User g. rodrigues - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:47:23Z http://mathoverflow.net/feeds/user/2562 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120729/cosheafification/124973#124973 Answer by G. Rodrigues for Cosheafification G. Rodrigues 2013-03-19T15:37:51Z 2013-03-21T02:48:40Z <p>Unless I have not botched things horribly, here is a general proof that cosheafification exists following the outline J. Curry gave.</p> <p>Fix a target category $\mathcal{A}$ that is complete, cocomplete and locally presentable. Examples include categories of algebraic objects like linear spaces, categories of sheaves, the category of Banach spaces and linear contractions, etc.</p> <p>Let $\Omega$ be a small site and $\mathbf{PCoShv}(\Omega, \mathcal{A})$ the category of precosheaves, that is, functors $\Omega\longrightarrow \mathcal{A}$. The category $\mathbf{PCoShv}(\Omega, \mathcal{A})$ is complete and cocomplete with limits and colimits computed pointwise and locally presentable by [AR, corollary 1.54].</p> <p>Cosheaves are defined dually to sheaves, that is, they send the cone induced by a sieve of the small site to a colimiting cone. Denote the full subcategory of cosheaves by $\mathbf{CoShv}(\Omega, \mathcal{A})$. By the interchange of colimits theorem, $\mathbf{CoShv}(\Omega, \mathcal{A})$ is cocomplete and the inclusion functor $\mathbf{CoShv}(\Omega, \mathcal{A})\longrightarrow \mathbf{PCoShv}(\Omega, \mathcal{A})$ is cocontinuous.</p> <p>proposition: $\mathbf{CoShv}(\Omega, \mathcal{A})$ is accessible. proof: The category of sheaves on a small site is limit-sketchable by [RP, pg. 331]. The same construction yields by duality that the category of cosheaves is the category of models of a colimit sketch, therefore it is accessible by [AR, corollary 2.61]. Q. E. D.</p> <p>proposition: $\mathbf{CoShv}(\Omega, \mathcal{A})$ is complete and locally-presentable. proof: we already know that $\mathbf{CoShv}(\Omega, \mathcal{A})$ is cocomplete and accessible, so the result follows from [AR, corollary 2.47]. Q. E. D.</p> <p>theorem: the full inclusion $\mathbf{CoShv}(\Omega, \mathcal{A})\longrightarrow \mathbf{PCoShv}(\Omega, \mathcal{A})$ has a right adjoint. proof: follows from the previous results and the fact that cocontinuous functors between locally presentable categories have right adjoints. This adjoint functor theorem can be pieced together via the concept of totality. More precisely, locally presentable categories are total by [MK, corollary 6.5 and remark 6.6] and total categories are compact, that is, every cocontinuous functor has a right adjoint -- this is [MK, theorem 5.6].</p> <p>Bibliography: [AR] J. Adamek, J. Rosicky - Locally presentable and accessible categories, Cambridge University Press (1994).</p> <p>[MK] Max Kelly - A survey of totality for enriched and ordinary categories, Cahiers de Top. et Géom. Diff. Catégoriques, 27 no. 2 (1986), p. 109-132</p> <p>[RP] R. Pare - Some applications of categorical model theory, in Categories in Computer Science and Logic, Contemporary Mathematics, vol. 92 (1989)</p> <p>edit: cleaned up, added a couple of references and made mention of the correct adjoint functor theorem.</p> http://mathoverflow.net/questions/35122/factorization-through-ell-1-and-operator-ideals Factorization through $\ell_{1}$ and operator ideals G. Rodrigues 2010-08-10T15:00:50Z 2012-03-09T16:59:06Z <p>Recently, I bumped into the class of operators that factor through $\ell_{1}(X)$ for some set $X$. For now, $X$ is a set with arbitrary cardinality but if it leads to a more concrete answer to my questions below, feel free to restrict $X$ to be countable. The restriction makes little to no difference to what it follows. I should also warn that I am fairly ignorant of operator ideals and Banach space theory, so please be gentle.</p> <p>First some definitions. Define,</p> <p><code>$$\|T\|_{\ell_{1}}= \inf\{\|R\|\|S\|\}$$</code></p> <p>where the infimum is taken over all the factorizations of $T$ as $\xrightarrow{S}\ell_{1}(X)\xrightarrow{R}$. Obviously, <code>$\|T\|\leq \|T\|_{\ell_{1}}$</code>. Define $\mathcal{L}(A, B)$ to be the linear space of operators that factor through some $\ell_{1}(X)$ with the above norm. The little bit of thought that I have dedicated to this has produced the following up to now (and please correct me, if I have fumbled somewhere):</p> <ol> <li><p>We have <code>$\|RTS\|_{\ell_{1}}\leq \|R\|\|T\|_{ell_{1}}\|S\|$</code>. Sketch: obvious.</p></li> <li><p>The normed space $\mathcal{L}(A, B)$ is complete. Sketch: If $(T_{n})$ is <code>$\|,\|_{\ell_{1}}$</code>-Cauchy then it has a uniform limit. To prove that this limit factors through some $\ell_{1}(X)$ note two things. First, if you have a factorization through $\ell_{1}(X)$ as $RS$ and $X\subseteq Y$ then, since $\ell_{1}(X)$ is a norm-1 complemented subspace of $\ell_{1}(Y)$, you can make the factorization to pass through the larger $\ell_{1}(Y)$ without altering $\|R\|\|S\|$. Second, one has the isometric isomorphism, $$\sum_{n}\ell_{1}(X_{n})\cong \ell_{1}(\coprod_{n} X_{n})$$ which allows to take a sequence of factorizations and push them all to a common space $\ell_{1}(X)$. Thus the uniform limit factors through some $\ell_{1}(X)$.</p></li> <li><p>Finite-rank operators factor through $\ell_{1}(X)$. Sketch: all finite-dimensional spaces are linearly homeomorphic to $\ell_{1}(n)$. These first three conditions taken together mean that $(A, B)\mapsto \mathcal{L}(A, B)$ is an operator ideal (or Banach ideal, I am uncertain of the official terminology).</p></li> <li><p>Each $T$ is completely continuous. Sketch: a sequence in $\ell_{1}(X)$ lives inside a copy of $\ell_{1}$. The Schur property of $\ell_{1}$ gives the result.</p></li> </ol> <p>Now for my first batch of questions: can this class of operators be characterized? Any more salient properties of these operators? And what about the norm <code>$\|,\|_{\ell_{1}}$</code>, is there some other more enlightening description of it? How far is it from the operator norm?</p> <p>The second batch of questions is related to what are the properties required of a full subcategory $C$ of the category of Banach spaces so that one obtains an operator ideal by factorizing operators through it. An obvious example is the ideal of weakly compact operators that by Davis-Figiel-Johnson-Pelczynski is the class of operators that factor through reflexive spaces. My guess is that something like $\omega_{1}$-filteredness of $C$ with $\omega_{1}$ the first uncountable ordinal, is enough for the argument to go through, but I am sure someone smarter and more knowledgeable has already thought about this.</p> <p>If you have appropriate references, that would be great; extra kudos if available online. Next September I will have access to a library and plan to get my hands on the Defant, Floret monograph <em>Tensor norms and operators ideals</em> -- not a very cheerful prospect actually, as the book looks rather daunting. The book <em>Absolutely summing operators</em> by Diestel, Jarchow and Tonge should also be useful, but alas, last time I checked it was not available.</p> <p>Regards, TIA, G. Rodrigues</p> http://mathoverflow.net/questions/64298/short-five-lemma-in-banach-spaces Short five lemma in Banach spaces G. Rodrigues 2011-05-08T13:16:38Z 2011-05-08T13:38:06Z <p>Denote by $\mathbf{Ban}$ the category of Banach spaces and bounded linear maps and by $\mathbf{Banc}$ the subcategory of Banach spaces and linear contractions. The isomorphisms of $\mathbf{Ban}$ are the bounded linear bijections (open mapping theorem) and in $\mathbf{Banc}$ they are the isometric (linear) isomorphisms (easy computation). $\mathbf{Ban}$ is additive and has all finite limits and finite colimits, so we can speak of (short) exact sequences.</p> <p>proposition: the <a href="http://en.wikipedia.org/wiki/Short_five_lemma" rel="nofollow">short five lemma</a> holds in $\mathbf{Ban}$. proof: courtesy of the open mapping theorem, the diagram-chase proof for abelian groups transfers verbatim.