2 added 2 characters in body

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.

1

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 $L(c) = (\coprod_{End_C(b,c)} B) \coprod (\coprod_{f \in End_C(a,c), f \text{ does not factor over b via } \phi} )$.

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.