User tetrapharmakon - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:46:44Z http://mathoverflow.net/feeds/user/7952 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76217/understanding-penrose-diagrammatical-notation Understanding Penrose diagrammatical notation tetrapharmakon 2011-09-23T16:23:12Z 2013-04-16T21:01:39Z <p>I arrived to Penrose's paper <em>Applications of negative dimensional Tensors</em> after reading some bits of Baez's <em>Prehistory</em> (<a href="http://math.ucr.edu/home/baez/history.pdf" rel="nofollow">link</a>) and the first two chapters of Turaev's <em>Quantum invariants of knots and 3-manifolds</em> (<a href="http://gen.lib.rus.ec/get?nametype=orig&amp;md5=5E9CC644D232C0D4698413A3344F4964" rel="nofollow">link</a>). The main result in the last one is a presentation of $\text{Rib}$ (the category of ribbon graphs) by a set of generators and relations (braiding, twist and their inverses, duality morphisms...).</p> <p>On the other hand Penrose seems to add to the natural braided ribbon structure also a <em>differential</em> one finding generators also for the (anti)symmetrization and covariant-derivative operations. Are there some easily-found references for a structure theorem similar to Lemma 3.1.1 in the book of Turaev?</p> <p><strong>Edit</strong>: I'm sorry I forgot the second question: I have some problems in figuring out the "contraction" in Penrose notation, because in Turaev exposition the juxtaposition of "coupons" seems to be reserved to the composition of morphisms (contraction contrarily implicates some sort of Einstein summation, am I right?)</p> <p><strong>This is the last edit, I promise:</strong> the following convention to contract the sum of two tensors <img src="http://img853.imageshack.us/img853/6413/screenshot2lw.png" alt="alt text"></p> <p>doesn't make any sense to me.. How can the "f" leg of $\chi^b_{fde}$ split into two?</p> http://mathoverflow.net/questions/112069/a-fibrant-objects-structure-on-top A fibrant-objects structure on Top tetrapharmakon 2012-11-11T10:43:20Z 2013-04-01T20:32:35Z <p>(Sorry for the <a href="http://math.stackexchange.com/questions/234108/a-fibrant-objects-structure-on-bf-top" rel="nofollow">crossposting</a>, but I'm really interested in this question).</p> <p>One can <a href="http://arxiv.org/abs/1011.2926" rel="nofollow">define</a> (Paragraph 1.5, page 10) a fibrant-object structure on a suitable cartesian closed category of topological spaces $\bf Top$, called the $\pi_0$-fibrant structure:</p> <ol> <li>A $\pi_0$-equivalence is a map inducing a bijection at the level of $\pi_0$</li> <li>A $\pi_0$-fibration is a continuous map $p\colon E\to B$ having the RLP with respect to the map ${0}\to [0,1]$ including the 0: [I'm not able to reproduce the diagram, the TeX engine seems not to accept the "array" environment]</li> </ol> <p>Every property defining a fibrant structure can be easily shown in the way you see.</p> <p>Now I'm interested in extending this. The natural definition for a $\pi_n$-equivalence is a map $A\to B$ inducing isomorphisms $\pi_i(A)\to \pi_i(B)$ for all $0\le i\le n$. </p> <blockquote> <p>What should a $\pi_n$-fibration be in order to define a fibrant structure $\pi_n\text{-}\bf Top$ for all $n\in\mathbb N$?</p> <p>What if we "go to the limit" (and can it be done?) $\varinjlim_n \big(\pi_n\text{-}\bf Top\big)$ of these fibrant structures? Do we recover a known fibrant structure, obtained forgetting cofibrations and mutual lifting properties of a suitable model structure, on $\bf Top$?</p> </blockquote> http://mathoverflow.net/questions/111386/homotopical-galois-theory-of-coverings Homotopical Galois theory of coverings tetrapharmakon 2012-11-03T15:39:10Z 2012-11-03T15:39:10Z <p>In the hope this won't turn into a trivial problem (I couldn't find a similar discussion here), here's my question.