</p> <p>Now for my question:</p> <p>Q: does the five short lemma also hold in $\mathbf{Banc}$?</p> <p>Since $\mathbf{Banc}$ is not additive, a few words about the definitions. The crucial thing to notice is that kernels and cokernels in $\mathbf{Ban}$ are also kernels and cokernels in $\mathbf{Banc}$. For the sake of illustration, take the case of cokernels. In $\mathbf{Ban}$ they are given by the quotient map $\pi:B\to B/M$ with $M$ a closed subspace of $B$. If $T$ is a bounded linear map whose kernel contains $M$, then it factors uniquely through $\pi$ and the norm of the factorization equals the norm of $T$. But this is precisely what is needed for $\pi$ to be a cokernel in $\mathbf{Banc}$.</p> <p>So now, we have two short exact sequences connected by maps $g$, $f$ and $h$ with $g$ and $h$ isometric isomorphisms and $f$ a linear contraction. The short five lemma holds if $f$ is an isometric isomorphism -- is there a way to draw diagrams? In any case, just look at <a href="http://en.wikipedia.org/wiki/Short_five_lemma" rel="nofollow">short five lemma</a>.</p> <p>Start by noticing that by the above proposition $f$ must be a bijection and thus it has an inverse. What we need to prove is that this inverse is contractive. Alternatively, since we already know that $f$ is injective, the following easy result offers another route.</p> <p>proposition: a map $f$ is a quotient iff it is surjective on open unit balls iff it is dense on unit balls.</p> <p>Applying the above and chasing down an element of the open unit ball of the codomain of $f:B\to B'$, we get is that there is a $b$ in the open unit ball of $B$ and an $m$ in the kernel of the quotient in the top short exact sequence such that $f(b + m) = b'$. But of course, this does not get us very far.</p> <p>At this point, I am beginning to suspect that the lemma does not hold, but I have been unable to rig a counterexample. The hypothesis is very strong and rules out the "usual" gang of suspects.</p> <p>regards, G. Rodrigues</p> http://mathoverflow.net/questions/45466/when-is-a-left-kan-extension-closed When is a left Kan extension closed? G. Rodrigues 2010-11-09T17:12:01Z 2010-11-09T23:57:47Z <p>Fix a complete, cocomplete, symmetric monoidal closed category $\mathcal{V}$. I will also assume that there is a forgetful functor $\mathcal{V}\to \mathbf{Set}$ with a left adjoint. By standard results, this adjunction lifts to a 2-adjunction between the 2-category of categories and the 2-category of $\mathcal{V}$-categories; IOW, we can speak of the free $\mathcal{V}$-category on a category.</p> <p>Now, let $\mathcal{A}$ be a (small) symmetric monoidal closed $\mathcal{V}$-category. Then the $\mathcal{V}$-presheaf category $\mathbf{PShv}(\mathcal{A})$ is symmetric monoidal closed when endowed with the Day convolution structure and the Yoneda embedding is symmetric monoidal closed. Furthermore, any symmetric monoidal $\mathcal{V}$-functor $F:\mathcal{A}\to \mathcal{B}$ with $\mathcal{B}$ cocomplete, lifts uniquely (up to unique isomorphism) to a cocontinuous symmetric monoidal functor $\mathbf{PShv}(\mathcal{A})\to \mathcal{B}$ by taking the left Kan extension $L_{Y}(F)$ of $F$ along Yoneda $Y$.</p> <p>I can prove all this to myself without great effort. What has left me stumped is the following question:</p> <p>Q: assume $\mathcal{B}$ and $F$ closed, is $L_{Y}(F)$ closed?</p> <p>Since Yoneda is closed the converse is trivially true. If the answer is negative in general, is there any criteria to have a positive answer? If it helps the answer, my use case is when $\mathcal{A}$ is the free $\mathcal{V}$-category on a Heyting algebra (e.g. the open sets of a topological space) with monoidal structure the intersection. Since in a lattice there is at most one arrow between two objects, Day convolution degenerates into the pointwise tensor product.</p> <p>Thanks in advance, regards, G. Rodrigues</p> http://mathoverflow.net/questions/43311/sheaves-and-cosheaves/43419#43419 Answer by G. Rodrigues for sheaves and cosheaves G. Rodrigues 2010-10-24T19:23:12Z 2010-10-24T19:23:12Z <p>If you have the stomach for hard topos theory a good reference is</p> <p>Singular coverings of toposes -- M. Bunge, J. Funk</p> <p>The first chapter is probably enough to answer your question. The upshot is that if you have a site, then the category of cosheaves can be identified with the category of cocontinuous functors on the category of sheaves. This should give you a pretty good idea of what operations you can perform on cosheaves.</p> <p>Btw, in this context, cosheaves are also called Lawvere distributions -- distributions because of the analogy with the Riesz representation theorem that identifies measures (cosheaves) with linear functionals (cocontinuous functors).</p> <p>Hope it helps, regards, G. Rodrigues</p> http://mathoverflow.net/questions/39626/is-there-a-general-setting-for-self-reference/39776#39776 Answer by G. Rodrigues for Is there a general setting for self-reference? G. Rodrigues 2010-09-23T17:24:37Z 2010-09-23T17:24:37Z <p>I am not quite sure if it fits the bill but you can also check out:</p> <p><a href="http://arxiv.org/abs/math/0305282" rel="nofollow">N. Yanofsky - A Universal Approach to Self-Referential Paradoxes, Incompleteness and Fixed Points</a></p> http://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view/39534#39534 Answer by G. Rodrigues for Integrals from a non-analytic point of view G. Rodrigues 2010-09-21T19:15:15Z 2010-09-21T19:15:15Z <p>My answer here is realy just a footnote to Paul Siegel's excellent answer, but it has become too long to fit in a comment box. Integrals are siamese brothers to measures; leaving them out seems rather perverse to me. Anyway, here is how I think of integrals. The objective here is to tackle the "categorical" part; the analytical viewpoint will necessarily obtrude. But bear with me a little, this is a somewhat long post, with a punchline at the end.</p> <p>Fix a Boolean algebra $\Omega$. A map $\nu: \Omega\to V$ with values on a linear space $V$ is finitely additive if $\nu(E\cup F)= \nu(E) + \nu(F)$ for every disjoint $E, F$. Denote the linear space of such maps by $\mathbf{A}(\Omega, V)$.</p> <p>Theorem 1: There is a linear space $\mathbf{S}(\Omega)$ and a finitely additive map $\chi:\Omega\to \mathbf{S}(\Omega)$ universal among all finitely additive maps. proof: just follow the universal property and do the obvious thing (yeah, I suppose you can use the adjoint functor theorem but why would you?).</p> <p>The universal property recast in terms of representability gives the natural isomorphism ($\mathbf{Vect}$ is the category of linear spaces)</p> <p>$$\mathbf{A}(\Omega, V)\cong \mathbf{Vect}(\mathbf{S}(\Omega), V)$$</p> <p>Before continuing, let me elucidate a little bit of the structure of $\mathbf{S}(\Omega)$.</p> <p>Theorem 2: Let $f$ be a non-zero element of $\mathbf{S}(\Omega)$. Then there are non-zero scalars $k_n$ and non-zero, pairwise disjoint $E_n\in \Omega$ such that $f= \sum_n k_n\chi(E_n)$. Furthermore, if $\nu$ is a finitely additive map and $\widehat{\nu}$ the map induced on $\mathbf{S}(\Omega)$ by universality, then $\widehat{\nu}(f)= \sum_{n}k_n \nu(E_n)$.</p> <p>To put it simply, $\mathbf{S}(\Omega)$ is the linear space of "simple functions on $\Omega$" and the map induced by universality is the integral. Now use theorem 2 to put a norm on $\mathbf{S}(\Omega)$:</p> <p>$$\|\sum_n k_n\chi(E_n)\|= \max\{|k_n|\}$$</p> <p>Denote the completion by $\mathbf{L}_{\infty}(\Omega)$. On the other hand, put on the linear subspace of the <em>bounded</em> finitely additive maps $\Omega\to B$ with $B$ a Banach space, the semivariation norm (which I am not going to define). Denote this space by $\mathbf{BA}(\Omega, B)$. Then:</p> <p>Theorem 3: There is a bounded finitely additive $\chi:\Omega\to \mathbf{L}_{\infty}(\Omega)$ universal among all bounded finitely additive maps.