Is the terminal object always cofibrant?

Question

Let $\mathcal{M}$ be a model category. Is the unique map from the initial to the terminal object $!: \mathbf{0} \rightarrow \mathbf{1}$ always a cofibration?
If it helps, we can even assume that $\mathcal{M}$ is combinatorial (i.e. locally presentable and cofibrantly generated).

The question arose from a discussion of who to define spheres in a general model category and we agreed that $S^0 = \mathrm{hocolim}(I \leftarrow \mathbf{0} \rightarrow I)$ for an "appropriate object" $I$, seemed like a natural definition (where $I = \mathbf{1}$ in most cases). This mostly boils down to $S^0 \cong \mathbf{1} \sqcup \mathbf{1}$ but it is tied to the fact that $! : \mathbf{0} \rightarrow \mathbf{1}$ is a cofibration. Thus the question.

So far, we observed that if there exists a cofibrant object with a global element, i.e. a map $* : \mathbf{1} \rightarrow X$ with $X$ cofibrant, then the unique map $!$ must already be a cofibration. Indeed we can write $!$ as a retract of this global element $$\require{AMScd} \begin{CD} \mathbf{0} @>>> \mathbf{0} @>>> \mathbf{0}\\ @V{!}VV @V{}VV @V{!}VV \\ \mathbf{1} @>{*}>> X @>>> \mathbf{1} \end{CD} \, .$$

We know that cofibrant objects always exists, and we can even construct global elements out of them by taking the coproduct $\operatorname{colim}( \mathbf{1} \leftarrow \mathbf{0} \rightarrow X) \cong \mathbf{1} \sqcup X$. Writing it as a pushout like this even tells us that the map $\mathrm{id} \sqcup* : \mathbf{1} \rightarrow \mathbf{1} \sqcup X$ itself is a cofibration, but we cannot guarantee anymore that $\mathbf{1} \sqcup X$ is cofibrant (at least we didn't know how to show this).

At this point we don't even have an educated guess if the map $!:\mathbf{0} \rightarrow \mathbf{1}$ is always a cofibration or not.

Answer 1

There's a model structure on $\mathbf{Set}$ where cofibrations are surjections, fibrations are injections and weak equivalences are all maps. Because $!:\mathbf{0} \rightarrow \mathbf{1}$ is not surjective, it is not cofibration.

Answer 2

There is a (combinatorial simplicial) model structure on the category of simplicial rings where the fibrations and weak equivalences are induced by the forgetful functor to the category of simplicial sets. In this model structure, the cofibrant objects are the retracts of free simplicial rings. The terminal object (the zero ring) is not a retract of any free simplicial ring, hence not cofibrant.

More explicitly, consider the following commutative diagram: $$\require{AMScd} \begin{CD} \mathbb{Z} @>>> A \\ @VVV @VVV \\ \{ 0 \} @>>> \{ 0 \} \end{CD}$$ Since every simplicial abelian group is a Kan complex, $A \to \{ 0 \}$ is always a fibration, and it is a trivial fibration if and only if $A$ is a contractible Kan complex. But there is a homomorphism $\{ 0 \} \to A$ if and only if $A \cong \{ 0 \}$, and there are certainly non-zero simplicial rings that are contractible as Kan complexes. Therefore $\mathbb{Z} \to \{ 0 \}$ is not a cofibration.