User finn lawler - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:03:16Z http://mathoverflow.net/feeds/user/4262 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106358/categorification-of-coends-and-ends/106364#106364 Answer by Finn Lawler for Categorification of coends and ends Finn Lawler 2012-09-04T16:42:19Z 2012-09-04T16:42:19Z <p>I don't think I have seen anything like this published before, but I have written up a similar definition <a href="http://ncatlab.org/finnlawler/show/2-end" rel="nofollow">here</a> (see <a href="http://ncatlab.org/finnlawler/show/2-extranatural+transformation" rel="nofollow">here</a> too).</p> <p>One thing in your definition I would take issue with is that your 2-coend's universal property needs more to deserve the '2-' in its name: what I would expect (and what my definition, which otherwise seems to be equivalent to yours, says) is a bicategorical universal property that says</p> <ul> <li>for $(C', \mathrm{gl'}, \ldots)$ as above, $\eta$ exists and is invertible, and</li> <li>for $\theta, \theta' \colon C \to C'$, any 2-cell (i.e. modification between extranaturals/wedges) $\theta \circ \mathrm{gl} \to \theta' \circ \mathrm{gl}$ is equal to $t \circ \mathrm{gl}$ for a unique $t \colon \theta \to \theta'$.</li> </ul> http://mathoverflow.net/questions/101644/fiction-books-about-mathematicians/101697#101697 Answer by Finn Lawler for Fiction books about mathematicians? Finn Lawler 2012-07-08T19:48:12Z 2012-07-08T19:48:12Z <p><em>Kepler</em> by John Banville is a sort of 'fictional biography'. Banville is a Booker prize winner, very highly regarded. His prose is some of the most beautiful, dense and lyrical I've ever read, and I'd recommend <em>Kepler</em> to anyone with an interest in mathematics and a taste for masterful writing.</p> <p>(Banville also wrote <em>Doctor Copernicus</em>, which I haven't read.)</p> http://mathoverflow.net/questions/85768/morita-equivalence-for-categories/85770#85770 Answer by Finn Lawler for morita equivalence for categories Finn Lawler 2012-01-15T21:32:11Z 2012-01-15T21:32:11Z <p>That their <a href="http://nlab.mathforge.org/nlab/show/Cauchy+complete+category" rel="nofollow">Cauchy completions</a> are equivalent.</p> http://mathoverflow.net/questions/85391/any-example-of-a-non-strong-monad/85449#85449 Answer by Finn Lawler for Any example of a non-strong monad? Finn Lawler 2012-01-11T21:27:48Z 2012-01-11T21:27:48Z <p>Here is a class of examples different to Tom's: if your underlying monoidal category C is closed, then a strong monad on C is the same as a C-enriched monad, i.e. one that respects the <a href="http://ncatlab.org/nlab/show/enriched+category" rel="nofollow">enrichment</a> of C given by its internal hom (this is why every monad on Set is strong, as Andrej points out). So one example would be the monad on Cat (considered as a Set-category) whose algebras are cartesian closed categories -- it is known that this is not a Cat-monad (although it does extend to Cat as a groupoid-enriched category). I would imagine that the same is true for the monad for monoidal closed categories, or in general for categories with any one sort of mixed-variance structure.</p> http://mathoverflow.net/questions/84636/codomain-fibration/85349#85349 Answer by Finn Lawler for Codomain fibration. Finn Lawler 2012-01-10T18:14:37Z 2012-01-10T18:14:37Z <p>I assume that by 'cartesian' you mean 'having cartesian products'. In that case the answer is no. You've more or less said this yourself: for $C^{\mathbf{3}} \to C^{\mathbf{2}}$ or the composite $C^{\mathbf{3}} \to C$ to be fibrations it is again necessary and sufficient that C have pullbacks.</p> <p>The sequence makes perfect sense as a functor, of course, and the fibres are the same no matter what C looks like: an object over c is a composable pair of morphisms $\bullet \to \bullet \to c$ and a morphism is a 'square on top of a triangle'.</p> http://mathoverflow.net/questions/83799/monadicity-and-sheaves/83818#83818 Answer by Finn Lawler for Monadicity and sheaves. Finn Lawler 2011-12-18T19:19:04Z 2011-12-18T19:19:04Z <p>The answer (to the first question) is yes: <a href="http://ncatlab.org/nlab/show/reflective+subcategory" rel="nofollow">reflections</a> are always monadic, and the associated monad is <a href="http://ncatlab.org/nlab/show/idempotent+monad" rel="nofollow">idempotent</a>.