User garlef wegart - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-25T15:01:22Zhttp://mathoverflow.net/feeds/user/1261http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/131255/objects-which-cant-be-defined-without-making-choices-but-which-end-up-independen/131277#131277Answer by Garlef Wegart for objects which can't be defined without making choices but which end up independent of the choiceGarlef Wegart2013-05-20T21:53:39Z2013-05-20T22:14:40Z<p>I think your question can be unasked (in the sense of Gödel-Escher-Bach): The answer depends on what you consider to be a definition - you can define a mathematical object by its properties ~ or give an explicit description.</p>
<p>As an example consider the real numbers. An explicit description would be to take the rationals and show the Dedekind cuts to be a field. Then call it $\mathbb{R}$ and continue to investigate it using the explicit description.</p>
<p>On the other side you can prove the following theorem:</p>
<blockquote>
<p>There is an ordered field
$\mathbb R$ such that every subset
with a lower bound has an infimum. Every other ordered field $\mathbb R'$ with this property is uniquely isomorphic to $\mathbb R$.</p>
</blockquote>
<p>Now the theorem consists of two parts: First a description of the desired properties (ordered, infima) and the uniqueness and second a statement about the existence. In order to show the existence one has to give an explicit description of an object with the desired properties. In our example take the Dedekind cuts.</p>
<p>For the uniqueness part one should be able to work without an explicit description, using only the properties. In our case the argument goes as follows: ordered fields have characteristic 0 so we can embed $\mathbb Q$ into both $\mathbb R$ and $\mathbb R'$. Now use the existence of infima of subsets with a lower bound to construct the mapping. There is no need to refer to the Dedekind cuts in this part of the proof.</p>
<p>Back to your question: The crucial distinction is between <strong>description</strong> and <strong>implementation</strong>. For example there are several algorithms to sort a list: They are all implemenations of the same function - "same" in the sense of extensionality. Likewise there are different ways to build the real numbers - yet they all result in the same object; "same" in the sense specified above.</p>
<p>If one constructs an object explicitely making some choices and later on shows that it is "independent" of these choices <strong>one has to say what "independent" means in this context. This can only be done by listing properties of the object that characterise it up to some notion of equivalence.</strong></p>
<p>I personally prefer definitions via descriptions: For me $\mathrm{Ext}$ "is" the derived $\mathrm{Hom}$-functor.</p>
<p>PS: I guess the uniqueness of an object up to [some notion of equivalence] can be seen as some sort of extensional equality (in the sense of the yoneda lemma?). I guess Homotopy Type Theory deals with this question but I don't know enough about it to say more in this direction.</p>
http://mathoverflow.net/questions/128387/composition-of-cat-valued-distributors-compatible-with-grothendieck-constructioComposition of Cat-valued distributors - compatible with grothendieck construction?Garlef Wegart2013-04-22T19:56:35Z2013-04-23T07:08:15Z
<p>Let $C$ be a category and $F\in[C^{op}, Cat]$ be a strong functor.</p>
<p>(1) There are functors</p>
<p>$$hom_C(c',c)\times F(c)\to F(c').$$</p>
<p>(2) The grothendieck construction gives a 2-equvalence</p>
<p>$$\int_C: [C^{op},Cat]\to Fib_C$$</p>
<p>(3) Fibrations over $C$ are categories over $C$ together with a left action of the morphism double category $Mor_C$. ($Mor_C$ is a 2-monoid in the 2-category of spans from $C$ to $C$; the 2-monoidal structure is given by the strong pullback. This induces a 2-monad on the 2-category of spans from $C$ to $pt$, the one-point category. Algebras for this monad are fibrations.)</p>
<p>(4) Let $\int_C^D$ denote the <a href="http://mathoverflow.net/questions/128393/grothendieck-construction-for-profunctors" rel="nofollow">grothendieck construction for distributors</a>. We have</p>