</p> <p>I'm studying a little homotopical algebra in <a href="http://www.math.uni-hamburg.de/home/schreiber/Abstract%2520homotopy%2520theory%2520and%2520generalized%2520sheaf%2520cohomology.pdf" rel="nofollow">this</a> article by Brown. You can easily notice that Theorem 3 (page 430) and Proposition 3 (in the following page) imply that one can internalize the notion of "$\pi_1$ acting on the fibers of a covering", idea which dates back, if I'm not wrong, to Quillen's "Homotopical Algebra".</p> <p>This could be the starting point for some natural (?) questions: the action of $\pi_1$ on the fibers of a covering is worth to be studied because of Galois' theory of coverings (in fact the philosophy is that of Grothendieck's Galois Theory: Galois groups "are" homotopy groups).</p> <p>Now allow me to state the 64 thousand dollar question: </p> <blockquote> <p>can we recover Galois' theory of coverings in a suitable model/fibrant category? </p> </blockquote> <p>I.e., can we classify subgroups(#) of the fundamental group(#) of the base space of a fiber space(#), finding an (anti-)monotone bijection(#) between the lattice of intermediate objects between the base and a suitable "universal"(##) covering?</p> <p>My two cents: classically, we know very well what to do and how do do it. Here we certainly have enough informations about how to internalize each ingredient (at least those marked with "#"):</p> <ol> <li>Subgroups of a group object are (iso classes of) group mono to that object;</li> <li>$\pi_1(B)=\Omega B$ = pullback obtained exploiting a path object for B, which is doing externally what $\pi_1(B)$ did internally (it is a group which acts on the fibers of a fibration, Omega(Omega(B)) is abelian, ...);</li> <li>A fiber/coverng space is a fibration (here and in (2) one needs a pointed fibrant/model category);</li> <li>Antimonotone bijections are Galois' equivalences: here one looks to the subobjects poset of Omega(B), and to the posetal category C_B, having as objects fibrations with base B (the order is defined by: X &lt; Y iff one fibers over the other - a choice we are rather forced to, just because classically it is so).</li> </ol> <p>The problem seems to be that we lack something forcing C_B to admit a top element.</p> <p>Another question which is still in the handwaving zone: In studying classical Galois theory, I found really bothering that the splitting field of a field is only a weak limit (any two splitting fields are isomorphic, but not with a unique iso). All the same, it is really annoying to notice that the universal covering of (even a good) space is a weak limit. What if the localization functor killed this ambiguity "contracting" the groupoid of isomorphisms between different universal coverings, in passing to the homotopy category? Is there a way to write it down without using theology?</p> <p>Try to meet up this challenge: example 1.1.1.1 in Higher Topos Theory by Lurie suggests (not so coincidentally?) that "being homotopic" in Grp means to be conjugate; now, any two splitting fields are conjugate, am I wrong?</p> http://mathoverflow.net/questions/2556/real-world-applications-of-mathematics-by-arxiv-subject-area/73420#73420 Answer by tetrapharmakon for Real-world applications of mathematics, by arxiv subject area? tetrapharmakon 2011-08-22T17:14:06Z 2012-06-29T06:10:32Z <p><strong>math.KT K-Theory and Homology</strong></p> <p>Used for the purposes (edit: image processing and computational dynamics) in <a href="http://www.amazon.com/Computational-Homology-Applied-Mathematical-Sciences/dp/0387408533" rel="nofollow">this</a> book and in the <a href="http://chomp.rutgers.edu/" rel="nofollow">Chomp</a> project.