to being exact and Mal'cev. </p> <p>By the way, the Goursat categories you mention are only slightly weaker: they have RSR=SRS rather than RS=SR. You can still prove your condition for Goursat categories if you suppose that at least one of $G\to L$ and $G\to R$ is a split epimorphism (i.e. has a section), and in fact this can be used to characterize Goursat categories. See <a href="http://www.springerlink.com/content/a6x8448227q13062/" rel="nofollow">link text</a></p> http://mathoverflow.net/questions/24722/is-there-a-notion-of-good-distributor-profunctor-for-model-categories/46474#46474 Answer by Steve Lack for Is there a notion of "good" distributor/profunctor for model categories? Steve Lack 2010-11-18T11:47:37Z 2010-11-18T11:47:37Z <p>There's a natural generalization of Quillen adjunction. </p> <p>Let $A$ and $B$ be model categories, and $M:A^{op}\times B\to Set$ be a distributor/profunctor/module. This should be Quillen if whenever $i:a\to a'$ is a cofibration in $A$ and $p:b\to b'$ is a fibration in $B$, with either $i$ or $p$ a weak equivalence, then the induced map from $M(a',b)$ to the pullback $M(a,b)\times_{M(a,b')}M(a',b')$ is surjective.</p> http://mathoverflow.net/questions/18744/is-model-structure-on-catset-unique/46471#46471 Answer by Steve Lack for Is model structure on CatSet unique? Steve Lack 2010-11-18T11:35:15Z 2010-11-18T11:35:15Z <p>As has been pointed out above, there are many possible model structures on Cat with different weak equivalences. This is the only <em>proper</em> model structure on Cat for which the weak equivalences are the categorical equivalences. </p> http://mathoverflow.net/questions/46230/distributivity-commutativity-of-pushouts-and-pullbacks/46426#46426 Answer by Steve Lack for Distributivity / commutativity of pushouts and pullbacks Steve Lack 2010-11-17T23:04:19Z 2010-11-17T23:04:19Z <p>The condition Martin and Todd mention is indeed a sort of distributivity condition. It is also often called {\em stability} of pushouts. I think that it should not be called commutativity. </p> <p>Let D and C be small categories, and A a category with D-shaped limits and C-shaped colimits. Then D-limits commute in A with C-shaped colimits when the functor $[D,A]\to A$ which calculates the limit preserves C-colimits. This is equivalent to saying that the functor $[C,A]\to A$ which calculates the colimit preserves D-limits. The most famous example is commutativity of finite limits with filtered colimits. </p> <p>Note, however, that commutativity of pullbacks and pushouts in this sense is rare, and is probably not what was meant. </p> http://mathoverflow.net/questions/45668/is-the-decomposition-of-an-algebra-into-irreducible-components-essentially-unique/46302#46302 Answer by Steve Lack for Is the decomposition of an algebra into irreducible components essentially unique? Steve Lack 2010-11-17T00:06:18Z 2010-11-17T00:06:18Z <p>This is not true as stated. If you take the empty signature, or any signature with no constants, then the empty set is an algebra, and this will mess things up.</p> <p>This issue was raised in the paper </p> <blockquote> <p>M. Barr, The point of the empty set, Cahiers 13:1-12, 1972.</p> </blockquote> <p>I think that this is all that can go wrong. If you restrict to non-empty signatures, or to non-empty algebras then things are ok; but if you are also considering multisorted theories, then you need to make sure that each sort is non-empty. This issues was also discussed in the Barr paper. See also </p> <blockquote> <p>G.M. Kelly and A. Pultr, On algebraic recognition of direct-product decompositions, Journal of Pure and Applied Algebra 12:207--224, 1978.