<p>$$\int_C^C hom_C = Mor_C$$</p>
<p>and</p>
<p>$$\int_C^{pt} F = \int_C F$$</p>
<p><strong>Question:</strong> How do these parts fit together?</p>
<p>The maps from (1) should result in a transformation</p>
<p>$$hom_C\otimes_C F \to F$$</p>
<p>and the action of $Mor_C$ is given by a functor</p>
<p>$$\int hom \times_C \int F \to \int F.$$</p>
<p>Now: If the grothendieck construction was laxely (or even strongly) compatible with the the products involved - that is, A lax (or strong) bifunctor - these two actions would naturally correspond to each other.</p>
<p><strong>Question 2</strong>: Is it? And if so: Can this be shown without using the explicit description of the grothendieck construction; only by using the characterisation as a weighted limit?</p>
<p><a href="http://mathoverflow.net/questions/128393/grothendieck-construction-for-profunctors" rel="nofollow">Related Question</a></p>
<p>Edit: I reformulated the question.</p>
http://mathoverflow.net/questions/128393/grothendieck-construction-for-profunctorsgrothendieck construction for profunctorsGarlef Wegart2013-04-22T20:12:59Z2013-04-22T21:25:35Z
<p>Given categories $X$ and $Y$ and a strong functor</p>
<p>$$D:X^{op}\times Y\to Cat$$</p>
<p>we can of course build the oplax colimit</p>
<p>$$\mathrm{colim}^{oplax}_{X^{op}\times Y}D$$</p>
<p>via the usual (covariant) grothendieck construction:</p>
<p><em>Objects</em> are triples $(x,d,y)$ with $d\in D(x,y)$.</p>
<p><em>Morphisms</em> $(x,d,y)\to (x',d',y')$ are triples $(f,\phi, g)$ with $f:x'\to x$, $g:y\to y'$ and $$\varphi: f^*x_*g\to x'.$$</p>
<p>There is however another possible construction that corresponds better to the slogan "presheaves are distributors into the one-point category and copresheaves are distributors out of the one-point category":</p>
<p><em>Objects</em> are the same triples as above.</p>
<p><em>Morphisms</em> however are now triples $(f,\varphi, g)$ with $f:x\to x'$ (the direction changed!) and $g:y\to y'$ and</p>
<p>$$\varphi:x_*g \to f^*x'.$$</p>
<p>Why does it correspond better to the slogan stated above? Taking distributors having at one side the one-point category then specialises to the usual grothendieck constructions yielding fibered and opfibered categories respectively.</p>
<p><strong>Question:</strong> What can be said about the second construction? Is it some kind of colimit as well?</p>
<p><a href="http://mathoverflow.net/questions/128387/composition-of-cat-valued-distributors-compatible-with-grothendieck-constructio" rel="nofollow">Related question</a></p>
http://mathoverflow.net/questions/83944/enriched-categories-metric-spaces-monoidal-endofunctors-and-lipschitz-continuouEnriched Categories: Metric Spaces, Monoidal Endofunctors and Lipschitz-Continuous Maps.Garlef Wegart2011-12-20T14:09:27Z2011-12-20T14:09:27Z
<p>In the introduction to the <a href="http://tac.mta.ca/tac/reprints/articles/1/tr1abs.html" rel="nofollow">reprint</a> of "Metric spaces, generalized logic and closed categories" Lawvere talks about the following situation:</p>
<p>Let $\mathbb R_+$ denote $\mathbb R_{\geq 0}^\infty$. Every lax monoidal endofunctor $\lambda:\mathbb R_+\to\mathbb R_+$ gives rise to an endofunctor $\lambda.:\mathrm{Met}\to\mathrm{Met}$ on the category of generalized metric spaces, again leading to the notion of $\lambda$-Lipschitz-continous maps
$$ Lip^\lambda(X,Y):=Lip^1(X,\lambda.Y).$$
These are exactly the maps $f:X\to Y$ satisfying $d(x,x')\geq \lambda(d(fx,fx'))$. Lawvere goes on and "<em>suggest a whole family of monoidal structures [on the category of metric spaces] interpolating between</em>" the sum- (aka tensor product) and the max- (aka cartesian product) metric on the product of the underlying sets. He finally relates this to the
$$\frac{1}{p}+\frac{1}{q}=1$$
business occuring in analysis.</p>
<p><strong>Now for the question</strong>: Has this been worked out somewhere (replacing $\mathbb R_+$ by an arbitrary moinoidal category $\mathcal V$)?</p>
