# Are cofibrant/fibrant replacements homotopically unique?

Let $\mathcal{M}$ be a model category in the old sense, i.e. with factorisations that are not necessarily functorial, and let $X$ be an object in $\mathcal{M}$. The category of cofibrant replacements for $X$, denoted by $\mathbf{Cof} (X)$, is the full subcategory of the slice category $\mathcal{M}_{/ X}$ spanned by objects $q : \tilde{X} \to X$ where $\tilde{X}$ is cofibrant and $q$ is a weak equivalence.

Question. Is the nerve of $\mathbf{Cof} (X)$ (weakly) contractible?

This is true and easy to show when $\mathcal{M}$ has functorial cofibrant replacements (in particular, when $\mathcal{M}$ is a model category in the new sense): see e.g. Theorem 14.6.2 in [Hirschhorn, Model categories and their localizations]. The idea is to build an explicit 2-step homotopy from the identity functor to a constant functor, however, we really do need functoriality to make this proof go through.

The problem, therefore, is to find an alternative proof that does not make use of functoriality. It looks like the answer should still be affirmative. First of all, we may assume without loss of generality that $X$ is a terminal object in $\mathcal{M}$. In that case, $\mathbf{Cof} (X)$ is isomorphic to the full subcategory of $\mathcal{M}$ spanned by the cofibrant objects whose unique morphism to $X$ is a weak equivalence. Let $B$ be the nerve of this category. With some work, one can show that any two edges of $B$ are homotopic as paths in $| B |$. Something similar can be said for spheres, using the 2-coskeletality of $B$. Unfortunately, that isn't enough to show that $B$ is contractible (unlike the groupoid case!).

I suppose it is possible that the answer to the question is negative, which would then be an answer to this question.