</p> <p>Once again, recasting the universal property in terms of representability, we have a natural isometric isomorphism ($\mathbf{Ban}$ is the category of Banach spaces and bounded linear maps).</p> <p>$$\mathbf{BA}(\Omega, B)\cong \mathbf{Ban}(\mathbf{L}_{\infty}(\Omega), B)$$</p> <p>It is illuminating to write down what does the naturality of the isomorphism implies: I will leave that as an exercise to the reader.</p> <p>Note that $\mathbf{L}_{\infty}(\Omega)$ is a Banach algebra in a natural way (use theorem 2 or juggle the universal property around. Or "cheat" all the way up and use Stone duality) and that $\chi$ is <em>spectral</em> or <em>multiplicative</em>, that is, $\chi(E\cap F)= \chi(E)\chi(F)$. Theorem 3 can now be extended by saying that $\chi$ is universal among all spectral measures (with values in Banach algebras). This extension is trivial given theorem 3.</p> <p>The case of <code>$\mathbf{L}_{\infty}(\Omega)$</code> does not need the introduction of measures but of course, this is not so with <code>$\mathbf{L}_{1}$</code>. So fix a finitely additive, positive $\mu:\Omega\to \mathbb{R}$. For the sake of simplification I will assume $\mu$ non-degenerate, that is, $\mu(E)= 0$ implies $E= 0$ (otherwise, you will have to take some quotient along the way). A finitely additive $\nu:\Omega\to B$ with $B$ a Banach space is $\mu$-Lipschitz if there is a constant $C$ such that $\|\nu(E)\|\leq C\mu(E)$ for all $E$. The infimum of all the constants $C$ in the conditions of the inequality gives a norm and a normed space I will denote by $\mathbf{LA}(\Omega, \mu, B)$. On the other hand, endow $\mathbf{S}(\Omega)$ with the norm</p> <p>$$\|\sum_n k_n\chi(E_n)\|= \sum_n |k_n|\mu(E_n)$$</p> <p>and denote the completion by $\mathbf{L}_{1}(\Omega, \mu)$.</p> <p>Theorem 4: There is a finitely additive, $\mu$-Lipschitz $\chi:\Omega\to \mathbf{L}_{1}(\Omega, \mu)$ universal among all such maps.</p> <p>Before the conclusion let me address a few points.</p> <ol> <li><p>Measurable spaces are not needed. If you really want them, use Stone duality (that is, points count for nothing in measure theory so why not leave them out, heh?).</p></li> <li><p>Finitely additive measures are really not that much more general than $\sigma$-additive ones. I will leave this cryptic comment as is, and just note that once again, Stone duality is the key here.</p></li> <li><p>I am <em>not</em> advocating this approach to be used in teaching (unless your goal is to flunk and befuddle as many undergrads as humanly possible). For one, you need some functional analysis under the belt (Banach spaces, completions, semivariation, etc.). Intuition is very hard to come by as I have thrown away the measurable spaces without which THE most important example, Lebesgue measure (arguably, the core of a first measure theory course) cannot be constructed. The whole logic of the approach only makes sense after you have seen other instances of categorical thinking at work. I am sure you can think of other objections.</p></li> </ol> <p>How categorical is this approach? Certainly, the universal properties of the respective spaces are central to the whole business and at least, they make clear that some results are really just a consequence of abstract nonsense. In the words of P. Freyd, category theory is doing what it was invented for: to make the easy things really easy (or some such, my memory is lousy). For example, the Bochner vector integral is obtained simply by taking the projective tensor product. Fubini and Fubini-Tonelli on the equality of iterated integrals are other notable cases of categorical thinking at work. Now pepper with Stone duality and a few more tools (e.g. Hahn-Banach and the compact-Hausdorff monad) and you can get (a slight variation of) the Riesz representation theorem for compact Hausdorff spaces. Use the proper compactifications and generalize to wider classes of topological spaces. Or use Loomis-Sikorski to get Vitali-Hahn-Saks in one line (but this is really "cheating" as the crucial step in establishing Loomis-Sikorski is essentially the same as the one to establish Vitali-Hahn-Saks: a Baire category-theorem application). And a few more.</p> <p>But once again, how categorical is this approach? Well, the argument is categorical enough to be generalized to symmetric monoidal closed categories. See R. Borger -- A categorical approach to integration, in the 23rd volume of TAC available online. For the modifications needed to internalize the arguments to a topos (and much more) see the delightful Phd thesis of Mathew Jackson "A Sheaf theoretic approach to measure theory" -- this is available online, just google for it. Oh, by the way, you can see (almost) everything I have explained above in volume 3 of D. Fremlin's measure theory 5-volume series, also available online.</p> http://mathoverflow.net/questions/35335/collapsing-objects-in-a-category/35352#35352 Answer by G. Rodrigues for Collapsing objects in a category G. Rodrigues 2010-08-12T13:37:21Z 2010-08-12T13:37:21Z <p>Short answer: yes. There is a special class of 2-categorical limits called iso-inserters that does the trick. The paper to check is M. Kelly's "Elementary observations on 2-categorical limits", Bull. Austr. Math. Soc. 39 (1989), 301-317.</p> <p>Instead of explaining what these are let me go about explaining how you would make two objects $a$ and $b$ isomorphic. First pass to the underlying graph of the category and insert two edges $a\to b$ and $b\to a$ then take the free category of this new graph. You have a graph morphism from the original category into this new graph. Quotient the category to force the graph morphism to be a functor. Now quotient the category again to make the edges you inserted to be mutually inverse. There is only a small snag to this construction: the category you end up may not be locally small, because inserting isomorphisms may create a proper class of new morphisms. This is where things like "calculus of fractions" come in.</p> <p>Hope it helps, regards, G. Rodrigues</p> http://mathoverflow.net/questions/5786/how-do-i-check-if-a-functor-has-a-left-right-adjoint/5820#5820 Answer by G. Rodrigues for How do I check if a functor has a (left/right) adjoint? G. Rodrigues 2009-11-17T14:36:13Z 2010-08-09T16:33:14Z <p>There are two parts to this answer.</p> <ol> <li><p>First, a functor must be continuous (cocontinuous) to have a left (right) adjoint. Most of the times, it is easy to check that a functor does <em>not</em> preserve (co)limits and thus it cannot have a a left (right) adjoint.</p></li> <li><p>(co)continuity is not enough to actually prove that a functor has the required adjoint, but it is <em>almost</em> good enough. Let me elaborate on this. If you have a functor $F:P\to Q$ between complete partial orders (and thus cocomplete) then it is an easy exercise to construct a left adjoint by taking a $\sup$ of an appropriate subset. This can be generalized in a straightforward way to any functor by taking an appropriate (co)limit. The bad news is that this (co)limit is in general over a large category so it may not exist. This is where the so-called solution-set conditions come in; they are way to trim down this large category to a small one.</p></li> </ol> <p>As many people already said there are various variations of this type of conditions, from the more general but also very cumbersome to check solution-set condition to easier conditions which combine some form of well-poweredness (each object has only a set of subobjects -- or quotients, whatever the case may be) with the existence of a small separating (or generating) set. One that guarantees the existence of a right adjoint and that sticks out particularly in my memory is the existence of a small dense subcategory -- check chapter V of Kelly's book on enriched category for the precise details. It is particularly memorable, because many categories come with god-given small dense categories like presheaf categories (courtesy of Yoneda) and sheaf categories (because dense composed with left adjoint is dense).