Evaluation functors and injective model structure on diagram categories

Question

If $M$ is a combinatorial model category, it's known by the experts that there is a ''natural'' model structure on diagram categories $Hom(C,M)$, which is the projective model structure. The fibrations and weak equivalences are defined point wise.

There is also the one called injective model structure, where the cofibrations and weak equivalences are defined point wise.

I would like to know if for a given a morphism $\alpha \in Arr(C)$, the evaluation at $\alpha$ can be a right Quillen functor with the injective model structures on each side:

$Ev_\alpha: Hom(C,M) \to M^2$

Thanks !

Edit: Here $M^2= Hom([0 \to 1], M)= Arr(M)$, sorry for the confusion.

Answer 1

As already observed, this is not always true, but I will give a more general sufficient condition, which I believe may also be necessary. The condition is that $\alpha$ is an epimorphism in $C$.

I will denote the one arrow category by $[1]$. The left adjoint to $\mathrm{ev}_\alpha : \mathcal{M}^C \to \mathcal{M}^{[1]}$ is the left Kan extension $\mathrm{Lan}_\alpha : \mathcal{M}^{[1]} \to \mathcal{M}^C$. It can be computed explicitly. If $X \in \mathcal{M}^{[1]}$ and $c \in C$, then $(\mathrm{Lan}_\alpha X)_c$ is the pushout of $C(\alpha_1, c) \times X_0 \to C(\alpha_1, c) \times X_1$ along $C(\alpha_1, c) \times X_0 \to C(\alpha_0, c) \times X_0$. If $\alpha$ is an epimorphism, then $C(\alpha_1, c) \to C(\alpha_0, c)$ is injective for all $c$ and the pushout in question is a pushout along a cofibration. It follows from the Gluing Lemma that $\mathrm{Lan}_\alpha$ preserves levelwise (acyclic) cofibration, so it is a left Quillen functor.

Answer 2

In general, this is wrong. Consider for example the category with 3 objects $a,b,c$, a morphism $\alpha: a \rightarrow b$, a morphism $\tau: a \rightarrow c$ and two morphisms $\phi, \psi: b \rightarrow c$. Composition is defined in the only possible way. We can now actually compute the left adjoint of evaluation at $\alpha$. Given a $2$-diagram $X$, we obtain by computing the Kan extension that the left adjoint $L$ of evaluation is given as follows: It has $LX(a) = X(a)$, $LX(b) = X(b)$ and $LX(c)$ is the pushout of $X(b) \leftarrow X(a) \rightarrow X(b)$, where both maps are $X(\alpha)$. $LX(\phi)$ and $LX(\psi)$ are obtained by the two maps from $X(b)$ into this pushout. This pushout does not usually respect pointwise cofibrations: if $X(a) \rightarrow Y(a)$ and $X(b) \rightarrow Y(b)$ are cofibrations, the induced map on the pushouts need not be. For example, in topological spaces set $X(a) = *$, $X(b) = Y(a) = Y(b) = S^1$ with all involved maps either the identity of $S^1$ or the inclusion of a fixed basepoint into $S^1$. The map induced on the pushouts $S^1 \vee S^1 \rightarrow S^1$ is not even injective, so cannot be a cofibration.

However, if $C$ is such that there is a unique morphism between each two objects, I think the answer is yes. Explicitly, if I'm not mistaken, we can describe the left adjoint $L$ in this case as follows: Let $\phi_A: A \rightarrow B$ be an element of $M^2$.Then we have $L(c) = (\coprod_{End_C(b,c)} B) \coprod (\coprod_{f \in End_C(a,c), f \text{ does not factor over b via } \phi} A )$.

Given $g: c \rightarrow c'$ in $C$, the structure map $L(c) \rightarrow L(c')$ is given as follows: On $\coprod_{End_C(b,c)} B$, we send the $B$-summand f corresponding to $f: b \rightarrow c$ to the $B$-summand in $L(c')$ corresponding to $g \circ f: b \rightarrow c'$ via the identity of $B$. The $A$-summand corresponding to a map $f: a \rightarrow c$ is either send via the identity of $A$ to the $A$-summand corresponding to $g \circ f$ or, if $g \circ f$ does factor over $\phi$, via the map $\phi_A$ to the $B$-summand corresponding to the map $b \rightarrow c$ over which $g \circ f$ factors. This map is indeed unique since $C$ has only one morphism from $b$ to $c$ anyway. It is then clear that $L$ preserves pointwise cofibrations.