# Evaluation functors and injective model structure on diagram categories

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.

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Do you really want to end up in MxM? Evaluation at a morphism in C should end up in the diagrams of the form A -> B in M. You could evaluate at source and target, but then it seems to me you might as well ask the question with evaluation at a single object of C, and then the answer should be clear. –  Fabian Lenhardt Mar 28 '12 at 19:38
Yes by $M^2$ I meant the arrow category. I used the notation $M^2=Hom(2, M)$ where $2$ is the interval category (or the preorder given by the ordinal $2$). –  The mathwalker Mar 28 '12 at 20:56
By the way, you can get the projective model structure with a much less restrictive hypothesis than $M$ being combinatorial. You just need $M$ to be cofibrantly generated (and $C$ to be small, of course). For the injective model structure it seems combinatorial is needed, though I have sometimes wondered if you can get away with less. –  David White Aug 5 '12 at 16:38

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.

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Your pushout is really a big coproduct under the stated condition, as the one in my answer. Then the epimorphism condition is actually precisely the condition I tacitly use; so we actually agree. –  Fabian Lenhardt Mar 29 '12 at 9:46
I guess so. It also seems that your counterexample should essentially work whenever $\alpha$ is not an epimorphism, so this condition is apparently necessary. –  Karol Szumiło Mar 29 '12 at 11:04
Thanks to both of you! I'm not sure however if being an epimorphism is necessary. I'm thinking about all the results on direct (Reedy) categories, but I don't have a counterexample in mind right now. –  The mathwalker Mar 29 '12 at 12:45
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.
Thanks again, that helps ! I think the type of categories you're describing correspond to either a posetal category or the indiscrete category associated to a set $S$; it's a groupoid like EG for a group G. In fact it's isomorphic to some EG. –  The mathwalker Mar 29 '12 at 12:49
Fabian, when you write $End_C(b,c)$ do you mean $Hom(b,c)$ ? If so, the hypothesis that there is a unique morphism between any two objects, means that $Hom(b,c)$ is just a one-element set so the coproduct reduces to one object. If C is like EG which is a groupoid, the image of C must lie in the biggest groupoid contained in $M$ (some people call it the interior of $M$). It's like a representation of G. So I don't know if your structure map $L(c) \to L(c')$ is invertible in $M$. What do you think ? –  The mathwalker Mar 29 '12 at 16:33