</p> <p>Later edit: many people have complained about the limited usefulness of the adjoint functor theorems in that in many cases there is a direct, and thus much more enlightening, construction. But there <em>are</em> situations where such a direct construction is not available. One that I came across recently is when studying P. Johnstone's book <em>Stone spaces</em>, more precisely chapter III and the section on Manes' theorem about the monadicity of the category of compact Hausdorff spaces. In the sequel, P. Johnstone proves another result due to Manes, the fact that category of algebras (in the sense of universal algebra) in the category of compact Hausdorff spaces is also monadic. He remarks that one has to use the GAFT (and Beck's monadicity theorem) in this case, because there is no easy direct description of the left adjoint. Later in the book (somewhere, I am quoting from memory and do not have the book by me), he argues why there is no simple recipe for the left adjoint.</p> http://mathoverflow.net/questions/5497/what-are-the-qualities-of-a-good-math-teacher/33000#33000 Answer by G. Rodrigues for What are the qualities of a good (math) teacher? G. Rodrigues 2010-07-22T20:17:26Z 2010-07-22T20:17:26Z <p>Initially, this was supposed to be a comment about point 3) in Peter Tingley's answer, about a math teacher being able to "convey the beauty of the subject", but it got too long so I turned it into an answer.</p> <p>While I generally agree with it, I think it is also a tad too idealistic. Dr. Johnson, in one of his invectives against the Scots, barked that "much may be made of a Scot if he be caught young". Contrapositively, if a student arrives at university without any appreciation for Mathematics, there is only the slimmest of chances that he will gain it there. And teenagers? Mathematics has to compete in a teenager's mind with Lady Gaga and porn sites; with mtv rap and online gaming; with entertainments that range from the irrational to the inane. Call me an elitist cynic, but the plain matter of fact is that the majority of students will be completely unmoved by the beauty of mathematics, much as great literature can only be understood and enjoyed by a relative minority within each generation. I guess what I am saying is that a quality that a good teacher must have is a thick-skin and somehow, by whatever means, keep the flame alive inside his own heart so that when that special, receptive student does come along, he will be able to kindle the fire and infect him with the rapturous love for Mathematics. Or in the words of Samuel Beckett, "Try Again. Fail again. Fail better."</p> http://mathoverflow.net/questions/32990/double-dual-space-of-a-c-algebra-a/32993#32993 Answer by G. Rodrigues for Double dual space of a C* algebra A G. Rodrigues 2010-07-22T19:06:56Z 2010-07-22T19:06:56Z <p>This has nothing to do with operator algebras as it is well-known that if $B$ is a linear topological space (local convexity and Hausdorff-ness hypothesis included), the topological dual of $B^{\ast}$ endowed with the wk*-topology is $B$ itself, that is, every wk*-continuous functional on $B^{\ast}$ is an evaluation functional. For the proof, see for example the first section of chapter V in J. B. Conway's Functional Analysis textbook.</p> <p>Regards, G. Rodrigues</p> http://mathoverflow.net/questions/32071/is-the-dual-notion-of-a-presheaf-useful/32358#32358 Answer by G. Rodrigues for Is the dual notion of a presheaf useful? G. Rodrigues 2010-07-18T13:19:59Z 2010-07-18T13:19:59Z <p>Let me try and take a stab at this question. I will give not so much as an answer to your question, but more of a rambling collection of remarks. The post turned out to be <em>much longer</em> than I wanted, because as B. Pascal once complained, I did not have the time to write a shorter one. Expect some inconsistencies, a lot of hand-waving, etc.</p> <p>To put some order in my thoughts, I will try to argue the following four points. Fix a category $\mathcal{A}$, which is the category "of interest". Then:</p> <p>(1) Functors into $\mathcal{A}$ are interesting.</p> <p>(2) Functors into $\mathcal{A}$ tell you everything about $\mathcal{A}$.</p> <p>(3) Part of the perceived asymmetry follows from the fact that in many cases the interesting category $\mathcal{A}$ is equivalent to a presheaf category (or some localization thereof).</p> <p>(4) There is not really an asymmetry between functors out of and functors into $\mathcal{A}$, but more of a duality.</p> <p>The first point is easy to settle, as Qiaochu Yuan already gave a class of interesting examples: for a category of interest $\mathcal{A}$ and any category $\mathcal{I}$, a functor $\mathcal{I}\to \mathcal{A}$ is a diagram of shape $\mathcal{I}$ in $\mathcal{A}$, so hell yeah, functors into $\mathcal{A}$ are important. You may think this a rather pedestrian example, but below I give more examples.</p> <p>To explain (2), instead of working with categories, that is, in the $2$-category of categories, let us go one level down to the category of sets. A set $X$ is determined completely by its elements, that is, maps $\ast\to X$ where $\ast$ is a singleton set (any one will do: for the sake of determinacy, take the singleton set comprised of the element $\emptyset$). Now in a general category, we cannot speak of elements, but we can (and do) speak of <em>generalized elements</em>.</p> <p><b>Definition:</b> Let $\mathcal{A}$ be a category and $a$ an object of $\mathcal{A}$. A <em>generalized element</em> of $a$ is a map $x\colon d\to a$. This is also written symbolically as $x\in_{d} a$.</p> <p>We do not need generalized elements to develop category theory, but personally, I found them useful to build upon the intuition gained from working with more "concrete" categories. A very nice discussion of generalized elements is in Awodey's book on category theory. Now the kicker: Yoneda's lemma tells us that if we know all the generalized elements of an object $a$ then we know everything about $a$, including of course, the arrows <em>out of</em> $a$ (note: and by duality, if we know all the arrows out of $a$ we will know everything about $a$). Two possible objections may be raised:</p> <ol> <li><p>The original example involves a $2$-category: does not make much of a difference, as my reply to Kevin's comment (see above) still applies. And besides, there are $2$-categorial versions of Yoneda lemma.</p></li> <li><p>Yoneda's lemma works ok, but you need the knowledge of all generalized elements and these range over a potentially proper class of objects: that is true, but in virtually every interesting category one can trim down this proper class to a (small) set, even a singleton set. I will not spell out the proper definitions; they are intimately tied with "smallness" conditions and the adjoint functor theorems (and yes, the category of (small) categories satisfies them).</p></li> </ol> <p>For (3), let $\mathcal{C}^{\mathcal{B}}$ be a functor category. For size reasons, $\mathcal{B}$ has to be small (note: if you have no scientifick problems with creation ex nihilo, you can always spawn a larger universe by invoking the axiom of universes and sidestep this particular size issue). Since functor categories inherit most of the good properties of the codomain category, you will want $\mathcal{C}$ to be as good behaved as possible (e.g. complete, cocomplete, abelian, symmetric monoidal closed, etc.). It is not true that the structure of $\mathcal{B}$ is irrelevant for the structure of $\mathcal{C}^{\mathcal{B}}$, but it is true that in general, $\mathcal{B}$ only needs a bare minimum of structure. This asymmetry between the domain and the codomain categories is reinforced by the fact that many of the interesting categories $\mathcal{A}$ are equivalent to functor categories, even presheaf categories (or some localization of them). Here are two examples.