Model category structure on Set without axiom of choice - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:53:10Z http://mathoverflow.net/feeds/question/10246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10246/model-category-structure-on-set-without-axiom-of-choice Model category structure on Set without axiom of choice Reid Barton 2009-12-31T07:07:40Z 2012-02-16T01:14:49Z <p>There is a model category structure on Set in which the cofibrations are the monomorphisms, the fibrations are maps which are either epimorphisms or have empty domain, and the weak equivalences are the maps $f : X \rightarrow Y$ such that $X$ and $Y$ are both empty or both nonempty.</p> <p>In order for the lifting axioms to hold we need the axiom of choice. Suppose we want to avoid the axiom of choice. One option seems to be to replace "epimorphism" with "map which has a section" everywhere. Can we instead leave the definition of fibration unchanged and change the definition of cofibration?</p> <p>Note that if $A$ is a cofibrant set in this hypothetical model structure then any surjection $X \rightarrow A$ has a section. So it would be necessary that every set admits a surjection from a set $A$ of this type, which seems rather implausible to me. Perhaps the notion of model category needs to be modified in a setting without the axiom of choice.</p> <p>(Apologies if this question turns out to be meaningless or trivial; I have not thought about it much nor do I often try to avoid using the axiom of choice.)</p> <hr> <p>Let me explain the motivation behind this question. I am trying to get a better picture of what category theory without the axiom of choice looks like.</p> <p>My rough understanding is that in the absence of the axiom of choice we should use <a href="http://ncatlab.org/nlab/show/anafunctor" rel="nofollow">anafunctors</a> in place of functors in some if not all contexts. An anafunctor from $C$ to $D$ is a span of functors $C \leftarrow E \rightarrow D$ in which the left leg $E \leftarrow C$ is a surjection on objects and fully faithful. The point is that even under these conditions $E \rightarrow C$ may not have a section. For instance, suppose $C$ is a category with all binary products. There may not be a product functor $- \times -: C \times C \rightarrow C$, because we cannot simply choose a distinguished product of each pair of objects. However, if we define $Prod(C)$ to be the category whose objects are diagrams of the shape $\bullet \leftarrow \bullet \rightarrow \bullet$ in $C$ which express the center object as the product of the outer two, there is a forgetful functor $Prod(C) \rightarrow C \times C$ remembering only the outer objects, and it is surjective on objects (because $C$ has binary products) and fully faithful. Furthermore there is another forgetful functor $Prod(C) \rightarrow C$ which remembers only the center object. Together these define a product anafunctor $C \times C \leftarrow Prod(C) \rightarrow C$.</p> <p>"Classically" (:= under $AC$) the following paragraph holds: There is a model category structure on $Cat$ in which the cofibrations are functors which are injective on objects, fibrations are the functors with the right lifting property w.r.t. the inclusion of an object into the contractible groupoid on two objects, and weak equivalences are equivalences of categories. In particular the acyclic fibrations are the functors which are surjective on objects and fully faithful, exactly the functors we allow as left legs of anafunctors. Since every category is fibrant in this model structure, we can view an anafunctor from $C$ to $D$ as a representative of an element of $RHom(C, D)$, i.e., a functor from a cofibrant replacement for $C$ to $D$. Of course, every category is cofibrant too so for our cofibrant "replacement" we can just take $C$, and we learn that anafunctors from $C$ to $D$ are the same as functors when we consider both up to natural isomorphism (homotopy).</p> <p>I would like to understand anafunctors as a kind of $RHom$ also without the axiom of choice. But I cannot use the same definition of the model category structure, because the lifting axioms require $AC$. I would like to keep the same acyclic fibrations, since they appear in the definition of anafunctor, and I would like every object to be fibrant. I cannot really imagine what a cofibrant replacement could look like in this model structure, but then I am not accustomed to working without $AC$.</p> <p>My original question is related to this one via the functor which assigns to a set $S$ the codiscrete category on $S$, and it seems to contain the same kinds of difficulties.