Why Quillen equivalence is used to define "equivalent" model categories?

573 Views Asked by At

Why is Quillen equivalence used as the notion of equivalence of model categories?

So given two model categories $C,D$, $$L \dashv R:C \rightarrow D$$ is called a Quillen adjunction if $L$ preserves cofibrations and $R$ preserves fibrations (or some other equivalent formulations).


This induces a map in the homotopy categores: $$\Bbb L L \dashv \Bbb R R: Ho(C) \rightarrow Ho(D)$$

this is a Quillen equivalence if this adjunction is in fact an equivalence of categories.


What properties capture the idea that this is the right notion of "equivalence" by passing to homotopy category: rather than simply requiring adjoint equivalence of $C,D$?

References would help - I am aware of nlab post but am not satisfied.

1

There are 1 best solutions below

0
On

In homotopy theory, we are interested to study objects up to weak equivalences. Recall that if $\mathcal{C}$ is a model category with a class of weak equivalence $\mathcal{W}$, then the associated homotopy category $\mathbf{Ho}(\mathcal{C})$ is the localization $\mathcal{C}[\mathcal{W}^{-1}]$ of $\mathcal{C}$ with respect to $\mathcal{W}$. So asking for an adjunction between two model categories $\mathcal{C}$ and $\mathcal{D}$ such that it induces an equivalence of categories in their localizations is a sensible thing to do.

For instance, Quillen showed that, up to weak equivalences, simplicial sets and topological spaces are "Quillen" equivalent (with suitable model categories). Topological spaces are far more complicated than simplicial sets and this results allows you to consider objects which are much more combinatorial, as long as you care about things up to homotopy.

However, this doesn't mean the definition is perfect. As you may have seen in the nlab post you linked, Dugger-Shipley have shown that equivalent homotopy categories doesn't necessarily imply there is a Quillen equivalence on the level of model categories.