</p> http://mathoverflow.net/questions/83437/the-main-theorems-of-category-theory-and-their-applications/83473#83473 Answer by Finn Lawler for The main theorems of category theory and their applications Finn Lawler 2011-12-14T22:04:41Z 2011-12-14T22:04:41Z <p><a href="http://ncatlab.org/nlab/show/monadicity+theorem" rel="nofollow">Monadicity theorems</a> like Beck's give sufficient conditions on a functor $U \colon B \to A$ with a left adjoint F to be equivalent (in the slice category Cat/A) to the forgetful functor out of the category $A^{U F}$ of algebras for the monad $U F$.</p> <p>Aside from their obvious usefulness in universal algebra, one important application is in descent theory: suppose you have a category C and a C-indexed category E (i.e. a pseudofunctor $E \colon C^{\mathrm{op}} \to \mathrm{Cat}$) such that</p> <ol> <li><p>Each $f^* = Ef$ has a right adjoint $f_!$, and</p></li> <li><p>The Beck--Chevalley condition holds: E takes any pullback square in C to an isomorphism in Cat whose <a href="http://ncatlab.org/nlab/show/mate" rel="nofollow">mate</a> is again an isomorphism.</p></li> </ol> <p>Then (this is due to Bénabou and Roubaud) for a morphism f in C, the category of descent data for f is equivalent to the category of algebras for the monad $f_!f^*$. In particular, f is of effective descent if $f_!$ is monadic. See the nLab page on <a href="http://ncatlab.org/nlab/show/monadic+descent" rel="nofollow">monadic descent</a> for details.</p> <p>I believe the original application had C the category of commutative rings and $E \colon R \mapsto R\mathrm{-Mod}$.</p> http://mathoverflow.net/questions/83437/the-main-theorems-of-category-theory-and-their-applications/83469#83469 Answer by Finn Lawler for The main theorems of category theory and their applications Finn Lawler 2011-12-14T21:17:05Z 2011-12-14T21:17:05Z <p>There was some discussion <a href="http://mathoverflow.net/questions/12511/what-is-yonedas-lemma-a-generalization-of" rel="nofollow">here</a> about special cases of Yoneda's lemma, including the usual examples of Cayley's theorem for groups and Dedekind's embedding theorem for posets.</p> <p>It also seems that a good chunk of Tannaka duality can be seen as an application of the (enriched) Yoneda lemma -- see the <a href="http://ncatlab.org/nlab/show/Tannaka+duality" rel="nofollow">nLab page</a> for discussion.</p> http://mathoverflow.net/questions/83363/is-there-a-nice-application-of-category-theory-to-functional-complex-harmonic-ana/83367#83367 Answer by Finn Lawler for Is there a nice application of category theory to functional/complex/harmonic analysis? Finn Lawler 2011-12-13T21:21:03Z 2011-12-14T19:29:58Z <p>First, a disclaimer: I am not even close to being an analyst. Second, I don't know of any applications of category theory to the areas of analysis that you mention. I don't think we have got to that point yet, for the reason given below. But here is an answer to a more general question that I hope you'll find illuminating.</p> <p>I think the thing to remember here is that category theory is 'structural mathematics'. That is, it seeks to understand mathematical objects and constructions purely in terms of abstract external structure, as opposed to internal details about how an object is put together. In areas like algebra and computer science, this sort of structure is already there and visible, so for example it's easy to define the notion of 'group object' or 'monoid action', and to discuss constructions like quotients and semidirect products and so on in purely structural terms.</p> <p>My impression of analysis is that the structures involved are less unequivocal and less clearly visible, and consequently it's harder to give a broad unified structural picture of the kind that we're used to in algebra. So the category-theoretic or structural understanding of analysis is a good deal less well-developed. But there are some interesting facts about certain structures found in (very elementary) analysis and topology:</p> <ul> <li>Metric spaces are a particular kind of enriched category, and the notion of Cauchy completeness has a very neat definition in that context.</li> <li>Compact Hausdorff spaces are the algebras for a monad on Set (the ultrafilter monad).</li> <li>Topological spaces are lax algebras for the same monad on the bicategory Rel of relations.</li> <li>Normed vector spaces can be viewed as enriched categories with duals (Lawvere), or as compact-closed ordinary categories equipped with a certain kind of functor (see Geoff Cruttwell's thesis).</li> <li>C-star algebras are monadic over Set (although I gather not much more is known about this).</li> <li>Non-standard analysis has a nice interpretation in terms of the filter-quotient construction on toposes.</li> <li>There is a notion of Fourier or z-transform for Joyal's 'structure types'.</li> </ul> <p>There are probably many more (and this answer is CW so passers-by are invited to add them).