</p> <ol> <li><p>Let us consider the category of groups $\operatorname{Grp}$, undoubtebly a category "of interest". There is a category $\operatorname{Th}(\operatorname{Grp})$ that has all finite products such that $\operatorname{Grp}$ is equivalent to the category of product-preserving functors $\operatorname{Th}(\operatorname{Grp})\to \operatorname{Set}$ where $\operatorname{Set}$ is the category of sets. This equivalence can be generalized to a <em>very large class</em> of categories of "algebraic flavor" and even some that at first sight do not bear the least resemblance to "algebraic categories". A few extra remarks about this example. First, the category $\operatorname{Th}(\operatorname{Grp})$ is a category constructed to make the equivalence work (it is the free category with products on a group object), in other words, it does not exactly fall within the class of categories "of interest". It's a similar to the example of diagram categories in (1), where the domain, a free category on a graph, is just a categorial construction to make the identification of diagrams with functors, not an interesting category by itself. Second, by replacing $\operatorname{Set}$ by another category $\mathcal{B}$ (with at least finite products), you can now speak of groups in $\mathcal{B}$. This gives another class of examples where functors into a category are interesting.</p></li> <li><p>If $(\mathcal{C}, \mathcal{J})$ is a site (a category with a Grothendieck topology), by first taking the category of presheaves and then a suitable localization, one obtains the category of sheaves. This produces a host of geometric categories, like manifolds and schemes.</p></li> </ol> <p>Much like in example 1, the interest is not so much on $(\mathcal{C}, \mathcal{J})$ and even less in the codomain, which is usually the category of sets (other categories for codomain also work, but the categorial requirements for everything to work smoothly are fairly strong), but in the (pre)sheaf category.</p> <p>For my last point (4), a lot could be said, but I will just point you to two articles by F. W. Lawvere in the TAC Reprints, "Metric Spaces, Generalized Logic and Closed Categories" and "Taking categories seriously" (google for them, they are available online). In them, Lawvere makes several remarks about the duality between spaces and algebras of functions which are directly relevant to your question. To show that there is not so much an asymmetry but a duality between functors into and out of, let me give you two examples.</p> <ol> <li><p>In your (that is, the OP) last post, you speak about stacks. A stack on a category $\mathcal{A}$ can be defined as a functor with values in $\mathcal{A}$ satisfying some conditions -- this is the fibered category approach to stacks. But a stack can also be defined as weak $2$-functor with values in the $2$-category of categories (satisfying some extra conditions). For reasons that I will not explain, the first approach is better, but nevertheless the point should be clear. There are actually many examples of this "duality", that identifies some category of functors into $\mathcal{A}$ with some category of functors out of $\mathcal{A}$.</p></li> <li><p>Let me end up with an example from physics that further illustrates this duality. Quantum field theories are notoriously hard objects to define (let alone study). Several years ago, V. Turaev defined the notion of a Homotopy Quantum Field Theory, HQFT for short (check his papers in the arxiv if you are interested) which is a very simple, "toy" example of a QFT. If $X$ is a topological space, we can define a category that has for objects manifolds $M$ equipped with a homotopy class of maps $g\colon M\to X$ and a morphism $(M, g)\to (N, h)$ is a cobordism $W\colon M\to N$ with a homotppy class of maps into $X$ extending $g$ and $h$. An HQFT is a monoidal functor from this category into another monoidal category (usually, the category of finite-dimensional complex linear spaces). I am omitting lots of details, but the gist is that an HQFT gives us invariants of manifolds $M$ by mapping $M$ into some fixed background space $X$. But we can turn things around, for the category of $X$-HQFT's is a (functorial) invariant of the homotopy type of $X$, an invariant cooked up by mapping manifolds into $X$.</p></li> </ol> <p>Hope it helps, regards, G. Rodrigues</p> http://mathoverflow.net/questions/32149/fatous-lemma-and-the-bounded-convergence-theorem/32156#32156 Answer by G. Rodrigues for Fatou's Lemma and the bounded convergence theorem. G. Rodrigues 2010-07-16T11:40:24Z 2010-07-16T12:11:46Z <p>I am not sure if the question fits MO standards as it is an elementary measure theory question (and if it's not, expect to be tazered by the MO police). Here goes an answer, anyway.</p> <p>To expand on Robin Chapman's comment, first, the theorem as stated is false withouth the assumption that $E$ has finite measure. The correct generalization is the Lebesgue dominated convergence where the sequence $f_{n}$ is such that there is an integrable $g$ such that $\|f_n{x}\|\leq g$.</p> <p>To see why, it fails without the boundedness condition consider the sequence of intervals $E_n= [0, 1/n]$ and take the sequence $(n\chi(E_n))$ where $\chi(E_n)$ is the characteristic function (or indicator functions) of $E_n$. This sequence converges pointwise to $0$ but</p> <pre><code>$\int n\chi(E_n) = 1$ </code></pre> <p>so that the sequence of integrals does not converge to $0$. What is happening is that you are shrinking the support of the functions but the same time increasing their "amplitude" so that the two cancel each other out and the integral stays constant while the functions themselves converge to zero. The uniform bound on the sequence, prevents their "amplitudes" of running off to infinity and screwing up the integrals.</p> <p>Examples can be concocted where the convergence is uniform instead of just pointwise. The idea is to do the reverse of the previous example: shrink the amplitude of the functions (to guarantee their uniform convergence) while enlarging their support. This will need a measure space of infinite measure. I will leave that as an exercise.</p> <p>Regards, G. Rodrigues</p> http://mathoverflow.net/questions/22990/choice-vs-countable-choice Choice vs. countable choice G. Rodrigues 2010-04-29T14:56:34Z 2010-07-04T05:09:43Z <p>This question arose after reading the answers (and the comments to the answers) to <a href="http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice" rel="nofollow">http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice</a>.</p> <p>First things first. In <em>my intuitive conception of the hierarchy of sets</em>, the axiom of choice is obviously true. I mean, how can the product of a family of non-empty sets fail to be non-empty? I simply cannot fathom it. Now, I understand that there are people who disagree with me; a mathematician of a (more) constructive persuasion would reply that mathematical existence is constructive existence. Well, we can agree to disagree. And besides, the distinction between constructive and non-constructive proofs is very much worth having in mind. First, because constructive proofs usually give more information and second, there are many contexts where AC is not available (e.g. topoi).