<p><strong>Remark</strong>: Concerning the monoidal structures my first idea was to define various adjoints to the hom-like functors
$$Lip^\lambda(X,-)$$
but as we don't have
$$Lip^\lambda\times Lip^\lambda\to Lip^\lambda$$
but rather
$$Lip^\lambda\times Lip^\mu\to Lip^{\mu\circ\lambda}$$
i suspect we should define $X\otimes_\lambda^\mu Y$ by an expression like
$$Lip^\lambda(X,Lip^\mu(Y,Z))=:Lip^\lambda(X\otimes_\lambda^\mu Y,Z)$$
(of maybe $Lip^1$ on the right hand side). So instead of various monoidal structures we'd get various tensor products, compatible in some way...</p>
http://mathoverflow.net/questions/79817/reference-requestenriched-categories-metric-on-lipschitz-continuous-functionsReference Request(Enriched Categories): Metric on Lipschitz Continuous FunctionsGarlef Wegart2011-11-02T11:11:36Z2011-11-02T12:13:06Z
<p>If we consider metric spaces to be categories enriched over $\mathbb R_{\geq 0}$, the object corresponding to presheaves should be lipschitz-continuous functions $\operatorname{Lip^ 1}(M, \mathbb R_{\geq 0})$. Now there should be an obvious metric on this set; making the Yoneda map $$x\mapsto \operatorname d(-,x)$$
an isometric embedding. <strong>What is this metric?</strong></p>
http://mathoverflow.net/questions/79102/definition-of-enriched-caterories-or-internal-homs-without-using-monoidal-categorDefinition of enriched caterories or internal homs without using monoidal categories.Garlef Wegart2011-10-25T18:43:07Z2011-10-25T18:50:07Z
<p>I know this question may seem nonsensical at first but let me exlain what i have in mind:</p>
<p>In enriched category theory we define categories enriched over a monoidal category $(\mathcal{V},\otimes, I)$. An enriched category then is given by a set/class of objects $\mathcal C$ and a rule assigning to every pair $X,Y$ of such objects a hom-object $[X,Y]$. Furthermore we define composition and identities using $\otimes$ and $I$, remodelling the definitions of usual category theory.</p>
<p><strong>Now for the question</strong>: Can we go the other way around?</p>
<p>Let's stick to internal homs for the beginning: Given a category $\mathcal V$ ; can say what additional data turn a functor $$[-,-]:\mathcal{V}^{\mathrm{op}}\times\mathcal V\to \mathcal V$$ into something like an internal hom?</p>
<p>In the case of $[X,-]$ having a left adjoint $-\otimes X$ for every $X$, these additional data should result in $(\mathcal V,\otimes)$ becomming a closed monoidal category with internal hom isomorphic to $[-,-]$.</p>
http://mathoverflow.net/questions/76443/enriched-categories-ideals-submodules-and-algebraic-geometryEnriched Categories: Ideals/Submodules and algebraic geometryGarlef Wegart2011-09-26T18:24:59Z2011-09-26T18:24:59Z
<p>While working through Atiyah/MacDonald for my final exams I realized the following:</p>
<p>The category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals</p>
<p>$$I\cdot J:=\langle{ab\mid a\in I, b\in J}\rangle$$</p>
<p>The internal Hom is given by the ideal quotient; let's denote it $[I, J]:=(J:I)=\lbrace a\in A\mid aI\subset J\rbrace$. We have</p>
<p>$$IJ\subset K \Leftrightarrow I\subset [J,K].$$</p>
<p>Furthermore the category(poset) of submodules $U(M)$ of an A-Module $M$ is enriched over the ideal category via</p>
<p>$$[U,V]:=\lbrace a\in A\mid aU\subset V\rbrace.$$</p>
<p>Where does this enrichment come from? More specifically: As in algebraic geometry we are dealing with the set of prime ideals ( wich is more or less the subcategory</p>
<p>$$\mathrm{Int^{op}}/A\subset\mathrm{CRing^{op}}/A$$</p>
<p>of ring-maps $A\to B$ with $B$ an integral domain), I'd like to see the category of Ideals as some "flattened" version of $\mathrm{CRing^{op}}/A$ by taking the kernel. Or more generally i'd like to see like to see it as a "flattened" version of the category of modules (by taking the annihilator?). So:</p>
<p><strong>Question:</strong> Is it possible to build the enriched categories $I(A)$ and $U(M)$ out of $\mathrm{CRing^{op}}/A$ and $\mathrm{Mod}_A/M$ in a nice way? So: Can we for example write down $I\cdot J$ as a kernel of some $A\to B$? Or $\mathrm{Ann}(M)\cdot \mathrm{Ann}(N)$ just by using $N,M$ and constructions in/on $\mathrm{Mod}_A$?</p>