</p> </blockquote> http://mathoverflow.net/questions/41386/equalizer-completion/46214#46214 Answer by Steve Lack for Equalizer completion Steve Lack 2010-11-16T08:55:29Z 2010-11-16T10:21:21Z <p>There are general results about how to freely add limits or colimits to a category. They are formally dual, but people normally state the colimit variety because they involve the category of presheaves. </p> <p>To freely add some class of colimits to a given category $C$, you form the closure of the representables in the presheaf category $[C^{op},\mathsf{Set}]$ under the given type of colimit.</p> <p>If you want to do this in such a way that certain existing colimits are preserved, then you form the closure not in the presheaf category, but in the full subcategory of those presheaves which send the existing colimits in C into limits in Set. </p> <p>So to freely complete a category $C$ with finite products to a category with finite limits, you should look at the Yoneda embedding of $C$ into the opposite $FP(C,Set)^{op}$ of the category $FP(C,\mathsf{Set})$ of finite-product-preserving functors from $C$ to $\mathsf{Set}$. Now take the closure of the representables under finite limits. That is, finite limits in $FP(C,\mathsf{Set})^{op}$, or finite colimits in $FP(C,\mathsf{Set})$. This then gives the value at $C$ of a left biadjoint to the forgetful 2-functor from the 2-category LEX of categories with finite limits to the 2-category FP of categories with finite products.</p> <p>There is an explicit, syntactic description, due to Andy Pitts. It can be found, for example, in the paper </p> <blockquote> <p>M. Bunge and A. Carboni, The symmetric topos, Journal of Pure and Applied Algebra, 1995.</p> </blockquote> <p>An object is a pair of maps $f,f':X \to X'$ equipped with a common retraction $r$. A morphism from such a pair to $g,g':Y \to Y',s$ consists of an equivalence class, under a suitably defined equivalence relation, of pairs $(a,a')$ where $a:X \to Y$, $a':X'\to Y'$, and the evident diagrams commute. </p> http://mathoverflow.net/questions/57099/why-do-filtered-colimits-commute-with-finite-limits Comment by Steve Lack Steve Lack 2013-03-11T23:55:25Z 2013-03-11T23:55:25Z Thanks, Dylan. I agree that J being filtered will have to be used. But what properties of Set should be used? http://mathoverflow.net/questions/85295/finitary-mono-preserving-functors-on-varieties-that-fail-to-preserve-intersection Comment by Steve Lack Steve Lack 2012-01-10T09:44:47Z 2012-01-10T09:44:47Z An empty intersection of subobjects is the whole object, so is always preserved. The example I gave involved failure to preserve a binary intersection. Perhaps it would be helpful if you explained what you were really after. http://mathoverflow.net/questions/85295/finitary-mono-preserving-functors-on-varieties-that-fail-to-preserve-intersection Comment by Steve Lack Steve Lack 2012-01-10T02:15:13Z 2012-01-10T02:15:13Z Rob, I'm not sure quite what this means. I take it that the modified version sends everything to 1? In that case the empty set will no longer possess an algebra structure. http://mathoverflow.net/questions/85295/finitary-mono-preserving-functors-on-varieties-that-fail-to-preserve-intersection Comment by Steve Lack Steve Lack 2012-01-09T23:03:54Z 2012-01-09T23:03:54Z The endofunctor T of Set sending a set X to a singleton 1 if it is non-empty, and to the empty set 0 if it is empty preserves monomorphisms, since it sends <i>every</i> map to a monomorphism. But it does not preserve the intersection of the two maps from a singleton 1 to a two-element set 2. http://mathoverflow.net/questions/41305/classification-of-quasi-topoi/67433#67433 Comment by Steve Lack Steve Lack 2011-06-11T05:48:38Z 2011-06-11T05:48:38Z it extends to Grothendieck toposes; not sure about more generally than that. http://mathoverflow.net/questions/64382/comparing-discrete-fibrations-and-their-duals/64600#64600 