</p> <p>Personally, I think that the structural and material viewpoints on mathematics complement each other very well, so I'd be delighted if someone could point out (or write!) a structural account of -- a sort of 'Mac Lane and Birkhoff' for -- even elementary analysis.</p> http://mathoverflow.net/questions/82231/categorical-brouwer-heyting-kolmogorov-interpretation/82233#82233 Answer by Finn Lawler for Categorical Brouwer-Heyting-Kolmogorov interpretation Finn Lawler 2011-11-30T00:41:24Z 2011-11-30T00:41:24Z <p>I don't have it with me, and I can't recall the exact details, but I'm pretty sure Lambek &amp; Scott's <em>Introduction to Higher-Order Categorical Logic</em> (<a href="http://books.google.com/books?id=6PY_emBeGjUC&amp;lpg=PP1&amp;pg=PR5#v=onepage&amp;q&amp;f=false" rel="nofollow">link</a>) is what you're looking for. In particular, they prove the equivalence between cartesian closed categories and simply-typed $\lambda$-calculi (so you get the Curry--Howard correspondence for free!).</p> http://mathoverflow.net/questions/81515/when-is-a-colimit-of-a-subcollection-the-same-as-the-overall-colimit/81520#81520 Answer by Finn Lawler for When is a colimit of a subcollection the same as the overall colimit? Finn Lawler 2011-11-21T16:20:48Z 2011-11-21T16:20:48Z <p>Dylan's comment is right. More generally, if $D \colon J \to C$ is a diagram and $L \colon J' \to J$ a functor, then the colimit of D is isomorphic to the colimit of DL if and only if L is (co)final. See Mac Lane, <em>Cats Work</em>, section IX.3.</p> http://mathoverflow.net/questions/79817/reference-requestenriched-categories-metric-on-lipschitz-continuous-functions/79823#79823 Answer by Finn Lawler for Reference Request(Enriched Categories): Metric on Lipschitz Continuous Functions Finn Lawler 2011-11-02T12:13:06Z 2011-11-02T12:13:06Z <p>It is the usual sup metric. See section 2 of Lawvere's original <a href="http://www.tac.mta.ca/tac/reprints/articles/1/tr1abs.html" rel="nofollow">article</a>.</p> http://mathoverflow.net/questions/79102/definition-of-enriched-caterories-or-internal-homs-without-using-monoidal-categor/79103#79103 Answer by Finn Lawler for Definition of enriched caterories or internal homs without using monoidal categories. Finn Lawler 2011-10-25T18:50:07Z 2011-10-25T18:50:07Z <p>This is exactly the notion of a <a href="http://en.wikipedia.org/wiki/Closed_category" rel="nofollow">closed category</a>. See Eilenberg and Kelly's article in the 1965 La Jolla proceedings (Springer 1966). I think they also describe categories enriched in a closed category.</p> http://mathoverflow.net/questions/78855/automorphisms-and-bicategories/78859#78859 Answer by Finn Lawler for Automorphisms and Bicategories Finn Lawler 2011-10-23T01:24:20Z 2011-10-23T01:24:20Z <p>I don't see bicategories coming into this in a useful way, but I think what you have is a consequence of two more general facts:</p> <ul> <li><p>The non-uniqueness of algebraic closures is a general fact about <a href="http://nlab.mathforge.org/nlab/show/injective+hull" rel="nofollow">injective hulls</a> -- they are 'unique' up to <em>non-unique</em> isomorphism.</p></li> <li><p>Every morphism in a groupoid yields an isomorphism of vertex groups by conjugation -- if G is a groupoid and x is an object of G, then the vertex group at x is G(x,x), and if $f \colon x \to y$ is a morphism then $g \mapsto f^{-1} g f$ is an isomorphism between G(x,x) and G(y,y).</p></li> </ul> <p>With the objects satisfying a universal property the comparison isomorphisms between them are unique, so that the groupoid of objects satisfying the universal property is codiscrete, i.e. there is exactly one morphism between any two objects, so in particular the vertex groups of this groupoid are trivial. For an object A in a concrete category (or in a category with a chosen class of 'embeddings') there is a groupoid of injective hulls of A that is not in general codiscrete, and so it can have non-trivial vertex groups. But this groupoid, though not codiscrete, is still connected, so that each vertex group is (non-uniquely!) isomorphic to every other via conjugation by a morphism (necessarily invertible) of injective hulls.</p> <p>Edit: The fundamental group of a space, as in JSE's analogy, bears much the same relationship with the fundamental <em>groupoid</em> of the space -- in particular, $\pi_1$s at points in the same path-component are isomorphic via conjugation in the same way.