http://mathoverflow.net/questions/52989/reference-request-2-monads-and-2-adjunctionsReference request: 2-Monads and 2-AdjunctionsGarlef Wegart2011-01-23T20:14:33Z2011-01-23T23:12:05Z
<p>Given a category $\mathcal C$ together with a monad $T$ on $\mathcal C$, we get an adjunction $$\mathcal C^T(T-,-)\cong \mathcal C(-,\mathrm{For}-).$$</p>
<p>Is the same true for 2-monads on a 2-category?</p>
http://mathoverflow.net/questions/47219/strong-colimits-of-categoriesStrong colimits of categories.Garlef Wegart2010-11-24T11:13:47Z2010-11-29T15:53:55Z
<p>Let $\mathcal C$ be a category and let $\mathcal F:\mathcal C\to\mathcal C\textrm{at}$ be a strong bifunctor. Given another category $\mathcal D$, let $\triangle_{\mathcal D}$ denote the constant functor $\mathcal C\to\mathcal C\textrm{at}$. Now define lax/strong limits and colimits as follows:</p>
<ul>
<li><p>A <strong>lax limit</strong> of $\mathcal F$ is a
category $\mathsf{lim}\mathcal F$
together with a natural equivalence
$$[\triangle_{(-)},\mathcal
F] \cong \mathcal
C\textrm{at}(-,\mathsf{lim}\mathcal
F).$$ Here $[\triangle_{(-)},\mathcal F]$ denotes the category
of lax natural transformations and modifications.</p></li>
<li><p>A <strong>lax colimit</strong> of $\mathcal F$ is a
category $\mathsf{colim}\mathcal F$
together with a natural equivalence
$$[\mathcal
F,\triangle_{(-)}] \cong \mathcal
C\textrm{at}(\mathsf{colim}\mathcal
F,-).$$</p></li>
<li>We define <strong>strong limits</strong> and <strong>strong colimts</strong> by replacing lax natural
transormations with strong natural transfomations.</li>
</ul>
<p>Now, if my calculations are correct, a lax colimit of such a functor $\mathcal F$
is given by the grothendieck construction $\mathcal C\int\mathcal F$ and a lax limit is given by the category of <strong>strict</strong> sections
$s:\mathcal C\to \mathcal C\int\mathcal F$, i.e. the category $\mathcal C\textrm{at}/\mathcal C(\operatorname{id}_\mathcal C,\pi)$, where $\pi:\mathcal C\int\mathcal F\to\mathcal C$ is the opfibration corresponding to $\mathcal F$.</p>
<p>If we consider only the category of opcartesian sections, that is sections that map every morphism in $\mathcal C$
to an opcartesian morphism, we get a strong limit.</p>
<p>Now for the question:</p>
<p><strong>Is there an explicit description of the strong colimit of a functor $\mathcal F:\mathcal C\to\mathcal C\textrm{at}$?</strong></p>
http://mathoverflow.net/questions/23794/kan-extensions-and-the-yoneda-embeddingKan extensions and the yoneda embedding.Garlef Wegart2010-05-06T22:33:06Z2010-11-18T20:48:09Z
<p>[Edit: For a category $C$ let $C^\wedge$ denote the category of presheaves on $C$.]</p>
<p>Let $f:C\to D$ be a functor. Precomposition with $f^{op}$ induces a functor
$f^\wedge:D^\wedge \to C^\wedge$. This functor has both a left- and a right adjoint, called <em>left-</em> and <em>right kan extension</em>:</p>
<p>$f_\wedge \dashv f^\wedge \dashv f_+$.</p>
<p>Now for $c\in C$ we get $D^\wedge(D(-,fc),Y)=Y(fc)=f^\wedge Y(c)=C^\wedge(C(-,c),f^\wedge Y)$. This gives us the restriction of $f_\wedge$ to $C$ along the yoneda embedding: It is $f$ (composed with the yoneda embedding).</p>
<p>Now here's my question:</p>
<blockquote>
<p>What is the restriction of $f_+$ to $C$ along the yoneda embedding?</p>
</blockquote>
<p>It seems not to agree with $f$ but:</p>
<blockquote>
<p>Is there a nice connection between $f_+C(-,c)$ and $D(-,fc)$?</p>
</blockquote>
http://mathoverflow.net/questions/24722/is-there-a-notion-of-good-distributor-profunctor-for-model-categoriesIs there a notion of "good" distributor/profunctor for model categories?Garlef Wegart2010-05-15T10:43:07Z2010-11-18T11:47:37Z
<p>When considering functors between model categories one possibility is to restrict ones attention to quillen adjunctions. But what about distributors?</p>
<blockquote>
<p>What are the natural distributors to consider between model categories?</p>
</blockquote>
http://mathoverflow.net/questions/40722/reference-request-2-grothendieck-constructionReference request: 2-Grothendieck ConstructionGarlef Wegart2010-10-01T08:10:49Z2010-10-01T14:21:13Z