</p> http://mathoverflow.net/questions/65895/2-tqft-are-to-frobenius-algebras-as-are-to-hopf-algebras 2-TQFT are to Frobenius Algebras as ??? are to Hopf Algebras tetrapharmakon 2011-05-24T20:06:54Z 2012-06-22T01:49:27Z <p>The question arose this morning during a seminar about HAs.</p> <p>In a few words: can the equivalence $2-TQFT_k \leftrightarrow Frob_k$ be "modified" in a sensible way to give a similar one between the category $HA$ of Hopf algebras and a suitable "topological" category (I mean: a -even functor- category made 'with' topological objects, hopefully in a sufficiently small neighborhood of $2-TQFT$)? In particular i would like to find a visual analogue for the antipode map $s:H\to H$.</p> <p>Bad thing is that it takes a while to discover there seem to be no way to define it as an arrow in $Cob(2)$: just try to draw in $Cob(2)$ the diagram</p> <p><img src="http://upload.wikimedia.org/wikipedia/commons/4/45/Hopf_algebra.svg" alt="alt text"></p> <p>...any sensible choice for $s$ leaves in the manifold one hole more than the minimum. Spending a couple of words about the "sensible choice", it seems to me the only way not to increase the genus of the surface is to take as cobordism a-cap-and-a-cup, namely the [Cob(2)-analogue of the] composition $\eta\circ \epsilon\colon H\to k\to H$ in the former diagram... But I'm not able to characterize it as a Frobenius-Algebra map in any sensible way.</p> <p>So, help me... (maybe the person I discussed with this morning is here? His website is <a href="http://www.mate.uncor.edu/~aigarcia/" rel="nofollow">this</a>.)</p> http://mathoverflow.net/questions/92800/monoids-in-a-barr-exact-category Monoids in a Barr-exact category tetrapharmakon 2012-04-01T08:56:26Z 2012-04-01T08:56:26Z <p>Is there a fast way to prove/disprove that the category of internal monoids in a Barr-exact category is itself Barr-exact? Direct computation seems sooo boring...</p> http://mathoverflow.net/questions/87849/does-bf-pset-admit-products Does $\bf pSet$ admit products? tetrapharmakon 2012-02-07T23:03:21Z 2012-02-07T23:24:36Z <p>The question is in the title. The category $\bf pSet$ of partial functions has sets as objects and $\hom(X,Y)$ is the set of all triples $(X,Y,f)$ such that there exists $D\subseteq X$ and $f\colon D\to Y$. Composition of arrows is composition of <a href="http://en.wikipedia.org/wiki/Composition_of_relations" rel="nofollow">relations</a>.</p> http://mathoverflow.net/questions/85541/the-open-problem-of-nth-quantization The open problem of nth quantization tetrapharmakon 2012-01-13T00:55:52Z 2012-01-13T00:55:52Z <p>In trying to explain a quote by E. Nelson, "First quantization is a mystery, but second quantization is a functor!" Baez points out what follows (full text available in <a href="http://math.ucr.edu/home/baez/categories.html" rel="nofollow">this</a> week find; I'm also reading <a href="http://math.ucr.edu/home/baez/nth_quantization.html" rel="nofollow">this</a> other -and mathematically clearer- week find)</p> <blockquote> <p>First quantization is a mystery. It is the attempt to get from a classical description of a physical system to a quantum description of the "same" system. Now it doesn't seem to be true that God created a classical universe on the first day and then quantized it on the second day. So it's unnatural to try to get from classical to quantum mechanics. Nonetheless we are inclined to do so since we understand classical mechanics better. So we'd like to find a way to start with a classical mechanics problem - that is, a phase space and a Hamiltonian function on it - and cook up a quantum mechanics problem - that is, a Hilbert space with a Hamiltonian operator on it. It has become clear that there is no utterly general systematic procedure for doing so.