</p> <p>A second (personal) reason for championing AC is a pragmatic one: it allows us to prove many things. And "many things" include things that physicists use without a blink. Analysis can hardly get off the ground without some form of choice. Countable choice (ACC) or dependent countable choice (ACDC) is enough for most elementary analysis and many constructivists have no problem with ACC or ACDC. For example, ACC and the stronger ACDC are enough to prove that the countable union of countable sets is countable or Baire's theorem but it is not enough to prove Hahn-Banach, Tychonoff or Krein-Milman (please, correct me if I am wrong). And this is where my question comes in. In one of the comments to the post cited above someone wrote (quoting from memory) that the majority of practicing mathematicians views countable choice as "true". I have seen this repeated many times, and the way I read this is that while the majority of practicing mathematicians views ACC as "obviously true", a part of this population harbours, in various degrees, some doubts about full AC. Assuming that I have not misread these statements, why in the minds of some people ACC is "unproblematic" but AC's validity is not? What is the intuitive explanation (or philosophical reason, if you will) why making countably infinite choices is "unproblematic" but making arbitrarily infinite choices is somehow "more suspicious" and "fraught with dangers"? I for one, cannot see any difference, but then again I freely confess my ignorance about these matters. Let me stress once again that I do not think for a moment that denying AC is "wrong" in some absolute sense of the word; I just would like to understand better what is the obstruction (to use a geometric metaphor) from passing from countably infinite choices to arbitrarily infinite ones.</p> <p>Note: some rewriting and expansion of the original post to address some of the comments.</p> http://mathoverflow.net/questions/27901/does-cauchy-continuity-imply-uniform-continuity-no/27948#27948 Answer by G. Rodrigues for Does Cauchy continuity imply uniform continuity? [No.] G. Rodrigues 2010-06-12T17:14:50Z 2010-06-12T17:14:50Z <p>Several people have already given examples to the effect that preservation of Cauchyness is not enough to prove that a map is uniformly continuous. It is still possible however, to characterize uniform continuity in terms of sequences. In case you are interested here goes the result (for metric spaces only, for uniform spaces you would need nets (or filters)).</p> <p>Theor: Let $f:X\to Y$ be a map between metric spaces (both metrics denoted by d). Then $f$ is uniformly continuous iff for every pair of sequences $(x_n)$ and $(z_n)$ in $X$ such that $d(x_n, z_n)$ converges to $0$, then $d(f(x_n), f(z_n))$ converges to zero.</p> <p>Proof: exercise to the reader.</p> http://mathoverflow.net/questions/25141/elementary-vector-measure-question-what-am-i-doing-wrong Elementary vector measure question: what am I doing wrong? G. Rodrigues 2010-05-18T15:15:17Z 2010-05-18T15:54:25Z <p>This is an edited post of a post I made on sci.math (e.g. to fit MO markup) with an elementary question on vector measures. Since it is almost a week and I have received no answers, I am trying here. Below, I will "prove" a theorem that is false (because it has a simple counter example) but I cannot find where the flaw in the proof is. This is an elementary question, so I am not sure if it fits MO standards. If it does not, feel free to close it down -- I will leave that judgement to the moderators. I am also certain that once someone points out the hole in my argument I am sure to slap myself on the forehead, cry out what a complete idiot I am and vow not to show my face in public for the next few years. But at this point, this problem is driving me bonkers, and I much prefer my sanity over my reputation (if I have any).</p> <p>WARNING: long post ahead.</p> <p>SOME BACKGROUND: All Banach spaces are over the real field. Let $\Omega$ be a Boolean algebra. In a harmless abuse of notation, the top element of $\Omega$ will be denoted by $\Omega$. By a measure on $\Omega$ I mean a <em>finitely additive</em> map $u$ on $\Omega$ with values in a Banach space. If the codomain $B$ of $u$ is the real field I will call it a scalar measure. Note in particular that (scalar) measures do <em>not</em> take infinite values. Neither $\sigma$-completeness nor $\sigma$-additivity will be used anywhere (I told you it was an elementary question).</p> <p>Recall that if $u$ is a measure, then its <em>variation</em> $|u|$ is the (possibly infinite) quantity,</p> <p>$$|u|: E\mapsto \sup {\sum_{F\in \mathcal{E}}\|u(F)\|}$$</p> <p>where the supremum is taken over the set of all finite partitions $\mathcal{E}$ of $E$. Assuming $|u|$ is finite for every $E$, then $|u|$ is a scalar positive measure (and therefore monotone and bounded).</p> <p>On the other hand, the <em>semivariation</em> $\|u\|$ of $u$ is defined to be the (possibly infinite) quantity,</p> <p>$$\|u\|: E\mapsto \sup{|b^{\ast}u|(E)}$$</p> <p>where the supremum is taken over all $b^{\ast}$ in the unit ball of the dual space $B^{\ast}$. To avoid any possible misunderstandings, $|b^{\ast}u|$ is the variation of the scalar measure $E\mapsto b^{\ast}(u(E))$.</p> <p>Assuming $u$ has finite (or bounded) semivariation, that is, $\|u\|(E)$ is finite for every $E$, then the semivariation is positive, monotone and subadditive. An application of Hahn-Banach yields that for every $E$:</p> <p>(A) $\|u(E)\| \leq \|u\|(E)$</p> <p>Less trivial is the fact that $u$ has finite semivariation iff it is bounded. This is a consequence of the fundamental inequality for the semivariation. Since I will not need this inequality, I will just direct you to proposition 11, pg. 4 of the Diestel-Uhl monograph Vector Measures. Another elementary fact is that if the codomain $B$ is finite-dimensional then the variation and the semivariation agree (note: this fails in every infinite-dimensional space by Dvoretzky-Rogers).</p> <p>EXAMPLE: Let $\Omega$ be a Boolean algebra with an infinite partition of unity (note: infinite cardinality of $\Omega$ is enough to guarantee this). Consider the map $\chi:\Omega\to \mathbf{L}^{\infty}(\Omega)$ given by $E \mapsto \chi(E)$, where $\chi(E)$ is the characteristic function of $E$. Then $\chi$ is finitely additive and bounded but it is easy to see that its variation is unbounded.</p> <p>If you are wondering what is $\mathbf{L}^{\infty}(\Omega)$ for a general Boolean algebra, suffice to say that such a space can indeed be constructed and with all the right properties, but to not tarry too long, just take $\Omega$ to be the power set of $\mathbb{N}$ and replace $\mathbf{L}^{\infty}(\Omega)$ with the Banach space of bounded real-valued functions on $\mathbb{N}$ with the supremum norm.</p> <p>Now, I am going to "prove" that every bounded measure $v$ has bounded variation in obvious contradiction with the above example. This will be done by constructing a control measure for $v$ in a very special way. Since what I need is for someone to tell me where and why I have gone astray, I am going to detail the argument, even to the point of pedantry.</p> <p>ARGUMENT: Denote by $\mathbf{BA}(\Omega)$ the space of bounded scalar measures on $\Omega$. One can introduce a partial order on $\mathbf{BA}(\Omega)$ by taking the pointwise order:</p> <p>$$u \leq v \mbox{ iff } u(E) \leq v(E) \mbox{ for all } E \mbox{ in }\Omega$$</p> <p>It is easy to see that with the pointwise order $\mathbf{BA}(\Omega)$ is a partially ordered linear space.</p> <p>Note: For partially ordered linear spaces, Banach lattices, etc. I will take as my reference chapter 5, volume 3 of Fremlin's 5-volume work on measure theory. It is available online, so googling will easily get you to it.</p> <p>We can also put a norm on $\mathbf{BA}(\Omega)$ by taking the total variation, that is, $\|u\| = |u|(\Omega)$.</p> <p>Note: the context should make clear when $\|,\|$ denotes the norm of an element of a Banach space or the semivariation.</p> <p>Lemma 1: The space $\mathbf{BA}(\Omega)$ with the pointwise order and total variation norm is a Banach lattice.</p> <p>Proof: This can also be found in the books. One proof proceeds by noting that $\mathbf{BA}(\Omega)$ is the dual of $\mathbf{L}^{\infty}(\Omega)$, which is a Banach lattice. A (sketch of a) more direct proof goes as follows. Completeness of $\mathbf{BA}(\Omega)$ is straightforward because only finite additivity is involved. By elementary results on partially ordered linear spaces, to prove the existence of arbitrary binary suprema and infima it suffices to prove the existence of the supremum $\sup{u, 0}$ for every $u$ (the <em>positive part</em> $u^{+}$ of $u$). This is given by $u^{+}: E \mapsto \sup{u(F): F\leq E}$. It can also be seen that the <em>absolute value</em> $|u|$ of u defined by $\sup{u, -u}$ is just the variation of u, so that first, there is no ambiguity in my notation, and second, the Banach lattice condition</p> <p>(B) if $|u| \leq |v|$ then $\|u\| \leq \|v\|$</p> <p>is trivially satisfied. Q. E. D.