</p> http://mathoverflow.net/questions/77885/2-completeness-analog-of-completeness-theorem/77893#77893 Answer by Finn Lawler for 2-completeness analog of completeness theorem Finn Lawler 2011-10-12T04:05:51Z 2011-10-12T04:05:51Z <p>Your suspicion is correct: in general, a V-category has all weighted V-limits if it has all conical V-limits <em>and</em> is cotensored over V (see Kelly's <em>Basic Concepts of Enriched Category Theory</em>, section 3.10). For V = Cat (and this is true for bicategories too), cotensors can be constructed from conical limits and cotensors with the arrow category 2, although I don't know the original reference for that. 'Finite' limits are a bit more complicated in the enriched case, but see Street's 'Limits indexed by category-valued 2-functors', JPAA 1972, for the 2-case.</p> http://mathoverflow.net/questions/70361/a-self-dual-adjunction/70366#70366 Answer by Finn Lawler for a “self-dual” adjunction Finn Lawler 2011-07-14T19:24:54Z 2011-07-14T19:24:54Z <p>(I'm a bit confused by your notation (what is $I$?), but if you mean what I think you mean...)</p> <p>I don't think there is an 'official' name for these things, but I've seen the term 'self-adjoint' used, sometimes qualified by 'on the left' or 'on the right' according to whether $U \dashv U^{op}$ or $U^{op} \dashv U$. See e.g. Mac Lane &amp; Moerdijk, <em>Sheaves in Geometry and Logic</em>, chapter IV, section 5.</p> <p>I believe it was Manes who observed that the power-object functor $P \colon E^{op} \to E$ of an elementary topos $E$ is not only self-adjoint on the right but monadic, with the corollary that toposes have finite colimits. (See loc. cit.)</p> <p>Hayo Thielecke has studied self-adjunctions as a way to understand the notion of 'continuations' in computer science. See his Edinburgh Ph.D. thesis.</p> http://mathoverflow.net/questions/68791/adjunctions-form-a-stack/68816#68816 Answer by Finn Lawler for Adjunctions form a stack Finn Lawler 2011-06-25T22:16:53Z 2011-06-25T22:16:53Z <p>Following up on my comment above, I think you do indeed get a lax transformation $G \to F$ that won't be pseudo in general. In fact, this seems to be a case of <a href="http://ncatlab.org/nlab/show/doctrinal+adjunction" rel="nofollow">doctrinal adjunction</a> for the 2-monad on $[\operatorname{ob} C, \mathrm{Cat}]$ whose algebras are Cat-valued functors $C \to \mathrm{Cat}$. Doctrinal adjunction says that if you have a 2-monad T on a 2-category K, and an adjunction $f \dashv g$ in K, then there is a bijection between 2-cells that make f a colax morphism of T-algebras, on the one hand, and 2-cells that make g a lax morphism on the other; moreover, the entire adjunction lives in T-Alg if and only if f is a pseudo morphism of algebras and the 'colax' part of f's structure 2-cell is the mate of u's (lax) structure 2-cell.</p> <p>In your situation, $F \to G$ is a pseudo morphism of algebras, and the adjunctions $F_U \to G_U \dashv G_U \to F_U$ form a single adjunction in $[\operatorname{ob} C, \mathrm{Cat}]$, so that by the above the right adjoints make up a lax transformation, but the data you have isn't enough to make it pseudo.</p> http://mathoverflow.net/questions/68728/name-of-slice-category-with-2-cells-as-morphisms/68743#68743 Answer by Finn Lawler for Name of "slice" category with 2-cells as morphisms ? Finn Lawler 2011-06-24T15:23:55Z 2011-06-24T15:23:55Z <p>See <a href="http://mathoverflow.net/questions/16554/the-urge-to-combine-1-and-2-morphisms-in-slicing-a-2-category/16559#16559" rel="nofollow">this answer</a> to much the same question. I would call this the 'lax' slice category, although it's not so common a notion that everyone would know what you meant, so maybe you should keep the scare quotes around 'lax'.</p> <p>A propos of Martin's comment, the correct notion of slice 2-category depends on what you're doing -- you might want the strict version, with strictly commuting triangles, or the pseudo version, with invertible 2-cells (this is the strictest one that makes sense for non-strict 2-categories), or this lax version. Or you might want to restrict to (discrete) (op)fibrations as objects.</p> http://mathoverflow.net/questions/67824/why-does-hom-need-an-identity-in-the-definition-of-the-category/67870#67870 Answer by Finn Lawler for Why does Hom need an identity in the definition of the category? Finn Lawler 2011-06-15T15:32:04Z 2011-06-15T16:40:02Z <p>As Fernando points out, you can't talk about isomorphisms in a semicategory, which means that they won't be as much use as categories in describing universes of mathematical objects. But the category of semicategories has a surprisingly interesting relationship to that of categories. There is of course a forgetful functor $\mathrm{Cat} \to \mathrm{Semicat}$, and as Scott says it has a left adjoint that does what you expect. But it also has a <em>right</em> adjoint, which takes a semicategory S to the category of <em>idempotents</em> in S: the objects are idempotents $e \colon a \to a$ and a morphism $e \to e'$ is a morphism $f \colon a \to a'$ in S such that $fe = f = e'f$. So we get a monad on Cat whose unit is the canonical functor from a category to its idempotent-splitting completion, or Cauchy completion, or Karoubi envelope.