Spahn HTT, A.2.2 (Rev #2)

Definition

Let XX be an object in a model category.

  1. A cylinder object is defined to be a factorization of the codiagonal map XXXX\coprod X\to X for XX into a cofibration followed by a weak equivalence.

  2. A path object is defined to be a factorization of the diagonal map XX×XX\to X\times X for XX into a weak equivalence followed by a fibration .

Proposition A.2.2.1

Let CC be a model category. Let XX be a cofibrant object of CC. Let YY be a fibrant object of CC. Let f,g:XYf,g:X\to Y be two parallel morphisms. Then the following conditions are equivalent.

  1. The coproduct map fgf\coprod g factors through every cylinder object for XX.

  2. The coproduct map fgf\coprod g factors through some cylinder object for XX.

  3. The product map f×gf\times g factors through every path object for YY.

  4. The product map f×gf\times g factors through some path object for YY.

Definition

(homotopy, homotopy category of a model category)

Let CC be a model category.

(1) Two maps f,g:XYf,g:X\to Y from a cofibrant object to a fibrant object satisfying the conditions of Proposition A.2.2.1 are called homotopic morphisms. Homotopy is an equivalence relation \simeq on hom C(X,Y)hom_C (X,Y).

(2) The homotopy category hCh C of CC is defined to have as objects the fibrant-cofibrant objects of CC. The hom objects hom hC(X,Y)hom_{hC}(X,Y) are defined to be the set of \simeq equivalence classes of hom C(X,Y)hom_C (X,Y).

Remark

The homotopy category hChC (more precisely the projection map Q:ChCQ:C\to hC) is couniversal in the following sense:

  • for any (possibly large) category AA and functor F:CAF : C \to A such that FF sends all wWw \in W to isomorphisms in AA, there exists a functor F Q:hCAF_Q : hC \to A and a natural isomorphism
C F A Q F Q hC \array{ C &&\stackrel{F}{\to}& A \\ \downarrow^Q& \Downarrow^{\simeq}& \nearrow_{F_Q} \\ hC }

The second condition implies that the functor F QF_Q in the first condition is unique up to unique isomorphism.

Revision on June 29, 2012 at 20:44:02 by Stephan Alexander Spahn?. See the history of this page for a list of all contributions to it.