User marc olschok - MathOverflow

Answer by Marc Olschok for isomorphism locus of functors on presentable categories

2012-12-07T17:20:32Z

Since $\{x\in C : \eta(x) \textrm{ is an isomorphism}\}$ is cocomplete (indeed closed under colimits in $C$) it remains to show that it is accessible.

First observe that the arrow category $D^{\cdot\to\cdot}$ is again locally presentable. Let $D^{\cdot\to\cdot}_\cong$ be its full replete subcategory determined by all isomorphisms. This subcategory is also locally presentable and closed under colimits in $D^{\cdot\to\cdot}$.

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 $\{x\in C : \eta(x) \textrm{ is an isomorphism}\}$ is exactly the full preimage of $D^{\cdot\to\cdot}_\cong$ under $H$. By Remark 2.50 of LPAC it is therefore also accessible.

Answer by Marc Olschok for Left determined model structure on delta-generated topological spaces

2012-07-06T17:41:50Z

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.

Given a map $f\colon X\to Y$ in $\mathrm{Top}_\Delta$, first build the following diagram in $\mathrm{Top}$:

$$ \matrix{ X & \mathop{\longrightarrow}\limits^{\tilde{f}} & N_f & \mathop{\longrightarrow}\limits^{p'} & X \cr {\scriptstyle f} \big\downarrow {\ }& & {\ } \big\downarrow {\scriptstyle f'} & & {\ } \big\downarrow {\scriptstyle f} \cr Y & \mathop{\longrightarrow}\limits_t & Y^I & \mathop{\longrightarrow}\limits_{p_0} & Y \cr & & {\ } \big\downarrow {\scriptstyle p_1} & & \cr & & Y{\ } & & } $$

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 $\mathrm{Top}$.

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 $f$.

The map $\hat{f}$ is always a fibratrion. This is the only appeal to classical algebraic topology.

Now apply the coreflection $k\colon \mathrm{Top}\to \mathrm{Top}_\Delta$ to that diagram.

Then we have:

(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.

(2) $k(p')$ 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.

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.

Answer by Marc Olschok for Gluing two graphs

2012-03-21T16:34:45Z

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.

For simplicity, I suppose you consider only undirected graphs.

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 $\forall a,a'\in A: (a,a')\in\alpha\Rightarrow (ha,ha')\in\beta$.

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$

Comment by Marc Olschok

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.