</p> <p>Böhm, Lack and Street use this framework <a href="http://arxiv.org/abs/1102.4931" rel="nofollow">here</a> to talk about weak Hopf algebras. They show that 'weak monoids' fall naturally out of the formal theory of monads if instead of working directly in a bicategory you Cauchy-complete the hom-categories first.</p> <p>Another application of semicategories and semifunctors is in computer science: Hayashi, <em>Adjunction of semifunctors: categorical structures in nonextensional $\lambda$-calculus</em>, TCS 41, shows how to describe $\lambda$-calculus without the $\eta$-law quite elegantly. I haven't worked it out, but it seems to me that this framework should also give a way of talking about 'weak limits' (the kind with not-necessarily-unique mediating morphisms) in terms of adjunctions.</p> http://mathoverflow.net/questions/67083/reference-request-lax-ends/67092#67092 Answer by Finn Lawler for Reference Request: Lax Ends Finn Lawler 2011-06-06T22:00:04Z 2011-06-06T22:00:04Z <p><a href="http://ehres.pagesperso-orange.fr/Cahiers/EsquissesM.htm" rel="nofollow">This page</a> says that you may be able to get a copy by emailing Andrée Ehresmann.</p> <p>I don't know the exact answer to your question, but if you can't find a reference then it may be worth recalling that:</p> <ul> <li>For Cat-valued F and G, $\mathrm{Nat}(F,G) \simeq \{F,G\}$, the limit of G weighted by F,</li> <li>ends are $\hom$-weighted limits, and</li> <li>there are lax morphism classifiers for 2-functors, meaning that $\mathrm{Lax}(F,G) \simeq \mathrm{Nat}(QF,G)$ for another 2-functor $QF$.</li> </ul> <p>So if you define the lax end $\oint_x T(x,x)$ to be the representative of $\mathrm{Lax}(\hom_K, L(1,T))$, then you get $\oint_x [F x, G x] \simeq \mathrm{Lax}(\hom_K, [F-, G-])$, which is not quite what you want, but it's close.</p> <p>Hope that helps.</p> http://mathoverflow.net/questions/66641/relation-between-monads-operads-and-algebraic-theories/66696#66696 Answer by Finn Lawler for Relation between monads, operads and algebraic theories Finn Lawler 2011-06-01T23:42:41Z 2011-06-01T23:42:41Z <p>Apart from Todd's recommendations, which I'd second, for monads and Lawvere theories there is Nishizawa and Power, <em>Lawvere theories enriched over a general base</em>, JPAA 213, 2009, and the references therein. I'd also recommend Linton's <em>An outline of functorial semantics</em>, LNM 80, 1969 (republished in TAC Reprints). Lots of people have defined 'generalized multicategories' of one sort or another, going back to Burroni in 1971: Cruttwell and Shulman's <em>A unified framework for generalized multicategories</em>, TAC 24(21), 2010, is a useful account.</p> http://mathoverflow.net/questions/66313/monad-arising-from-operad/66324#66324 Answer by Finn Lawler for Monad arising from operad Finn Lawler 2011-05-28T23:19:51Z 2011-05-28T23:19:51Z <p>The answer is no, and it is explained in Appendix C of Leinster's book <em>Higher Operads, Higher Categories</em>.</p> <p>Briefly, a 'plain operad', for Leinster, is a non-symmetric operad in Set; a T-operad for T the free-monoid monad. Leinster shows that if T is a cartesian monad then a T-operad is the same thing as a cartesian monad S equipped with a cartesian transformation $S \to T$. So a plain operad is the same thing as a cartesian monad on Set that is 'augmented over' the free-monoid monad. But not every cartesian monad on Set is so augmented.</p> http://mathoverflow.net/questions/36830/is-there-a-tricategory-of-bicategories-and-biprofunctors/66237#66237 Answer by Finn Lawler for Is there a tricategory of bicategories and biprofunctors? Finn Lawler 2011-05-27T21:59:59Z 2011-05-27T21:59:59Z <p>If you're still interested, I've worked this out on my personal web at nLab <a href="http://ncatlab.org/finnlawler/show/biprofunctor" rel="nofollow">here</a>, with supporting material linked to from <a href="http://ncatlab.org/finnlawler/show/2-categorical+miscellany" rel="nofollow">this page</a>.</p> http://mathoverflow.net/questions/64382/comparing-discrete-fibrations-and-their-duals Comparing discrete fibrations and their duals Finn Lawler 2011-05-09T12:51:01Z 2011-05-11T12:12:19Z <p>I'm not sure if this is the right place to ask this question, but I'll ask it anyway, in the hope that some kindly Australian (true or honorary) is passing by and takes pity on me...