<p>Hi Folks,</p>
<p>i'm looking for a reference on the 2-grothendieck construction for a functor $F:\mathcal{I}\to \mathcal{B}\mathrm{icat}$ from a bicategory $\mathcal{I}$ to the tricategory of bicategories. Actually for my purposes it would be sufficient to consider functors going only to $\mathcal{C}\mathrm{at}$.</p>
http://mathoverflow.net/questions/38993/co-limits-and-fibrations-of-dg-categories(Co-)Limits and fibrations of DG-Categories?Garlef Wegart2010-09-16T16:53:32Z2010-09-16T17:05:52Z
<p>First of all, let me see if i got the 1-categorical version right:</p>
<ul>
<li>Let $\mathcal F:C\to Cat $ be a
(pseudo-) functor. The 2-colimit
$\mathrm{colim}_C\mathcal F$ is then
given by the grothendieck
construction $\int_C \mathcal F$ and
the 2-limit is given by the category
of cartesian sections of the
fibration $\int_C \mathcal F\to C$, right?</li>
</ul>
<p>Can this be transported to the setting of dg$_k$-categories? So: </p>
<ol>
<li>Is there a notion of fibration of dg
categories? I would imagine them to
be algebras for a dg-monad
$(\mathrm{id}_C,-)$ arising from
forming dg-comma categories with the
identity-span on $C$.</li>
<li>What about a
grothendieck construction for
functors $\mathcal F: C\to \mathrm{dgCat}$ from
a category $C$ to dg-Categories?</li>
<li>Cartesian sections should then be
defined as algebra-morphisms from
the identity on $C$ to $\int_C
\mathcal F$.</li>
</ol>
<p>I think there are some problems with what i just said: What are dg-comma categories? What are the right functors $C\to\mathrm{dgCat}$? (I guess one should build a dg-category $C'$ out of $C$ by taking the free $k$-category and then consider it as a dg-category, concentrated in degree 0, and then consider $\mathrm{dgCat}(C',\mathrm{dgCat})$) Same goes for the definition of an algebra morphism: What are the coherences to consider?</p>
<p>So i guess the core question is:</p>
<ul>
<li>What is the right notion of limit for dg categories? (And why?)</li>
</ul>
http://mathoverflow.net/questions/36766/adjunctions-algebras-of-the-induced-monad-vs-coalgebras-of-the-induced-comonadAdjunctions: Algebras of the induced monad VS. Coalgebras of the induced comonad.Garlef Wegart2010-08-26T14:00:39Z2010-08-26T17:41:01Z
<p>Given an adjunction, we get a monad on one side and a comonad on the other side. What is the connection between their algebra and coalgebra categories? Are they allways equivalent?</p>
<p>The example i have in mind is the starting point of algebraic geometry (or more general: The fundamental theorem of formal concept analysis) ~ the relation between polynomials and points in affine space given by fRx iff f(x)=0 induces an order preserving isomorphism between radical ideals and "algebraic sets".</p>
http://mathoverflow.net/questions/21318/is-assigning-the-endomorphism-object-in-some-sense-functorialIs assigning the endomorphism object in some sense functorial?Garlef Wegart2010-04-14T10:15:39Z2010-04-14T12:38:06Z
<p>Let $\mathcal V$ be a monoidal category and let $\mathcal C$ be a $\mathcal V$-category. Let's denote the $\mathcal V$-valued hom-functor $[-,-]$. Now for every object $X\in\mathcal C$ we have it's endomorphism object $\mathcal End(X):=[X,X]$ - it is actually a monoid in $\mathcal V$. Can the assignment
$$X\mapsto \mathcal End(X)$$
be considered functorial in some way? Is there a language that captures the relations between $\mathcal End(X)$ when $X$ varies? What happens on the level of module categories $\mathcal V^{\mathcal End(X)}$?</p>
<p>I've allready thought about this a bit but i don't want to reinvent the wheel. [Edit] So here's what i've been thinking of:</p>
<p>For every object $X\in\mathcal C$ we get a functor $[X,-]:\mathcal C\to\mathcal End(X)-\operatorname{mod}$. I think these functors are connected in a vaguely functorial way - hopefully by adjunctions between the module categories (adjunction in the 2-category of categories under $\mathcal C$).</p>
<p>My idea:</p>
<ol>