</p> <p>The cone of positive elements of $\mathbf{BA}(\Omega)$ will be denoted simply by $P$. Note that if $u$ is positive then it coincides with its variation.</p> <p>Next, a lemma relating boundedness in the norm with boundedness for the pointwise order.</p> <p>Lemma 2: Let $A$ be a subset of $P$. Then $A$ is order-bounded iff it is norm-bounded.</p> <p>Proof: The direct implication follows from the Banach lattice condition (B). For the converse implication, we prove the contrapositive. So suppose $A$ is not order-bounded. Pick a non-zero positive $v$ and consider the sequence of positive measures $v_n = (n v)/\|v\|$. Since $A$ is not order-bounded there is $u_n$ in $A$ such that $u_n \geq v_n$ for every $n$. By the Banach lattice condition (B) it follows that $\|u_n\| \geq \|v_n\| = n$, so that $A$ is not norm-bounded. Q. E. D.</p> <p>The next two lemmas amount to a proof that norm-bounded subsets of the positive cone have a supremum. The structure of the proof is fairly standard and is patterned after proofs of similar facts in other contexts (e.g. the fact that a category has all coproducts if it has finite coproducts and filtered colimits).</p> <p>Lemma 3: Let $(u_i)$ be a norm-bounded, monotone net of positive measures. Then the pointwise limit</p> <p>(C) $E \mapsto \lim_i u_i(E) = \sup{u_i(E)}$</p> <p>exists and defines a map $u$ that is bounded, finitely additive and the supremum of $(u_i)$.</p> <p>Proof: Since each $u_i$ is positive (and therefore monotone) and $(u_i)$ is norm-bounded, by a constant $C$ say, then, $u_i(E) \leq u_i(\Omega) \leq C$ and the supremum $\sup {u_i(E)}$ exists. Since the net is monotone, this supremum is just $\lim_i u_i(E)$ from which it follows that $u(E) = \lim_i u_i(E)$ is finitely additive. Since it is positive, it is monotone and therefore bounded by $u(\Omega) = \lim_i u_i(\Omega) \leq C$. Since the order is the pointwise order, $u$ is the supremum of $(u_i)$. Q. E. D.</p> <p>Lemma 4: If $A$ is a norm-bounded subset of $P$ then it has a supremum.</p> <p>Proof: By lemma 2, $A$ is order-bounded, by $v$ say. Consider the net $J \mapsto \sup J$ where $J$ runs over the filtered partial order of the finite subsets of $A$. The existence of $\sup J$ for every finite $J$ is guaranteed by lemma 1. Since $v$ bounds $A$, it follows that $v$ bounds $\sup J$ for every $J$. Since the net is monotone and norm-bounded by $\|v\|$, lemma 3 gives us its supremum. A simple check proves that this supremum is precisely the supremum of $A$. Q. E. D.</p> <p>Now, let $v:\Omega\to B$ be a bounded measure. Since it is bounded, it has finite semivariation. This implies that the set of positive measures ${|b^{\ast}v|}$, where $b^{\ast}$ ranges over the unit ball of $B^{\ast}$, is norm-bounded. By lemma 4, it has a supremum $u$. By the very definition of the order structure, we have for every $E$, $|b^{\ast}v|(E) \leq u(E)$ and thus by definition of the semivariation:</p> <p>$$\|v\|(E) \leq u(E)$$</p> <p>But by inequality (A) we have,</p> <p>$$\|v(E)\| \leq u(E)$$</p> <p>and this implies that $v$ has bounded variation!!!!!! For if ${E_n}$ is a finite partition of $E$ then</p> <p>$$\sum_n \|v(E_n)\| \leq \sum_n u(E_n) = u(E) \leq u(\Omega)$$</p> <p>Can anyone help me out here and point out where and why is my argument screwed?</p> <p>Regards and thanks in advance, G. Rodrigues</p> http://mathoverflow.net/questions/12129/category-of-graphs/12300#12300 Answer by G. Rodrigues for Category of graphs. G. Rodrigues 2010-01-19T11:26:05Z 2010-01-19T11:26:05Z <p>Consider the category G that has exactly two distinct objects (call them a and b) and two distinct arrows a->b. A graph is precisely a functor G -> Set. This means that the category of graphs is a presheaf category. The rest follows.</p> http://mathoverflow.net/questions/10576/completion-of-category-is-idempotent/10695#10695 Answer by G. Rodrigues for completion of category is idempotent G. Rodrigues 2010-01-04T11:37:22Z 2010-01-04T11:37:22Z <p>The answer to your question is: No.</p> <p>Let A be a complete category and M be a subcategory. Assume it full and isomorphism closed to simplify things. Form the subcategory C(M) as the full iso-closed subcategory generated by the objects of M together with the limits of all small diagrams in M. Is C(M) complete? Not necessarily. I do not have an example by me, so you just have to take my word for it. Do the next best thing: form C(C(M)) = C^2(M). Is C^2(M) complete? Not necessarily (C(M) wasn't so why should C^2(M) be, right?). You should see where this is going. One keeps iterating the C construction going into the transfinite range until it eventually stabilizes (e.g. when the cardinal of the ambient category is reached).</p> <p>Transfinite constructions of completions are... ugly (to put it mildly). There are all sorts of technical complications. For a taste, read the chapter on locally presentable categories in Borceux's second volume. The completion of categories under classes of (co)limits has been studied intensively by some category theorists, most notably Max Kelly. His book on enriched categories has some material on this. He has also several articles on the subject, just google them.</p> http://mathoverflow.net/questions/7197/how-can-category-theory-help-my-research-in-set-theory/7424#7424 Answer by G. Rodrigues for How can category theory help my research in set theory? G. Rodrigues 2009-12-01T12:50:46Z 2009-12-01T12:50:46Z <p>This is not a direct answer to the question but more a small note on the difference between set theory as usually done and set theory viewed through the lenses of category theory.</p> <p>For me, the most striking aspect of topos theory was the unearthing of vast families of categories that appear in "ordinary" mathematics and that behave very much like the category of sets (denoted by Set).</p> <p>One can reverse the point of view and ask what is special from a categorial point of view about the category of sets? Well, one partial answer is that Set is well-pointed, that is, the terminal is a separator (or generator). This means that every arrow $f:a\to b$ is determined uniquely by the family of points $fx:t\to b$, with $t$ the terminal and $x:t\to a$ a (global) point of $a$. This is a precise categorial version of our view of sets as "discrete, structureless piles of sand".</p> <p>If we did set theory this way, what would be different? There is a (at least one) major philosophical difference. There is no global $\in$. Since the equality predicate of sets is defined by</p> <p>$X=Y &lt;=> (\forall x, x\in X&lt;=> x\in Y)$</p> <p>it follows that there is no global equality predicate for sets. This may be unintuitive from our conception of sets, but it is the Right Thing in category land, because equality between objects is <em>not preserved by equivalences</em>. The Freyd / Scedrov book Categories, allegories contains a formal treatment of a language invariant under equivalences and a considerable part of M. Makkai's work on n-categories is building this insight right into the foundation of higher categories.</p> <p>Hope it helps a little, regards.</p> http://mathoverflow.net/questions/7247/measure-spaces-as-presheaves/7307#7307 Answer by G. Rodrigues for measure spaces as presheaves? G. Rodrigues 2009-11-30T19:28:00Z 2009-11-30T19:28:00Z <p>For a sheaf-theoretical interpretation of measure theory, measure spaces are the wrong objects, you want <em>measure algebras</em> and then consider certain Grothendieck topologies on a Boolean algebra.</p> <p>For measure algebras, check out volume 3 of FRemlin's 5-volume opus dedicated to measure theory. For more sheaf-theoretical stuff, google for Mathew Jackson's phd thesis on measure theory in the context of topos theory.