</p> <p>In <em>Fibrations in bicategories</em>, Street shows that V-profunctors are exactly the codiscrete cofibrations in the 2-category V-Cat (i.e. the discrete 2-sided fibrations in the opposite of V-Cat). Rosebrugh and Wood later generalized this to well-behaved proarrow equipments.</p> <p>When V=Set, so that V-Cat is Cat, then, codiscrete cofibrations turn out to be essentially the same thing as discrete fibrations. My question is</p> <blockquote> <p>Why is this true? That is, for which bicategories K is DFib(B,A) equivalent to CodCofib(B,A) for all objects A and B?</p> </blockquote> <p>I ask because (aside from curiosity) I'd like to know whether I can expect a 'biprofunctor' $L^{\mathrm{op}} \times K \to \mathrm{Cat}$ to be the same as a discrete fibration in Bicat, or even whether this is true in the strictly Cat-enriched case. In my specific case L and K are locally discrete, if that makes a difference.</p> http://mathoverflow.net/questions/64237/limits-in-functor-categories/64240#64240 Answer by Finn Lawler for Limits in functor categories Finn Lawler 2011-05-07T23:01:34Z 2011-05-07T23:01:34Z <p>What you're asking is whether every limit in a functor category $[B,C]$ is a pointwise limit. The answer is yes if C is complete, but not always otherwise. Kelly gives an example in <em>Basic Concepts of Enriched Category Theory</em>, section 3.3, of a limit in a functor category that is not a pointwise limit.</p> <p>I don't know of any striking examples of always-pointwise properties. One obvious example is that of being an absolute limit or colimit (i.e. one preserved by every functor). The latter are characterized by the fact that a category is Cauchy complete if and only if it has all absolute colimits.</p> http://mathoverflow.net/questions/57460/possible-semantics-for-categorical-co-constness/57470#57470 Answer by Finn Lawler for Possible semantics for categorical co-constness Finn Lawler 2011-03-05T16:55:50Z 2011-05-04T15:37:18Z <p><em>Edit</em>: This is not quite right -- the two definitions of constantness are not equivalent even in the category of sets, which means that the representability argument doesn't quite work. See the <a href="http://ncatlab.org/nlab/show/constant+morphism" rel="nofollow">nlab page</a> for details, but note that they are equivalent if the hom sets $C(X,A)$ are inhabited for every X, or equivalently if A admits a global section.</p> <p>This doesn't answer all of your questions, but...</p> <p>A morphism is constant in your sense iff it is representably constant in (what you say is) Lawvere's sense, that is if $k_* \colon C(X,A) \to C(X,B)$ factors through the terminal object in Set. So if C itself has a terminal object then the two are equivalent.</p> <p>A morphism c is co-constant iff $c^* \colon C(B,Y) \to C(A,Y)$ is constant, and it's easy to see that if C has an initial object then it's equivalent to ask that c factor through it. But in Set the initial object $\emptyset$ is strict, meaning that any morphism into $\emptyset$ is an isomorphism. So there are no non-trivial co-constant morphisms in Set, or indeed in any category with a strict initial object (such as (by a result of Joyal) any cartesian closed category with initial object, like a topos).</p> <p>I can't tell off the top of my head whether your definition would be useful elsewhere, but the above does put some restrictions on what sorts of categories non-trivial co-constant morphisms might turn up in.</p> http://mathoverflow.net/questions/61009/compact-elements-and-continuous-functors/61012#61012 Answer by Finn Lawler for compact elements and continuous functors Finn Lawler 2011-04-08T01:36:53Z 2011-04-08T01:36:53Z <p>Have you tried <a href="http://ncatlab.org/nlab/show/compact+object" rel="nofollow">nLab again</a>?</p> <p>I'm not sure continuous functors are what you're looking for, incidentally. It seems more likely that a 'Scott-continuous functor' should be one that preserves filtered colimits.</p> <p>Replacing the preorders (and metric spaces) of domain theory with (enriched) categories is not a new idea. Have a look at <em>Categories for fixpoint semantics</em> (1978) by Daniel Lehmann, and <em>Solving recursive domain equations with enriched categories</em> (1994) by Kim Ritter Wagner.