<li>Let $f:X\to Y$ be a morphism in $\mathcal C_0$. On the one hand $[Y,X]$ is a bimodule from $\mathcal End(Y)$ to $\mathcal End(X)$. On the other hand $[Y,X]$ becomes a bimodule in the other direction - from $\mathcal End(X)$ $\mathcal End(Y)$ - by pre- and postcomposition with $f$ i.e. pulling back the module structure along $[f,f]:[Y,X]\to[X,Y]$. So assuming $\mathcal V$ is nice enough we have two functors $[Y,X]\otimes_{\mathcal End(X)}$ and $[Y,X]\otimes_{\mathcal End(Y)}$. However i can think of no canditate for a unit of a supposed adjunction between these two.</li>
<li>As in (2) $[Y,X]$ also becomes a semigroup object in $\mathcal V$ that has a (left/right) unit - and thus is a monoid - precisely when $f$ has an (left/right) inverse.</li>
</ol>
<p>My vague guess is that the framework where this question could be handled is that of extranatural transformations.</p>
http://mathoverflow.net/questions/21003/polynomial-bijection-from-qxq-to-q/21004#21004Answer by Garlef Wegart for Polynomial bijection from QxQ to Q?Garlef Wegart2010-04-11T12:49:24Z2010-04-11T13:00:35Z<p>Edit[ The following is wrong ~ see comments]</p>
<p>I don't think so.</p>
<p>Suppose $f$ to be surjective. Let $x\mapsto 0$ and $y\mapsto 1$. Now consider two distinct paths $\gamma,\eta:[0,1]\to\mathbb Q\times\mathbb Q$ from $x$ to $y$. Since $f$ is continuous it maps these paths surjectively onto $[0,1]$ (more exactly $[0,1]\subset f\gamma([0,1])$ and $[0,1]\subset f\eta([0,1])$). Thus, $f$ cannot be injective.</p>
http://mathoverflow.net/questions/20651/motivation-for-equivariant-sheavesMotivation for equivariant sheaves?Garlef Wegart2010-04-07T17:54:31Z2010-04-07T21:15:12Z
<p>Hello everyone;</p>
<p>i'm looking for a motivation for equivariant sheaves (see <a href="http://ncatlab.org/nlab/show/equivariant+sheaf" rel="nofollow">http://ncatlab.org/nlab/show/equivariant+sheaf</a>) ~ <strong>Why are we interested in them?</strong></p>
<p>More explicitely: Can I think of G-equivariant sheaves on a space X as a quotient of the category of sheaves (by some action? in a more general sense) by G?</p>
http://mathoverflow.net/questions/19190/category-groupoid-x-posetCategory = Groupoid x Poset?Garlef Wegart2010-03-24T13:29:11Z2010-03-24T23:05:07Z
<p>Is it possible to split a given category $C$ up into its groupoid of isomorphisms and a category that resembles a poset?</p>
<p>"Splitting up" should be that $C$ can be expressed as some kind of extension of a groupoid $G$ by a poset $P$ (or "directed category" $P$ the only epimorphisms in $P$ are the identities, all isomorphisms in $P$ are identities).</p>
http://mathoverflow.net/questions/18593/what-are-the-worst-notations-in-your-opinion/18607#18607Answer by Garlef Wegart for What are the worst notations, in your opinion ?Garlef Wegart2010-03-18T15:43:30Z2010-03-18T15:43:30Z<p>I think composition of arrows $f:X\to Y$ and $g:Y\to Z$ should be written $fg$ not $gf$. First of all it would make the notation $\hom(X,Y)\to\hom(Y,Z)\to \hom(X,Z)$ much more natural: $\hom(E,X)$ should be a left $\hom(E,E)$ module because $E$ is on the left :) Secondly, diagrams are written from left to right (even stronger: Almost anything in the western world is written left to right). And i think the strange (-1) needed when shifting complexes is an effect of this <strong>twisted</strong> notation.</p>
http://mathoverflow.net/questions/18514/closed-monoidal-structure-on-the-derived-category-of-sheavesClosed monoidal structure on the derived category of sheavesGarlef Wegart2010-03-17T18:52:01Z2010-03-18T13:11:51Z
<p>Given a topological space X, i'd like to find Der X - the derived category of sheaves of abelian groups on X - to be a closed monoidal category. Hom should be cohomological and the internal-hom should be triangulated.</p>
<ol>
<li>Is this possible in full generality? (Unbounded complexes, no restrictions on X)</li>
<li>Consider a sheaf of rings R or equivalently a ring of sheaves. This gives us two things: An abelian category of left R modules that we can derive; let's call this one Der R; A monoid R in Der X whose category of modules we denote dMod R. Is Der R = dMod R'? Ff not: how do they relate?</li>
