Model categories: $\text{Ho}$ and $\cal C_{cf}/\sim$

Question

I have asked this question about model categories:

Why $\text{Ho} \ \cal C$ is $\cal C_{cf}/\sim$ and not $\cal C/\sim$ and I got this answer: take for cofibratiobns Iso, weak equivalences all arrows. Then $\text{Ho}\ \cal C$ is $1$. Then $f \sim g$ for all f,g where $f \sim g$ means that $f$ and $g$ are homtopic. If $\cal C$ is poset, $\cal C/\sim$ is $\cal C$.

Now I wonder why cofibrations should be Iso, is this 
(and even the notion of cofibration) used in any way in the above?

Answer 1

The point of the choice of model category is that

1. all morphisms are weak equivalences, so that $\operatorname{Ho}\mathcal C$ is readily equivalent to $1$ (note that this relies on $\mathcal C$ having for example an initial object, since this fails for "disconnected" categories as Zhen Lin pointed out), but also all pairs of parallel morphisms are homotopic (to see this, note that a cylinder object $X\amalg X\rightarrowtail\operatorname{Cyl}X\twoheadrightarrow X$ is necessarily isomorphic to $X\amalg X$ and thus any pair of parallel morphisms admit a left homotopy between them), showing that $\mathcal C/\sim$ will not necessarily be equivalent to $1$ (the explicit example they give is to take $\mathcal C$ to be a posetal category, in which case homotopy of maps is just given by equality and therefore $\mathcal C/\sim$ is exactly the same as $\mathcal C$)
2. the only cofibrant objects are initial in $\mathcal C$, so that $\mathcal C_{c}$ (and thus $\mathcal C_{cf}/\sim$) is also readily equivalent to $1$, showing that this is correct in creating the homotopy category of $\mathcal C$

The answer you were given was emphasising how taking $\mathcal C/\sim$ is wrong (which doesn't require the cofibrations). Seeing how $\mathcal C_{cf}/\sim$ is right requires the cofibrations essentially by definition.

While this answer certainly provides a concrete counterexample for your question, I'll provide a softer reason why one would want to restrict their attention to $\mathcal C_{cf}$. Note that the notion of homotopy is not that well-defined over general objects: there are two notions of homotopy (left homotopy with cylinder objects and right homotopy with path objects), and we don't know a priori whether they coincide. Perhaps you could remedy this by taking $\sim$ to be the equivalence relation generated by both kinds of homotopy, but such a thing would be incredibly difficult to study. However, when you restrict your attention to nicer objects, you get a cleaner notion of homotopy theory:

If $X$ is cofibrant and $Y$ is fibrant, then the notions of left and right homotopy on $\operatorname{Hom}_{\mathcal C}(X,Y)$ coincide and define an equivalence relation.

(for a proof, see here) Therefore, if we restrict our attention to $\mathcal C_{cf}$, the relation $\sim$ becomes much easier to work with: if $f\sim g$, then it admits a left and a right homotopy, and you are free to use either one to reason with it! What's most important, though, is that we get an analogue of Whitehead's theorem:

If $X$ and $Y$ are cofibrant fibrant, then a weak equivalence $X\to Y$ is automatically a homotopy equivalence $X\simeq Y$.

Showing that $\mathcal C_{cf}/\sim$ will convert weak equivalences to isomorphisms.

It's also important to note: you can technically make sense of a homotopy category just using the weak equivalences (see here), defining it by being universal with the property that the image of weak equivalences in $\mathcal C\to\operatorname{Ho}\mathcal C$ are isomorphisms. We can even define it explicitly, by taking $\operatorname{Ho}\mathcal C$ to be the category with the same objects as $\mathcal C$, but whose morphisms $X\to Y$ are given by zigzags $$ X\gets\gets\to\gets\to\dots\to\to\gets Y $$ where all left-pointing arrows are given by weak equivalences and are thought of as formal inverses to the weak equivalence. However, at this level of generality, the homotopy category could be extremely large, and extremely difficult to control. Part of the motivation for studying a model category is that the auxiliary data provided by cofibrations and fibrations makes it easier to study the homotopy category ($\mathcal C_{cf}/\sim$ is a much better-behaved category). Therefore, one would expect that the cofibrations and fibrations would play a significant role when defining $\operatorname{Ho}\mathcal C$; if we just used $\mathcal C/\sim$, then what good were the cofibrations and fibrations?