</p> <p>Hope it helps, regards, G. Rodrigues</p> http://mathoverflow.net/questions/3165/presheaves-as-limits-of-representable-functors/5821#5821 Answer by G. Rodrigues for Presheaves as limits of representable functors? G. Rodrigues 2009-11-17T14:48:08Z 2009-11-17T14:48:08Z <p>It follows by the strong form of Yoneda lemma. If the base category $V$ is symmetric monoidal closed, complete and cocomplete, then any presheaf $F:A^{\ast}\to V$ has a left Kan extension along the Yoneda embedding. The coend formula for left Kan extensions then yields</p> <p>$Fa\cong \int^{b}A(b, a)\otimes Fb$</p> <p>Note that the theorem says that every presheaf is a colimit of representables in a <em>canonical</em> way, or that the wedge $w_{b}:A(b, a)\otimes Fb\to Fa$ is universal. Or that the Yoneda embedding is dense.</p> http://mathoverflow.net/questions/120729/cosheafification/124973#124973 Comment by G. Rodrigues G. Rodrigues 2013-03-21T02:45:24Z 2013-03-21T02:45:24Z @Justin: No; no need to invoke any set-theoretical principle. http://mathoverflow.net/questions/124998/was-the-early-calculus-inconsistent/125026#125026 Comment by G. Rodrigues G. Rodrigues 2013-03-20T15:22:43Z 2013-03-20T15:22:43Z If we identify &quot;Divinity&quot; with theology, then no theologian I know disputes that it is not a science in the modern sense of the empirical sciences. But then, neither is mathematics. They will however claim, or some will, that it is a science in the Aristotelian sense. As far as dealing &quot;with the things that do not exist&quot;, your opinion is duly noted, but unless you are a Platonist, mathematicians inevitably deal with objects with no extra-mental existence. And since mathematical objects are abstract, not localized in space-time, empirical considerations are of little avail to him. http://mathoverflow.net/questions/117274/fixed-objects-of-the-m-endofunctor-on-category-meas/117283#117283 Comment by G. Rodrigues G. Rodrigues 2012-12-26T22:13:33Z 2012-12-26T22:13:33Z Isn't this the Giry monad? http://mathoverflow.net/questions/104777/what-are-the-algebras-for-the-double-dualization-monad Comment by G. Rodrigues G. Rodrigues 2012-08-17T12:20:56Z 2012-08-17T12:20:56Z @Yemon Choi: Linton himself provided a proof in a paper in the &quot;Proceedings of the conference on integration, topology and geometry in linear spaces&quot;. He goes on characterizing dual spaces by introducing a suitable equivalence relation on the possible algebra structures for a given dual. The proof itself can be simplified further by using Beck's monadicity theorem and the following three facts: 1. the dual functor preserves finite limts (it is representabe) 2. the dual functor preserves finite colimits (standard functional analysis results) 3. Linton's lemma. http://mathoverflow.net/questions/101420/music-mathematical-point-of-view-revised/101425#101425 Comment by G. Rodrigues G. Rodrigues 2012-07-06T15:20:29Z 2012-07-06T15:20:29Z @David Feldman: &quot;from an empirical body of value judgements by a population of listeners, deduce structural correlates&quot; That could constitute an interesting exercise in a theory of taste but is hardly relevant for the consideration of the total, aesthetic musical order. More general, from what little you wrote, methinks we would violently disagree on some general philosophical points: on the nature of the Good or about the reduction of the mind to the brain. But this is off-topic, so I will just shut-up. http://mathoverflow.net/questions/101420/music-mathematical-point-of-view-revised/101425#101425 Comment by G. Rodrigues G. Rodrigues 2012-07-05T18:48:59Z 2012-07-05T18:48:59Z @David Feldman: &quot;They are mostly failures...mathematics won't tell you what is good and what is bad art because the question isn't well-defined&quot;. Probably nitpicking, but saying the question is &quot;ill-defined&quot; is itself an answer of sorts to the question of what is good or bad art. A more correct answer is simply to say that the question, like all general philosophical questions, is not amenable to a mathematical treatment because what counts as &quot;good&quot; or &quot;bad&quot; art is not quantifiable or measurable -- words taken in the broadest possible sense. http://mathoverflow.net/questions/47278/is-there-such-a-thing-as-the-sigma-completion-of-a-boolean-algebra/58788#58788 Comment by G. Rodrigues G. Rodrigues 2012-03-20T20:45:41Z 2012-03-20T20:45:41Z The unit of the Stone adjunction $\beta$ is not in general $\sigma$-continuous, so why is the factorization $\overline{h}$ $\sigma$-continuous? http://mathoverflow.net/questions/59282/sums-compact-objects-f-g-objects-in-categories-of-modules Comment by G. Rodrigues G. Rodrigues 2011-11-20T12:05:55Z 2011-11-20T12:05:55Z @a-fortiori: right, but a <i>finite</i> coproduct is not a filtered colimit since the indexing category is a finite set which is not filtered. Taking the poset of finite subsets does not buy you anything because this poset has a top element so the colimit is trivially computed. So little-ness is still not weaker (or stronger) than compactness, right? (I have the strange feeling, that I am just about to have one of those doh moments...) http://mathoverflow.net/questions/59282/sums-compact-objects-f-g-objects-in-categories-of-modules Comment by G. Rodrigues G. Rodrigues 2011-11-19T14:00:06Z 2011-11-19T14:00:06Z Maybe I am missing something obvious, but since when is a coproduct a filtered colimit? In other words, why is &lt;i&gt;little-ness&lt;/i&gt; weaker than compactness? http://mathoverflow.net/questions/70361/a-self-dual-adjunction Comment by G. Rodrigues G. Rodrigues 2011-07-14T20:49:44Z 2011-07-14T20:49:44Z If I am parsing your notation correctly, another example of such an adjunction is given by the dual functor on the category of Banach spaces and linear contractions. That it is monadic (the induced monad is given on objects by the bidual) is due to Linton. http://mathoverflow.net/questions/64298/short-five-lemma-in-banach-spaces/64300#64300 Comment by G. Rodrigues G. Rodrigues 2011-05-09T01:18:11Z 2011-05-09T01:18:11Z Simple, short and sweet. Thanks. http://mathoverflow.net/questions/45466/when-is-a-left-kan-extension-closed/45499#45499 Comment by G. Rodrigues G. Rodrigues 2010-11-10T11:12:44Z 2010-11-10T11:12:44Z Yes. I meant &quot;strong&quot;. The lax case is easily disposed of and not very interesting for what I am pursuing. And I agree with Mike Shulman, great strategy! One of those knowledge tidbits that needs to be pointed out before we realize how &quot;obvious&quot; it is. http://mathoverflow.net/questions/43311/sheaves-and-cosheaves/43419#43419 Comment by G. Rodrigues G. Rodrigues 2010-10-26T23:46:44Z 2010-10-26T23:46:44Z @Neha: unfortunately, cannot help you with quasi-coherence. My knowledge about it resumes to knowing where I can find the definition if I ever need it. http://mathoverflow.net/questions/43690/whats-a-mathematician-to-do Comment by G. Rodrigues G. Rodrigues 2010-10-26T23:44:10Z 2010-10-26T23:44:10Z With my idealist hat on, I have always loved the following G. K. Chesterton quote. From memory: &quot;A man must love a thing very much if he not only practices it without any hope of fame or money, but even practices it without any hope of doing it well.&quot; http://mathoverflow.net/questions/41057/categories-first-or-categories-last-in-basic-algebra/41387#41387 Comment by G. Rodrigues G. Rodrigues 2010-10-19T11:52:25Z 2010-10-19T11:52:25Z +1 (and also +1 to Todd Trimble): your observation that category theory lowers the cognitive load is spot on. But what I liked most was what you have <i>not</i> mentioned: all the sloppy talk of category theory as a &quot;language&quot; or &quot;toolbox&quot; which at best is a tautology and at worst is nothing but a prejudice that denies category theory its relative autonomy and makes it the servant maid of whatever one ranks best in mathematics (independently of how objective such value judgements can be). And I better stop before I go off on a rant.