Clarify on Lemma 1.2.2, Hovey's Model Categories

Question

Let $\mathscr C$ be a model category, and denote as usual by $\operatorname{Ho}\mathscr C$ the localization of $\scr C$ with respect to the weak equivalences.

For any category $\scr D$, let $\mathrm{Fun}'(\mathscr C,\mathscr D)$ be the class of functors $\scr C\to D$ sending weak equivalences to isomorphisms. The property defining $\operatorname{Ho}\mathscr C$ is the following: $$ (*)\ \ \mathrm{Fun}(\operatorname{Ho}\mathscr C,\mathscr D)\cong \mathrm{Fun}'(\mathscr C,\mathscr D)$$ naturally in $\scr D$. Lemma 1.2.2 (iii) on Hovey's Model Categories says that, viewing $\mathrm{Fun}'(\mathscr C,\mathscr D)$, $\mathrm{Fun}(\operatorname{Ho}\mathscr C,\mathscr D)$ as categories instead of mere classes, the bijection ($*$) extends to an isomorphism of categories.

It is easy to see this when $\operatorname{Ho}\mathscr C$ is constructed as a quotient of the free category over $\scr C$, but I don't know how to prove it just by using the universal property of $\operatorname{Ho}\mathscr C$: I obtain a functor $\mathrm{Fun}(\operatorname{Ho}\mathscr C,\mathscr D)\to \mathrm{Fun}'(\mathscr C,\mathscr D)$, bijective on objects of course, which I can't show to be isomorphic. However, the universal property should contain all the information about $\operatorname{Ho}\mathscr C$; is there a way to prove Lemma 1.2.2 (iii) only by the universal property, or maybe, the fact that we are dealing with $2$-categorical structure in some way, requires to choose a certain representation for $\operatorname{Ho}\mathscr C$? Thank you for your help.

Answer 1

You need the following facts:

• For every category $\mathcal{D}$, we have a natural isomorphism $$\textrm{Fun} (\operatorname{Ho} \mathcal{C}, [\mathbf{2}, \mathcal{D}]) \cong \operatorname{mor} {[\operatorname{Ho} \mathcal{C}, \mathcal{D}]}$$ where $[-, -]$ is the functor category.

• For every category $\mathcal{D}$, we have a natural isomorphism $$\textrm{Fun}' (\mathcal{C}, [\mathbf{2}, \mathcal{D}]) \cong \operatorname{mor} {[\mathcal{C}, \mathcal{D}]_\mathrm{h}}$$ where $[-, -]_\mathrm{h}$ is the full subcategory of functors sending weak equivalences to isomorphisms.

It is then a formal consequence that the natural bijection $$\textrm{Fun} (\operatorname{Ho} \mathcal{C}, \mathcal{D}) \cong \textrm{Fun}' (\mathcal{C}, \mathcal{D})$$ can be promoted to a natural isomorphism $$[\operatorname{Ho} \mathcal{C}, \mathcal{D}] \cong [\mathcal{C}, \mathcal{D}]_\mathrm{h}$$ as claimed.

Answer 2

I don't think that you can deduce the isomorphism of categories $$\operatorname{Fun}(\operatorname{ho}\mathcal{C},\mathcal{D}) \cong \operatorname{Fun}'(\mathcal{C},\mathcal{D})$$ just knowing the isomorphism of classes $$\operatorname{Ob Fun}(\operatorname{ho}\mathcal{C},\mathcal{D}) \cong \operatorname{Ob Fun}'(\mathcal{C,D})$$ since taking objects is not a conservative functor. If you have an explicit construction of $\operatorname{ho}\mathcal{C}$, however, you can use it to prove the isomorphism of categories and deduce from it the isomorphism of the level of object classes.

You are right that a good universal property should give you all the information about an object, but the point here is that by taking $\operatorname{Ob}$ you truncate the universal property, so you lose information...