User marc olschok - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:28:51Z http://mathoverflow.net/feeds/user/22141 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114860/isomorphism-locus-of-functors-on-presentable-categories/115730#115730 Answer by Marc Olschok for isomorphism locus of functors on presentable categories Marc Olschok 2012-12-07T17:20:32Z 2012-12-07T17:20:32Z <p>Since <code>$\{x\in C : \eta(x) \textrm{ is an isomorphism}\}$</code> is cocomplete (indeed closed under colimits in $C$) it remains to show that it is accessible.</p> <p>First observe that the arrow category $D^{\cdot\to\cdot}$ is again locally presentable. Let <code>$D^{\cdot\to\cdot}_\cong$</code> be its full replete subcategory determined by all isomorphisms. This subcategory is also locally presentable and closed under colimits in $D^{\cdot\to\cdot}$.</p> <p>Now consider the functor $H:C\longrightarrow D^{\cdot\to\cdot}$ given by $H(x) = \eta(x)$ on objects and $H(u:c\to c') = (Fu,Gu):\eta(c)\to\eta(c')$ on morphisms. Then $H$ is again a cocontinuous functor between locally presentable categories and <code>$\{x\in C : \eta(x) \textrm{ is an isomorphism}\}$</code> is exactly the full preimage of <code>$D^{\cdot\to\cdot}_\cong$</code> under $H$. By Remark 2.50 of LPAC it is therefore also accessible.</p> http://mathoverflow.net/questions/100153/left-determined-model-structure-on-delta-generated-topological-spaces/101519#101519 Answer by Marc Olschok for Left determined model structure on delta-generated topological spaces Marc Olschok 2012-07-06T17:41:50Z 2012-07-06T18:02:09Z <p>Philippe's argument works well. I just wanted to remark that in this special case one does not need to invoke the homotopy exchange property explicitely as long as one already accepts that a model structure with the given cofibrations and fibrant object exists for the category $\mathrm{Top}_\Delta$ of $\Delta$-generated spaces.</p> <p>Given a map $f\colon X\to Y$ in $\mathrm{Top}_\Delta$, first build the following diagram in $\mathrm{Top}$:</p> <p>$$\matrix{ X &amp; \mathop{\longrightarrow}\limits^{\tilde{f}} &amp; N_f &amp; \mathop{\longrightarrow}\limits^{p'} &amp; X \cr {\scriptstyle f} \big\downarrow {\ }&amp; &amp; {\ } \big\downarrow {\scriptstyle f'} &amp; &amp; {\ } \big\downarrow {\scriptstyle f} \cr Y &amp; \mathop{\longrightarrow}\limits_t &amp; Y^I &amp; \mathop{\longrightarrow}\limits_{p_0} &amp; Y \cr &amp; &amp; {\ } \big\downarrow {\scriptstyle p_1} &amp; &amp; \cr &amp; &amp; Y{\ } &amp; &amp; }$$</p> <p>Here $(Y^I,p_0,p_1,t)$ is the usual path object on $Y$, $N_f$ is the pullback of $p_0$ and $f$, and $\tilde{f}$ is the map induced by $id_X$ and $f.t\colon X \to Y^I$. Observe that the $p_0$ and $p_1$ are trivial fibrations in $\mathrm{Top}$.</p> <p>Define $\hat{f}\colon N_f\to Y$ as the composite $f'.p_0\colon N_f\to Y^I\to Y$. Then $f = \tilde{f}.\hat{f}\colon X\to N_f\to X$ is called the 'glueing factorization' of $f$.</p> <p>The map $\hat{f}$ is always a fibratrion. This is the only appeal to classical algebraic topology.</p> <p>Now apply the coreflection $k\colon \mathrm{Top}\to \mathrm{Top}_\Delta$ to that diagram.</p> <p>Then we have:</p> <p>(1) $k(\hat{f})$ is a fibration in $\mathrm{Top}_\Delta$ and the maps $k(p_0)$ and $k(p_1)$ are trivial fibrations in $\mathrm{Top}_\Delta$ because $k$ preserves fibrations and trivial fibrations.</p> <p>(2) $k(p')$ is a trivial fibration in $\mathrm{Top}_\Delta$ because it is the pullback (in <code>$\mathrm{Top}_\Delta$</code>) of $k(p_0)$ along $k(f)$. Therefore $k(\tilde{f})$ lies in the smallest localizer by the 2-for-3-property.</p> <p>Now suppose that $f$ is a weak equivalence in $\mathrm{Top}_\Delta$. Then the same holds for $k(f)$ and the 2-for-3 property yields that $k(f')$ is also a weak equivalence. Consequently $k(\hat{f}) = k(f').k(p_1)$ is a trivial fibration and $k(f) = k(\tilde{f}).k(\hat{f})$ is in the smallest localizer.</p> http://mathoverflow.net/questions/90965/gluing-two-graphs/91833#91833 Answer by Marc Olschok for Gluing two graphs Marc Olschok 2012-03-21T16:34:45Z 2012-03-21T16:34:45Z <p>Since I cannot add comments, I put this as an answer. I do not know about specific graph-theoretical literature about the operation you describe, but it seems that it can be exhibited as a pushout in a suitable category.</p> <p>For simplicity, I suppose you consider only undirected graphs.</p> <p>Take the category where objects are sets equipped with a reflexive and symmetric relation, and where a morphism $h: (A,\alpha)\to(B,\beta)$ is an (ordinary) map of sets $h:A\to B$ that satisfies $\forall a,a'\in A: (a,a')\in\alpha\Rightarrow (ha,ha')\in\beta$.</p> <p>Then the objects of that category correspond to (undirected simple) graphs and therefore also give a notion of graph homomorphism. Let $1$ be the graph with exactly one vertex and write $n$ for the discrete graph with $n$ vertices. Then, given two graphs $G$ and $H$, choosing $n$ vertices in $G$ and $H$ is the same as giving monomorphisms $n\to G$ and $n\to H$, and your gluing operation is then given as a pushout of $G\leftarrow$n$\rightarrow H$</p> http://mathoverflow.net/questions/100153/left-determined-model-structure-on-delta-generated-topological-spaces/101519#101519 Comment by Marc Olschok Marc Olschok 2012-07-06T17:52:54Z 2012-07-06T17:52:54Z Sorry, I have know idea why the rendering software suddenly trips over line (2) above. The text should read $k(p')$ is a trivial fibration in $\mathrm{Top}_\Delta$ because it is the pullback (in $\mathrm{Top}_\Delta$) of $k(p_0)$ along $k(f)$. Therefore $k(\tilde{f})$ lies in the smallest localizer by the 2-for-3-property.