</p> <p>Mathematically, if quantization were "natural" it would be a functor from the category whose objects are symplectic manifolds (= phase spaces) and whose morphisms are symplectic maps (= canonical transformations) to the category whose objects are Hilbert spaces and whose morphisms are unitary operators. Alas, there is no such nice functor. So quantization is always an ad hoc and problematic thing to attempt. A lot is known about it, but more isn't. That's why first quantization is a mystery.</p> <p>(By the way, <em>I have seen many "no-go" theorems concerning quantization but I have never seen one phrased quite like the above. "There is no functor from the symplectic category to the Hilbert category such that ... holds."</em> Is anyone up to the challenge?? If this hasn't been done yet it would clarify the situation.) </p> </blockquote> <p>I find quite interesting the highlighted part: is anyone of you more acquainted with this topic? Is anyone <em>really</em> up to this challenge?</p> <p>From a really naive point of view, and with my really narrow knowledge about QM, I think that if we call $CS$ (resp., $QS$) the category of classical (resp., quantum) physical systems (deliberately avoiding to give a geometrical shape to these categories), then any "reasonable" functor $CS\to QS$ must encode the "canonical quantization" procedure, sending the phase-space of a classical system in a suitable Hilbert space, and "deforming" the canonical Poisson structure on the former space into a noncommutative one in the latter (even more naively, exchanging Poisson brackets with commutators of operators).</p> <p>I likely believed that any <em>no-go theorem</em> in this setting should be formulated in this way ("$\not\exists$ canonical quantization functors") even before reading Baez's note: is anyone out there trying to walk this path?</p> http://mathoverflow.net/questions/83437/the-main-theorems-of-category-theory-and-their-applications/83607#83607 Answer by tetrapharmakon for The main theorems of category theory and their applications tetrapharmakon 2011-12-16T10:03:33Z 2011-12-16T10:03:33Z <p><a href="http://en.wikipedia.org/wiki/Mitchell_embedding_theorem" rel="nofollow">Freyd-Mitchell</a> and <a href="http://www.anstuocmath.ro/mathematics/pdf21/17.pdf" rel="nofollow">Gabriel-Popescu</a> theorems, and also the characterization of <a href="http://www.ams.org/bull/1969-75-06/S0002-9904-1969-12431-6/S0002-9904-1969-12431-6.pdf" rel="nofollow">co-Grothendieck cats</a>.</p> http://mathoverflow.net/questions/58753/center-of-the-algebra-of-bounded-operators/78126#78126 Answer by tetrapharmakon for center of the algebra of bounded operators tetrapharmakon 2011-10-14T12:27:40Z 2011-10-14T12:27:40Z <p>What about the center of the algebra of possibly unbounded operator? (sorry for posting another question but I think it's not worth a new post)</p> http://mathoverflow.net/questions/65895/2-tqft-are-to-frobenius-algebras-as-are-to-hopf-algebras/65972#65972 Answer by tetrapharmakon for 2-TQFT are to Frobenius Algebras as ??? are to Hopf Algebras tetrapharmakon 2011-05-25T16:39:40Z 2011-05-25T16:39:40Z <p>This is my "geometric freshman explanation":</p> <p>the problem is to put something instead of the "?" doing the job of $s$ in</p> <p><img src="http://img841.imageshack.us/img841/9651/outy.png" alt="alt text"></p> <p>...but the composition $\eta\circ\epsilon$ is disconnected, and there is no way to obtain a disconnected manifold starting gluing something to that. :( such a pity.</p> http://mathoverflow.net/questions/65579/what-if-i-change-field-in-a-topological-quantum-field-theory What if I change field in a Topological Quantum Field Theory? tetrapharmakon 2011-05-20T18:55:09Z 2011-05-20T20:08:45Z <p>Of course I'm talking about the algebraic notion of <em>field</em>.