In a closed model category, is the identity $\textrm{id}: A \to A$ a cofibration? Does it only hold on some special cases? Or is it never true?



Is the identity a cofibration?
Yes: it is always a cofibration. Fix a trivial fibration $f : X \to Y$. It suffices to show that $\hom(A,X) \to \hom(A,Y) \times\_{\hom(A,Y)} \hom(A,X)$ is surjective (informally: $\mathsf{id}$ lifts against $f$). Inspecting the codomain of this map, we see that it's isomorphic to $\hom(A,X)$ so the map is not only surjective, it's an isomorphism.


The same argument shows that any isomorphism is a cofibration and even---as weak equivalences contain all isomorphisms---a trivial cofibration. This argument also does not require the model category to be closed or satisfy any additional properties. One consequence is that this fact then dualizes to show that isomorphisms are also (trivial) fibrations.