<li>Given monoids R,S,T in Der X do we get the usual adjunctions in two variables between their categories of bimodules?</li>
<li>Given rings R,S,T in Sh X do we get the usual adjunctions in two variables between their derived categories of bimodules?</li>
</ol>
<p>Now for the question: What is the right setting to do this? As i understand it, there's no suitable model structure that gives 4 in full generality.</p>
http://mathoverflow.net/questions/16672/sheaves-as-full-reflective-subcategoriesSheaves as full reflective subcategoriesGarlef Wegart2010-02-28T12:26:41Z2010-03-04T20:16:11Z
<p>Hello everyone.</p>
<p>My question is concerned with the following statement.</p>
<p><em>"Having a grothendieck topology on a category C is equivalent to having a full reflective subcategory Sh(C) in the category PSh(C) of presheaves, whose reflection is left exact."</em></p>
<p>What i need is a reference for this containing a proof. I tried google but could not find anything besides citations of this result.</p>
http://mathoverflow.net/questions/3165/presheaves-as-limits-of-representable-functorsPresheaves as limits of representable functors?Garlef Wegart2009-10-28T23:11:34Z2009-11-17T14:48:08Z
<p>If i remember correctly, i read that given a presheaf P:C^op -> Set it is possible to describe it as a limit of representable presheaves. Could someone give a description of the construction together with a proof?</p>
http://mathoverflow.net/questions/1083/do-good-math-jokes-exist/3500#3500Answer by Garlef Wegart for Do good math jokes exist?Garlef Wegart2009-10-31T01:23:33Z2009-10-31T01:23:33Z<p>Here's a legend we have at our institute:</p>
<p>Prof: "Give an example of a vector space."</p>
<p>Student: "V"</p>
http://mathoverflow.net/questions/3340/equivariant-derived-categories-via-their-propertiesEquivariant Derived Categories via their properties.Garlef Wegart2009-10-29T21:28:00Z2009-10-30T06:29:33Z
<p>There are some ways to define equivariant derived categories of all sorts. But all the ways i know of involve giving a concrete construction. Is the other way around possible? Is there some universal property that characterizes them?
More explicitly i'm thinking of a bifibration D -> T where T is for example some nice subcategory of Top and D are the derived k-sheaves. For every group object G in T there should be a equivariant bifibration D_G -> T^G and these fibrations should again be bifibered over the category of group objects in T.</p>
<p><strong>So here's the question</strong>: What properties do we actually want in such a situation?</p>
<p>For example i have in mind the "induction equivalence" and "quotient equivalence" as described in the book by Bernstein and Lunts.</p>
http://mathoverflow.net/questions/131255/objects-which-cant-be-defined-without-making-choices-but-which-end-up-independen/131278#131278Comment by Garlef WegartGarlef Wegart2013-05-21T19:29:05Z2013-05-21T19:29:05ZIn representation theory one can often get the multiplicities as dimensions of $\mathrm{hom}(L,X)$ - $L$ being simple. Is there an analogous description for finite groups?http://mathoverflow.net/questions/131255/objects-which-cant-be-defined-without-making-choices-but-which-end-up-independen/131277#131277Comment by Garlef WegartGarlef Wegart2013-05-21T18:51:31Z2013-05-21T18:51:31Z"This can only be done by listing properties of the object that characterise it up to some notion of equivalence." <-- I realised this to be not true - example: The fundamental group.
Then again; in all the cases I can think of there is also a description via universal properties. Maybe this can be turned into an actual theorem?http://mathoverflow.net/questions/131255/objects-which-cant-be-defined-without-making-choices-but-which-end-up-independen/131258#131258Comment by Garlef WegartGarlef Wegart2013-05-20T23:14:11Z2013-05-20T23:14:11ZYes ~ the talk i linked does not deal with numerical computations but rather with type theoretic ones: contructive proofs. And it seems we are looking for a constructive proof of the existence of square roots using the description of the reals via cauchy sequences - without choosing representatives.