</p> <p>In a few words, if a TQFT consists of a functor $Z\colon Cob(n)\to \mathbf{Vec}_k$, I'm wondering if there are sensible relations among different choices of $k$... Some example coming to my mind:</p> <ol> <li><p>Fix a real valued TQFT. Can I turn the functor sending objects to $Z(M)\otimes_\mathbb R\mathbb C$ (the "complexification" of the real vector space associated to $M$) into a complex valued TQFT?</p></li> <li><p>What if I consider, in general, a $k$-valued TQFT and a $K$-valued TQFT (with $K$ an extension of $k$)? What if $K$ is finite, infinite, algebraic, Galois or not, positive charateristic or not, etc etc.?</p></li> </ol> <p>(Feel free to retag the question).</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/62825#62825 Answer by tetrapharmakon for Theorems with unexpected conclusions tetrapharmakon 2011-04-24T09:48:05Z 2011-04-24T09:48:05Z <p><a href="http://en.wikipedia.org/wiki/Eckmann-Hilton_argument" rel="nofollow">Eckmann-Hilton argument</a>. I mean, WHY?</p> http://mathoverflow.net/questions/62464/rings-with-all-modules-projective/62465#62465 Answer by tetrapharmakon for Rings with all modules projective ? tetrapharmakon 2011-04-20T22:03:56Z 2011-04-20T22:03:56Z <p>They're called "semisimple artinian" rings. Prove that a ring $R$ (no commutativity is required) is semisimple artinian iff (equivalently)</p> <p>0) (definition is most books in Ring Theory) $R$ is right artinian and has no nonzero nilpotent right ideals.</p> <p>1) Any right R-module is projective.</p> <p>2) Any right R-module is injective.</p> <p>3) Any simple right R-module is projective.</p> <p>4.1) Any right R-module is semisimple</p> <p>4.2) R is a semisimple right module over itself (if you want, $R_R$ equals its socle).</p> <p>5) $R$ consists of the sum of (finitely many) right ideals.</p> http://mathoverflow.net/questions/60457/elementaryshortuseful/60837#60837 Answer by tetrapharmakon for Elementary+Short+Useful tetrapharmakon 2011-04-06T17:22:35Z 2011-04-06T17:22:35Z <p>Yoneda Lemma. :D</p> http://mathoverflow.net/questions/58769/orthogonal-and-parallel-morphisms "orthogonal" and "parallel" morphisms? tetrapharmakon 2011-03-17T17:05:07Z 2011-03-17T17:05:07Z <p>Let $\mathbf C$ a category with an initial object named $0$.</p> <p>Is there a name for the pair of arrows $f,g\colon A\to B$ such that the unique arrow $0\to A$ is their equalizer? And dually, is there a special name for $f,g\colon A\to B$ such that the coequalizer is $B\to 1$, when $1$ is the terminal object of $\mathbf C$? Finally, is it useful to name them? :)</p> <p>I can figure out how it works in case $\mathbf C$ is concrete: I want to map the fact that a couple of arrows is "everywhere equal" (if "coker(f,g)=the whole") or "nowhere equal" (if "ker(f,g)=nothing unnecessary"). </p> <p>No ideas for general situation + I'm searching references (something make me think about Lawvere but I'm not able to recover anything).</p> <p>Thanks a lot!</p> http://mathoverflow.net/questions/52846/prove-natural-transformation-is-epimorphism/52855#52855 Answer by tetrapharmakon for prove natural transformation is epimorphism tetrapharmakon 2011-01-22T19:20:16Z 2011-01-22T19:20:16Z <p>What is the definition of <em>epimorphism</em> you are using here? Right-cancellable presheaf-morphism lead to right-cancellable functions on component "by definition", I think..</p> http://mathoverflow.net/questions/51169/flux-through-a-mobius-strip Flux through a Mobius strip tetrapharmakon 2011-01-05T01:03:42Z 2011-01-09T03:04:32Z <p>A friend of mine asked me what is the flux of the electric field (or any vector field like $$\vec r=(x,y,z)\mapsto \frac{\vec r}{|r|^3}$$ where $|r|=(x^2+y^2+z^2)^{1/2}$) through a Mobius strip. It seems to me there are no way to compute it in the "standard" way because the strip is not orientable, but if I think about the fact that such a strip <em>can</em> indeed be built (for example using a thin metal layer), I also think that an answer <em>must</em> be mathematically expressible.