</p> http://mathoverflow.net/questions/10246/model-category-structure-on-set-without-axiom-of-choice/10299#10299 Answer by Reid Barton for Model category structure on Set without axiom of choice Reid Barton 2009-12-31T18:02:55Z 2009-12-31T18:02:55Z <p>I came across the nlab page for the axiom <a href="http://ncatlab.org/nlab/show/COSHEP" rel="nofollow">COSHEP</a> (category of sets has enough projectives) which seems to be just what's needed to obtain a model category structure, as usually understood, on Set with the weak equivalences and fibrations I wanted. This is a partial answer to my question, but not totally satisfying from the point of view of my motivation, since anafunctors are supposed to work without any such axiom, as far as I know. So I need to think more about what my question really should be. (But if you think you know what it should be, please tell me!)</p> http://mathoverflow.net/questions/10246/model-category-structure-on-set-without-axiom-of-choice/10307#10307 Answer by Mike Shulman for Model category structure on Set without axiom of choice Mike Shulman 2009-12-31T18:51:02Z 2009-12-31T19:01:09Z <p>Indeed, COSHEP (more traditionally called the "presentation axiom" by constructivists) does seem to be what you need in order to get a model structure on Set, or Cat. That's true in a lot of similar cases: cofibrant objects are always "projective" in some sense, and you can't expect to get many projective objects in a category constructed from sets unless you started with enough projective objects in Set itself. So I don't expect that ordinary model category theory will be much good for anything at all if you don't have at least COSHEP. But, thankfully, model categories are not the be-all and end-all of homotopy theory, so we can still formally invert the weak equivalences (functors that are fully faithful and essentially surjective) to obtain anafunctors as a derived hom.</p> http://mathoverflow.net/questions/10246/model-category-structure-on-set-without-axiom-of-choice/10368#10368 Answer by Denis-Charles Cisinski for Model category structure on Set without axiom of choice Denis-Charles Cisinski 2010-01-01T13:22:22Z 2012-02-16T01:09:44Z <p>For any elementary topos $T$, there is a model category structure on $T$, whose cofibrations are the monomorphisms, and whose weak equivalences are the maps $X\to Y$ such that, either $Y$ is empty, either $X$ is non-empty (the fibrations are the split epimorphisms). In a topos, there exists an internal Hom (written here $\underline{Hom}$), as well as a subobject classifier $\Omega$: it comes with a map $t:1\to \Omega$ which is the universal subobject, in the sense that, for any object $X$ in $T$, pulling back along $t$ induces a bijection $$\text{ { subobjects of X} }\simeq \ Hom(X,\Omega)$$ Using the universal property of $\Omega$, the diagonal $X\to X\times X$ defines a map $X\times X\to \Omega$, whence an embedding of $X$ into its power object: $X\to P(X)=\underline{Hom}(X,\Omega)$. This gives you injective resolutions. We thus have a functorial fibrant replacement $X\mapsto I(X)$, where $I(X)=X$ if $X$ is empty, and $I(X)=P(X)$ otherwise (note that an object of $T$ is fibrant if and only if it is either empty, either injective). It remains to factor maps into a trivial cofibration (resp. cofibration) followed by a fibration (resp. trivial fibration). Let $f:X\to Y$ be a map in $T$. Such a factorization for $f$ is given by the map $X\to F\times Y$ followed by the projection $F\times Y\to Y$, for $F=I(X)$ (resp. $F=P(X)$). Hence we get a model structure on $T$, while we never used the axiom of choice.</p> <p>Edit: and now, some genuine non sense (not abstract):</p> <p>This model category structure on $Set$ is rather degenerate though. It seems that, to get model categories in general (on $Cat$ or on simplicial sets), we should change the definition of a model category by asking for lifting properties only internally: given maps $i:A\to B$ and $p:X\to Y$, $i$ has the (internal) left lifting property with respect to $p$ is the map $$Hom(B,X)\to Hom(X,A)\times_{Hom(Y,A)}Hom(Y,B)$$ is an epimorphism. In this way $Cat$ and $SSet$ remain model categories without the axiom of choice.</p> <p>Edit: here, it seems I have been rather optimistic (see Mike comment below). I was thinking about working above an arbitrary topos (instead of sets), but, if we work externally in a setting without axiom of choice, then the lifting property suggested above is just the usual one, so that my remark seems to be (and is) silly. But what follows still makes sense.</p> <p>Note that it is easy to get a structure of category of fibrant objects (in the sense of Brown, <em>Abstract homotopy theory and generalized sheaf cohomology</em>, Trans. Amer. Math. Soc. 186 (1974), 419–458) on $Cat$, which is sufficient to have calculus of fractions (up to homotopy), and gives you the abstract tools to explain the good behaviour of anafunctors. As for the size problems when we avoid the axiom of choice, maybe it would be worth looking for an "internal notion of size" to get that the Hom's of any homotopy category of a model category are small in some sense (but I had never thought seriously about this before).</p> http://mathoverflow.net/questions/10246/model-category-structure-on-set-without-axiom-of-choice/45295#45295 Answer by goli for Model category structure on Set without axiom of choice goli 2010-11-08T11:40:25Z 2010-11-08T11:40:25Z <p>why in the category of sets every object is projective?</p>