</p> http://mathoverflow.net/questions/60875/conditional-equality-symbol/60966#60966 Answer by Finn Lawler for conditional equality symbol Finn Lawler 2011-04-07T16:36:46Z 2011-04-07T16:36:46Z <p>Freyd and Scedrov, in their book <em>Categories, Allegories</em>, use for this 'directed equality' a peculiar symbol that they call a <em>Venturi tube</em> and that looks a bit like $\mathrel{>=}$, so that $x \mathrel{>=} y$ means <em>if $x$ is defined then so is $y$ and $x=y$</em>. You can find some discussion at the nLab page on <a href="http://ncatlab.org/nlab/show/Kleene%20equality" rel="nofollow">Kleene equality</a> (the symmetric version of this) and in <a href="http://www.math.ntnu.no/~stacey/Mathforge/nForum/comments.php?DiscussionID=1416" rel="nofollow">this nForum thread</a>.</p> http://mathoverflow.net/questions/60064/condition-of-possibility-co-implication/60071#60071 Answer by Finn Lawler for Condition of possibility = Co-Implication Finn Lawler 2011-03-30T14:30:04Z 2011-03-30T14:30:04Z <p>You've answered your own question, in a way: if $q \hookrightarrow p$ is equivalent to $p \to q$ then the difference is only a matter of notation.</p> <p>You say that</p> <blockquote> <p>in logic many concepts are treated as strongly related (= inter-definable) but each in its own right</p> </blockquote> <p>but until Gentzen came along it was common in logic (and still is in some quarters) to try to get away with as few connectives as possible, so that even implication would be defined away and not studied in its own right. In intuitionistic logic, however, and in non-classical logics generally, it's often not possible to define the usual connectives in terms of a subset of them. In particular, $p \to q$ intuitionistically implies $q \hookrightarrow p$ but not the other way round.</p> <p>I don't know if intuitionistic logicians have studied $\lnot q \to \lnot p$ as a connective in its own right. I do know, though, that I wouldn't call it co-implication -- I would reserve that name for the category-theoretic dual, say $\leftarrow$, of implication, where $p \mapsto p \leftarrow q$ would be the left adjoint to $r \mapsto q \vee r$. Andrzej Filinski studied this in <em>Declarative continuations</em> (LNCS, I forget which volume) and Tristan Crolard in <em>Subtractive logic</em> (Theoretical Computer Science, again I don't have the exact reference to hand).</p> http://mathoverflow.net/questions/28486/an-explicit-description-of-lawveres-segment-in-the-category-of-simplicial-sets/28499#28499 Answer by Finn Lawler for An explicit description of Lawvere's segment in the category of simplicial sets Finn Lawler 2010-06-17T13:22:37Z 2011-03-13T17:40:14Z <p>I've never heard this called 'Lawvere's segment' before, but your $L$ is the subobject classifier in the presheaf topos $[A^{\mathrm{op}},\mathrm{Set}]$. In presheaf toposes generally, the subobject classifier $\Omega$ is the presheaf that sends an object $a$ to the set of all sieves on $a$, i.e. the set of subobjects of $\hom(-,a)$. This means in this case that simplicial subsets $S' \subset S$ are in bijection with simplicial maps $S \to L$.</p> <p>Off the top of my head, I can't be exactly sure what $L$ looks like, but I'd guess it's the constant simplicial set with $L_n = 2$ (the set of truth values) and all maps identities. Then the characteristic map $\chi_{S'} \colon S \to L$ will be the usual $s \mapsto 1$ if $s \in S'$ and $0$ otherwise, while the simplicial subset corresponding to some $\phi \colon S \to L$ will be given by the fibres $\phi_n^{-1}(1)$ in each degree. $S'$'s being a <em>simplicial</em> subset should correspond to $\phi$'s being a simplicial map.</p> <p>I could easily be wrong about that last paragraph, though. There might be more about this in Mac Lane--Moerdijk.</p> <p>Edit: I was wrong -- see the comments below.</p> http://mathoverflow.net/questions/116383/this-is-not-a-category-what-is-it/116431#116431 Comment by Finn Lawler Finn Lawler 2012-12-15T11:54:58Z 2012-12-15T11:54:58Z Link: <a href="http://ncatlab.org/nlab/show/simplicial+T-complex" rel="nofollow">ncatlab.org/nlab/show/simplicial+T-complex</a> http://mathoverflow.net/questions/110057/lost-soul-loneliness-in-pursing-math-advice-needed Comment by Finn Lawler Finn Lawler 2012-10-19T03:24:15Z 2012-10-19T03:24:15Z You have my sympathies too. Many, many people go through something like this, so remember that you're not alone. I will second Gerhard's advice: your university will almost certainly have available, or know where you can find, a counsellor and/or an occupational therapist to help you sort through your problems and improve your work/leisure balance. http://mathoverflow.net/questions/99358/terminology-changing-the-codomain-in-nested-maps-e-g-in-multilinear-algebra Comment by Finn Lawler Finn Lawler 2012-06-12T17:42:05Z 2012-06-12T17:42:05Z (Well, no, not <i>exactly</i> that statement, but part of it anyway: that the mapping acts on morphisms in a way that respects its action on objects.) http://mathoverflow.net/questions/99358/terminology-changing-the-codomain-in-nested-maps-e-g-in-multilinear-algebra Comment by Finn Lawler Finn Lawler 2012-06-12T14:52:12Z 2012-06-12T14:52:12Z This is exactly the statement that $-^A$ is a functor: <a href="http://en.wikipedia.org/wiki/Functor" rel="nofollow">en.wikipedia.org/wiki/Functor</a> ; <a href="http://ncatlab.org/nlab/show/functor" rel="nofollow">ncatlab.org/nlab/show/functor</a> http://mathoverflow.net/questions/93262/which-colimits-commute-with-which-limits-in-the-category-of-sets Comment by Finn Lawler Finn Lawler 2012-04-06T01:23:24Z 2012-04-06T01:23:24Z I don't know the answer to this, but if you come to Cambridge the weekend after next then Marie Bjerrum might be able to tell you: <a href="https://www.dpmms.cam.ac.uk/~jg352/PSSL93.html" rel="nofollow">dpmms.cam.ac.uk/~jg352/PSSL93.html</a> http://mathoverflow.net/questions/88460/group-extensions-and-actions-on-categories Comment by Finn Lawler Finn Lawler 2012-02-14T22:40:21Z 2012-02-14T22:40:21Z (Also, I think H-Set is equivalent not to BH but to [BH, Set].) http://mathoverflow.net/questions/88460/group-extensions-and-actions-on-categories Comment by Finn Lawler Finn Lawler 2012-02-14T22:38:29Z 2012-02-14T22:38:29Z John Baez wrote about Schreier theory in his TWF series: <a href="http://math.ucr.edu/home/baez/week223.html" rel="nofollow">math.ucr.edu/home/baez/week223.html</a> -- he points out a good reference for the Schreier theory of groupoids too: <a href="http://arxiv.org/abs/math.CT/0410202" rel="nofollow">arxiv.org/abs/math.CT/0410202</a> . http://mathoverflow.net/questions/84636/codomain-fibration/85597#85597 Comment by Finn Lawler Finn Lawler 2012-01-13T18:40:15Z 2012-01-13T18:40:15Z 'Simple' is the term that Jacobs uses in his book -- he writes s(C) for your $C^{pr}$. http://mathoverflow.net/questions/83363/is-there-a-nice-application-of-category-theory-to-functional-complex-harmonic-ana/83367#83367 Comment by Finn Lawler Finn Lawler 2011-12-14T19:27:07Z 2011-12-14T19:27:07Z @Todd: I didn't mean to suggest that analysis takes a material approach where, say, algebra takes a structural approach -- all I meant by the last paragraph was that, as Paul Garrett points out in his answer, texts on analysis typically eschew the structural point of view, and that it would be nice if there were more structurally-oriented accounts of analysis to complement the traditional ones. http://mathoverflow.net/questions/83363/is-there-a-nice-application-of-category-theory-to-functional-complex-harmonic-ana/83367#83367 Comment by Finn Lawler Finn Lawler 2011-12-13T22:50:23Z 2011-12-13T22:50:23Z @Yemon: yes, that's the one I was thinking of. http://mathoverflow.net/questions/81982/what-is-the-intuition-of-connections-for-cubical-sets/81987#81987 Comment by Finn Lawler Finn Lawler 2011-11-28T12:21:30Z 2011-11-28T12:21:30Z I got a 404 error when I clicked on it yesterday, but now it works. Odd. Never mind; lots of nice stuff there to get distracted by! http://mathoverflow.net/questions/81982/what-is-the-intuition-of-connections-for-cubical-sets/81987#81987 Comment by Finn Lawler Finn Lawler 2011-11-27T19:04:38Z 2011-11-27T19:04:38Z That URL doesn't seem to work, Tim -- do you mean this one: <a href="http://pages.bangor.ac.uk/~mas010/publicfull.htm" rel="nofollow">pages.bangor.ac.uk/~mas010/publicfull.htm</a> ? http://mathoverflow.net/questions/79825/two-kinds-of-equivalence-conjugate-vs-isomorphic-objects Comment by Finn Lawler Finn Lawler 2011-11-02T15:49:19Z 2011-11-02T15:49:19Z @Hans: the former -- [A is conjugate to B in the category X] iff [(A,X) is isomorphic to (B,X) in */Cat]. http://mathoverflow.net/questions/79825/two-kinds-of-equivalence-conjugate-vs-isomorphic-objects Comment by Finn Lawler Finn Lawler 2011-11-02T13:04:08Z 2011-11-02T13:04:08Z The two notions are not unconnnected -- A and B are conjugate if and only if they are isomorphic in the category of pointed categories (i.e. categories equipped with a distinguished object, with morphisms functors that strictly preserve these). http://mathoverflow.net/questions/79467/dense-subcategory-of-an-algebraic-category-consisting-of-free-objects Comment by Finn Lawler Finn Lawler 2011-10-29T22:37:01Z 2011-10-29T22:37:01Z What A&amp;R show in that remark is that for a finitary theory, the free algebras on the <i>finite</i> sets are dense among all algebras. I'm suggesting that their argument, for the bound $\aleph_0$, might work just as well for your finite bound $N$.