</p> <p>Searching on wikipedia I found that</p> <p><a href="http://en.wikipedia.org/wiki/Mobius_resistor" rel="nofollow">http://en.wikipedia.org/wiki/Mobius_resistor</a></p> <p>A Möbius resistor is an electrical component made up of two conductive surfaces separated by a dielectric material, twisted 180° and connected to form a Möbius strip. <em>It provides a resistor which has no residual self-inductance, meaning that it can resist the flow of electricity without causing magnetic interference at the same time.</em></p> <p>How can I relate the highlighted phrase to some known differential geometry (physics, analysis?) theorem?</p> <p>Thanks a lot!</p> http://mathoverflow.net/questions/46771/fundamental-group-of-r2-minus-the-irrationals Fundamental group of R^2 minus the (ir)rationals tetrapharmakon 2010-11-20T19:58:38Z 2010-11-20T22:44:47Z <p>Let $$E =\{(x,0) \in \mathbb{R}^2 \colon x \in \mathbb{Q} \}$$ $$F = \{(x,0) \in \mathbb{R}^2 \colon x \in \mathbb{R} \setminus \mathbb{Q}\}$$ compute the fundamental group of $\mathbb R^2\setminus E$ and $\mathbb R^2\setminus F$. How can I start?</p> <p>(I don't know why the { symbols don't appear)</p> http://mathoverflow.net/questions/45351/does-pi-1-have-a-right-adjoint Does $\pi_1$ have a right adjoint? tetrapharmakon 2010-11-08T20:13:18Z 2010-11-08T22:30:38Z <p>Eilenberg and Mac Lane showed that given a group $G$ there exists a pointed topological space $X_G$ such that $\pi(X_G,\bullet)\cong G$. It is obviously a way to "invert direction" to the functor $\pi_1\colon \mathbf{Top}^\bullet\to \mathbf{Grp}$ to a functor $\mathcal K\colon \mathbf{Grp}\to \mathbf{Top}^\bullet$ such that $\pi_1(\mathcal K(G),\bullet)\cong G$ (almost by definition). This is equivalent to say that there exists a natural transformation (equivalence, in this case) between $\pi_1\circ\mathcal K$ and $\mathbf{1}_{\mathbf{Grp}}$, which turns out to resemble some sort of <em>counity</em>.</p> <p>It would be wonderful if I could define an adjunction between the two categories in exam, given by the two functors. Everytime I try to think about some sort of <em>unity</em> to this hypotetical adjunction I poorly fail: considering the vast literature in the field of algebraic topology, I believe in only two possible cases. The first, nothing interesting arises from this adjunction. The second, there is no sort of adjunction.</p> <p>The key point, quite trivial, to answer is: it is well known that an adjunction is uniquely determined by one among <em>unity</em> and <em>counity</em>, provided the one is universal. But $\boldsymbol\varepsilon\colon \pi_1\circ\mathcal K\to \mathbf{1}_{\mathbf{Grp}}$ is an <em>equivalence</em>: can I conclude that it is universal?</p> http://mathoverflow.net/questions/44718/double-direct-systems-and-switching-of-direct-limits/44803#44803 Answer by tetrapharmakon for Double Direct Systems and Switching of Direct Limits tetrapharmakon 2010-11-04T10:49:53Z 2010-11-04T10:49:53Z <p>I think you will find an answer in par. 2.4 of this</p> <p><a href="http://people.math.jussieu.fr/~schapira/lectnotes/AlTo.pdf" rel="nofollow">http://people.math.jussieu.fr/~schapira/lectnotes/AlTo.pdf</a></p> <p>In particular I think one always have $$\varprojlim_I\;\varprojlim_J A_{ij} \cong \varprojlim_J\; \varprojlim_I A_{ij}$$ because $\varprojlim$ is right adjoint in an adjoint couple, and right adjoints preserve projective limits...</p> http://mathoverflow.net/questions/33638/is-the-subobject-functor-really-a-presheaf Is the subobject functor really a presheaf? tetrapharmakon 2010-07-28T09:53:26Z 2010-09-03T10:24:55Z <p>I refer to "Sheaves in Geometry and Logic", by S. MacLane.