Which seens to be the same question as asking if infima can be described without chosing represenatives.http://mathoverflow.net/questions/131255/objects-which-cant-be-defined-without-making-choices-but-which-end-up-independen/131258#131258Comment by Garlef WegartGarlef Wegart2013-05-20T22:36:28Z2013-05-20T22:36:28ZAgain; the distinction is between characterisation and implemention. Characterising the square root is trivial (it's the unique real number that results in the original number, when squared). Implementing it using the description of the reals i gave above is pretty easy.
Implementing it for the implementation of the reals via cauchy sequences seems to be nontrivial ~ at least to me :D.
There is a series of videolectures concerning questions like this, if i remember correctly:
<a href="http://videolectures.net/aug09_spitters_oconnor_cvia/" rel="nofollow">videolectures.net/aug09_spitters_oconnor_cvia</a>http://mathoverflow.net/questions/131255/objects-which-cant-be-defined-without-making-choices-but-which-end-up-independen/131258#131258Comment by Garlef WegartGarlef Wegart2013-05-20T22:21:29Z2013-05-20T22:21:29ZToink is right - to state it differently: The set of Cauchy sequences inherits the addition from $\mathbb Q$. The addition is compatible with the equivalence relation at hand and thus carries over to the quotient $\mathbb R$.http://mathoverflow.net/questions/128387/composition-of-cat-valued-distributors-compatible-with-grothendieck-constructioComment by Garlef WegartGarlef Wegart2013-04-23T07:05:58Z2013-04-23T07:05:58ZYeah; It seems i was a bit vauge. I reformulated the question to make it more comprehensible.http://mathoverflow.net/questions/128393/grothendieck-construction-for-profunctors/128396#128396Comment by Garlef WegartGarlef Wegart2013-04-22T21:29:26Z2013-04-22T21:29:26ZThanks for the refrerence. This seems to be the characterisation i was looking for; let's see if this helps with my other question. :)http://mathoverflow.net/questions/38993/co-limits-and-fibrations-of-dg-categoriesComment by Garlef WegartGarlef Wegart2013-04-22T19:02:07Z2013-04-22T19:02:07ZHm... wether it is the lax- or oplax-colimit depends on how you define the grothendieck construction: you chose the vertical part of a morphism to be either $f_*x\to x'$ or $x' \to f_*x$; both work equally and we have
$\int'_C F = \int_C op \circ F$ http://mathoverflow.net/questions/79102/definition-of-enriched-caterories-or-internal-homs-without-using-monoidal-categor/79103#79103Comment by Garlef WegartGarlef Wegart2011-10-25T19:23:45Z2011-10-25T19:23:45Zthanks for the super-fast answer!http://mathoverflow.net/questions/52989/reference-request-2-monads-and-2-adjunctions/53007#53007Comment by Garlef WegartGarlef Wegart2011-01-24T07:30:50Z2011-01-24T07:30:50ZThank you ~ this was already very helpful. The specific case i am dealing with is that of pseudoalgebras for a strict 2-monad.http://mathoverflow.net/questions/47219/strong-colimits-of-categories/47689#47689Comment by Garlef WegartGarlef Wegart2010-12-05T18:23:36Z2010-12-05T18:23:36ZYes, this is the right answer. In order to prove it one can either follow the construction given by Tom Fiore or split it into two parts: First, we establish the grothendieck construction as a lax colimit. Then, we realize that among all cones the strong cones correspond to functors that send every morphism $X\to Pm(X)$ and thus every (op)cartesian morphism to an isomorphism. These functors however correspond to functors out of the localisation of the grothendieck construction along all (op)cartesian morphisms.http://mathoverflow.net/questions/47219/strong-colimits-of-categoriesComment by Garlef WegartGarlef Wegart2010-11-25T16:11:11Z2010-11-25T16:11:11ZThis does not make the question obsolete, right?
(I fixed the question, btw.)http://mathoverflow.net/questions/47219/strong-colimits-of-categoriesComment by Garlef WegartGarlef Wegart2010-11-24T12:39:03Z2010-11-24T12:39:03ZTank you. This was a point i did not really consider.http://mathoverflow.net/questions/40722/reference-request-2-grothendieck-construction/40727#40727Comment by Garlef WegartGarlef Wegart2010-10-01T13:03:15Z2010-10-01T13:03:15ZThank you; This is exactly the article i was looking for.http://mathoverflow.net/questions/23794/kan-extensions-and-the-yoneda-embedding/23802#23802Comment by Garlef WegartGarlef Wegart2010-05-07T08:39:22Z2010-05-07T08:39:22ZThank you very much Martin, this is a good example.