Comment by Steve Lack Steve Lack 2011-05-13T23:42:57Z 2011-05-13T23:42:57Z Yes, Mike, quite right - the equivalence of discrete fibrations and codiscrete cofibrations is an exactness condition which will carry over to Cat-valued presheaves. And I'd guess Grothendieck 2-toposes should work as well, although I haven't checked the details. http://mathoverflow.net/questions/64382/comparing-discrete-fibrations-and-their-duals/64600#64600 Comment by Steve Lack Steve Lack 2011-05-11T23:35:48Z 2011-05-11T23:35:48Z First question: yes (sorry, that was unclear). Second question: no (but if I think of one I'll let you know). http://mathoverflow.net/questions/61195/functoriality-of-the-center-of-a-category Comment by Steve Lack Steve Lack 2011-04-29T21:41:33Z 2011-04-29T21:41:33Z sorry, I meant to say for each fully faithful <i>F:D-&gt;C</i>. http://mathoverflow.net/questions/61195/functoriality-of-the-center-of-a-category Comment by Steve Lack Steve Lack 2011-04-29T11:40:32Z 2011-04-29T11:40:32Z you also get a functor <i>Z(F):Z(C)-&gt; Z(D)</i> for each fully faithful <i>F:C-&gt; D</i>. http://mathoverflow.net/questions/58769/orthogonal-and-parallel-morphisms Comment by Steve Lack Steve Lack 2011-03-18T12:19:47Z 2011-03-18T12:19:47Z I agree with Mike's comment about the intuition &quot;everywhere equal&quot;, and that the former is something to do with disjointness. In fact Diers calls a pair with the latter property &quot;codisjointed&quot;. http://mathoverflow.net/questions/58010/on-a-corollary-in-mitchells-book/58110#58110 Comment by Steve Lack Steve Lack 2011-03-11T10:50:59Z 2011-03-11T10:50:59Z Ralph, you can still construct the 3x3. Factorize $pj:B'\to C$ as an epi $p':B'\to C'$ followed by a mono $k:C'\to C$. Form the cokernel $r:C\to C''$ of $k$, and the kernel $i':A'\to B'$ of $p'$. Then $ji':A'\to B$ factorizes through $i$ as some mono $h:A'\to A$, and $rp:B\to C''$ factorizes through $q$ as some epi $p'':B''\to C''$. Form the cokernel $s:A\to A''$ of $h$. Then $qi:A\to B''$ factorizes through $s$ as some $i'':A''\to B''$. Now (i) check that $i''$ is mono, and (ii) check that $p''$ is the cokernel of $p''$. http://mathoverflow.net/questions/57941/pull-backs-which-are-push-outs/57950#57950 Comment by Steve Lack Steve Lack 2011-03-10T22:22:33Z 2011-03-10T22:22:33Z Fernando, pretoposes do not have the first property. In fact even the category Set does not have this property: consider the case where $C=0$. http://mathoverflow.net/questions/58010/on-a-corollary-in-mitchells-book/58029#58029 Comment by Steve Lack Steve Lack 2011-03-10T09:38:58Z 2011-03-10T09:38:58Z OK, thanks Ralph, I've now given a fix. http://mathoverflow.net/questions/58010/on-a-corollary-in-mitchells-book/58029#58029 Comment by Steve Lack Steve Lack 2011-03-10T08:09:12Z 2011-03-10T08:09:12Z Ralph: yes, of course, you're quite right, this argument needs the category to be enriched over abelian groups, not just over pointed sets. Is this part of the Mitchell definition? http://mathoverflow.net/questions/57099/why-do-filtered-colimits-commute-with-finite-limits/57124#57124 Comment by Steve Lack Steve Lack 2011-03-02T22:27:06Z 2011-03-02T22:27:06Z Thanks, Buschi. I certainly agree that if the category is locally cartesian closed then we have $colim_i(X_i\times_Y B_i)\cong (colim_i X_i)\times_Y (colim_i B_i)$: this is the same argument I gave applied to the slice category ${\mathcal C}\downarrow Y$. But I don't follow the first paragraph. For one thing, the triple diagonal $I\to I\times I\times I$ does not seem to be directly relevant. You can form the product $X_i\times B_j$ for any $i$ and $j$, but the pullback $X_i\times_{Y_k} B_j$ only makes sense if we have first chosen maps $i\to k$ and $j\to k$.