</p> <p>Let <strong>C</strong> be a category. Dealing with a <em>subobject</em> of an object $D \in \text{Ob}_{\mathbf C}$, one defines an equivalence relation between morphisms towards <em>D</em>:</p> <blockquote> <p>Two monomorphisms $f:A\to D$, $g:B\to D$ with a common codomain <em>D</em> are called <em>equivalent</em> if there exists an isomorphism $h\colon A\to B$ such that <em>gh</em>= <em>f</em>. A <em>sbobject</em> of <em>D</em> is an equivalence class of monos towards <em>D</em>. The collection Sub<sub><strong>C</strong></sub>(<em>D</em>) of subobject of <em>D</em> carries a natural partial order [...]. Then Sub<sub><strong>C</strong></sub>(<em>D</em>) is <strong>the set</strong> of all subobjects of <em>D</em> in the category <strong>C</strong>.</p> </blockquote> <p>I can't figure out <em>why</em> Sub<sub><strong>C</strong></sub>(<em>D</em>) is a set, rather than a proper class! Indeed, we are considering something like an qeuivalence relation on</p> <p>$\displaystyle \coprod_{A\in \text{Ob}} \text{Hom}_{\bf C}(A,D)$</p> <p>which is not a set, as soon as <strong>C</strong> isn't small.</p> <p>So, how can I avoid the problem?</p> http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial/34063#34063 Answer by tetrapharmakon for If $f$ is infinitely differentiable then $f$ coincides with a polynomial tetrapharmakon 2010-07-31T22:35:58Z 2010-07-31T22:35:58Z <p>Maybe unuseful, but it remains true if you consider $f\in C^\infty(\mathbb R,\mathbb R)$.</p> <p>Try showing that</p> <p><strong>Lemma.</strong> Let $I\subseteq \mathbb R$ be a nonempty interval and $f\in C^{\infty}(I)$. If $f$ is not a polynomial on $I$, then there exists a compact subset $J\Subset I$ in which $f$ is not a polynomial. Moreover, $f(x)\neq 0\;\forall x\in J$.</p> http://mathoverflow.net/questions/33995/is-the-only-known-universe Is {Ø,{Ø},{Ø,{Ø}}, ... } the only known universe? tetrapharmakon 2010-07-31T09:08:38Z 2010-07-31T17:09:13Z <p>In the first pages of SGA4 I read</p> <blockquote> <p>[...] <em>Cependant le seul univers connu est l'ensemble des symboles du type</em> {&Oslash;,{&Oslash;},{&Oslash;,{&Oslash;}}, ... } <em>etc. (tous les éléments de cet univers sont des ensembles finis et cet univers est dénombrable). En particulier, on ne connaît pas d'univers qui contienne un élément de cardinal infini.</em> [...]</p> </blockquote> <p>(the sole known universe is like {&Oslash;,{&Oslash;},{&Oslash;,{&Oslash;}}, ... }, and we don't know any universe with a infinite cardinal).</p> <p><em>Mais, c'est vrai?</em> I wonder if during all these years somebody discovered a universe "bigger" than that exhibited by Grothendieck.</p> http://mathoverflow.net/questions/112069/a-fibrant-objects-structure-on-top/114916#114916 Comment by tetrapharmakon tetrapharmakon 2012-12-18T22:27:32Z 2012-12-18T22:27:32Z (which is the submitted version, but it's strange, isn't it?) http://mathoverflow.net/questions/112069/a-fibrant-objects-structure-on-top/114916#114916 Comment by tetrapharmakon tetrapharmakon 2012-12-18T22:26:42Z 2012-12-18T22:26:42Z It's obviously $I^{p-1}$; I checked and the error is repeated verbatim in the version of the paper you linked me. http://mathoverflow.net/questions/112069/a-fibrant-objects-structure-on-top/114916#114916 Comment by tetrapharmakon tetrapharmakon 2012-12-17T20:01:04Z 2012-12-17T20:01:04Z ...I'm totally uncomfortable with the definition of $V^{p-1}$. Can you write explicitly $V^{-1}$ (if it can be defined), $V^0,V^1, V^2$? http://mathoverflow.net/questions/112069/a-fibrant-objects-structure-on-top/114916#114916 Comment by tetrapharmakon tetrapharmakon 2012-12-17T19:51:52Z 2012-12-17T19:51:52Z maybe you mean &quot;$V^{p-1}$ = the union of all faces of $I^p$ except for $I^{p-1}\times 1$&quot;? http://mathoverflow.net/questions/111386/homotopical-galois-theory-of-coverings