# If weak-equivalences in a model category are generated by an equivalence relation $\sim$ on hom-sets…

Suppose that $C$ is a bicomplete category, and $\sim$ is an equivalence relation defined on every hom-set of $C$, compatible with composition. Then we can define the quotient category $C/\sim$ whose objects are the same as $C$ and whose morphisms are classes of morphisms of $C$ with respect to $\sim$; let's denote with $\pi:C\twoheadrightarrow C/\sim$ the associated projection functor.

We say that a map $f:X\rightarrow Y$ is an equivalence if $\pi(f)$ is an isomorphism, i.e. if there is $g:Y\rightarrow X$ such that $gf\sim 1_X$ and $fg\sim 1_Y$; let's denote with $E$ the class of such equivalences.

Now let's suppose that $E$ can be completed to a model structure on $C$. Then there is a projection-localization functor $L:C\twoheadrightarrow HoC$ to the homotopy category of $C$ with respect to such a model structure. In particular the functor $\pi$ uniquely descends to $D\pi:HoC\twoheadrightarrow C/\sim$ such that $D\pi\circ L=\pi$, and thus $f\approx g \Rightarrow f\sim g$, where $\approx$ denotes the homotopy relation defined within the model structure.

Which hypothesis do I have to require so that the converse statement is also true (i.e. so that $f\sim g \Rightarrow f\approx g$ and thus $HoC\simeq C/\sim$)? Is it sufficient for example that every object in $C$ is bifibrant (so that $HoC$ is just the classical homotopy category of $C$) or do I need more (or maybe less) subtle properties?

• I doubt bifibrancy is enough (or even necessary). You should think about how the proof works for the case $\mathcal{C} = \mathbf{Top}$ and the classical model structure – then it will be clear that you should ask for some notion of path object or cylinder object with respect to $\sim$. – Zhen Lin Aug 18 '15 at 7:16
• Yes, I though that bifibrancy had little to do with this, it was just an example. I'm quite new to these concepts, and I have only seen algebrical examples (I was thinking in particular to the category of modules over a frobenius ring, for which I can't prove that statement). I'll gite Top a try, do you have some particular reference to suggest? – LK512 Aug 18 '15 at 8:15