Comment by tetrapharmakon tetrapharmakon 2012-11-07T12:52:17Z 2012-11-07T12:52:17Z Thank you, mr. Shulman! Can you provide me some references about this? Is the existence of a universal fibration automatically provided? http://mathoverflow.net/questions/111386/homotopical-galois-theory-of-coverings Comment by tetrapharmakon tetrapharmakon 2012-11-03T16:44:48Z 2012-11-03T16:44:48Z Can you explain the downvote? http://mathoverflow.net/questions/92800/monoids-in-a-barr-exact-category Comment by tetrapharmakon tetrapharmakon 2012-04-03T08:54:04Z 2012-04-03T08:54:04Z :) @Mike: any clue to begin? http://mathoverflow.net/questions/92800/monoids-in-a-barr-exact-category Comment by tetrapharmakon tetrapharmakon 2012-04-02T09:33:45Z 2012-04-02T09:33:45Z @Martin: you're right, sorry. I couldn't find much informations about the claim in the OP, so I tried to prove that statement on my own. Let C be Barr-exact; Mon(C) inherits completeness of C, and if I'm not wrong any internal-monoid can be factored into $m\circ e$, with $m$ a mono, $e$ a regular (hence extremal) epimorphism. Here I stopped, but trying to examine a special case seems a good idea... it sounded so natural to me that I wanted to obtain it once and for all in full generality. @Mike Shulman: your guess sounds good too. Unfortunately I'm not really keen on internal-manipulations.. http://mathoverflow.net/questions/14815/is-there-a-finitely-complete-category-with-terminal-object-but-no-subobject-class/14820#14820 Comment by tetrapharmakon tetrapharmakon 2012-02-29T14:46:21Z 2012-02-29T14:46:21Z Typo: if $1\to \Omega$ classyfies everything, it classifies <i>concrete</i> categories http://mathoverflow.net/questions/14815/is-there-a-finitely-complete-category-with-terminal-object-but-no-subobject-class/14820#14820 Comment by tetrapharmakon tetrapharmakon 2012-02-29T14:45:32Z 2012-02-29T14:45:32Z @Steven Gubkin: It seems to me (but this adaption is mainly due to @Salvatore Tringali) that one can repeat word-by-word your proof to see that $\bf cat$ (small categories) is not a topos (choose 0 = terminal category, $A$ any concrete one -if $1\to\Omega$ classyfies everything, it classifies small categories-, argue in the same way because monos in $\bf cat$ are faithful-injective on objects functors, conclude that $Ob(A)\le Ob(\Omega)$, contradiction). Can you please confirm this? Thanks a lot! http://mathoverflow.net/questions/45832/are-there-examples-of-non-orientable-manifolds-in-nature/45898#45898 Comment by tetrapharmakon tetrapharmakon 2012-02-13T11:43:41Z 2012-02-13T11:43:41Z This is AWESOME! http://mathoverflow.net/questions/87849/does-bf-pset-admit-products Comment by tetrapharmakon tetrapharmakon 2012-02-08T10:27:42Z 2012-02-08T10:27:42Z @Martin: no, the idea is much more simpler: in $\bf Set$ you take as $\hom(X,Y)$ the set of functions everywhere defined on $X$. In $\bf pSet$ you relax this taking any function defined on <i>any</i> $D\subset X$... http://mathoverflow.net/questions/87849/does-bf-pset-admit-products Comment by tetrapharmakon tetrapharmakon 2012-02-08T09:49:06Z 2012-02-08T09:49:06Z @you: let me please understand if I'm wrong in saying that &quot;naif&quot; products doesn't work... http://mathoverflow.net/questions/87849/does-bf-pset-admit-products/87851#87851 Comment by tetrapharmakon tetrapharmakon 2012-02-08T09:47:02Z 2012-02-08T09:47:02Z Thank you Owen and Qiaochu! I suspected there was a link between the two categories ($\bf pSet$ admits a zero objects, which seems quite strange if you ignore that equivalence). http://mathoverflow.net/questions/87849/does-bf-pset-admit-products/87851#87851 Comment by tetrapharmakon tetrapharmakon 2012-02-07T23:21:59Z 2012-02-07T23:21:59Z Can <i>any</i> partial function be seen as a map between &quot;punctured&quot; sets?