Derivator of localizations of spectra

Question

Since my question is very specific let me introduce the context. I am trying to apply derivator theory to stable homotopy theory. Theorem 3 of the following preprint by Franke https://pdfs.semanticscholar.org/0178/462ac5c2e39ef23d3ed80d927133322248c4.pdf
states that given a strong, stable derivator $\mathscr{D}$ defined on finite posets if we take a triangulated subcategory $E$ of the underlying category $\mathscr{D}(1)$ (where $1$ is the terminal category with just one object) there exists a subderivator $\mathscr{E}$ of $\mathscr{D}$ such that $\mathscr{E}(1)=E$ and for any P finite poset $\mathscr{E}(P)$ is the full subcategory of $\mathscr{D}(P)$ generated by the objects which are pointwise in $E$, i.e. $X \in \mathscr{E}(P)$ iff $\forall p \in P$ we have $p^*X \in E$.

I set $\mathcal{S}p$ to be the derivator of spectra, then I take $\mathscr{D}=\mathcal{S}p^{2}$ and consider as subcategory $E$ the full subcategory generated by the morphisms $f \colon X \rightarrow Y$ which are $F$-equivalences and $Y$ is $F$-local for $F$ a fixed spectrum. I call the associated derivator $\mathscr{E}$ the derivator of $F$-localizations.

Also I can define on the category $\mathcal{S}p(P)$ the notions of $F$-equivalence, $F$-acyclic and $F$-local diagrams of spectra by asking that they are pointwise such in the stable homotopy category. It is relatively easy to see that they have the same properties as the objects with the same name in the theory of Bousfield localization of spectra, for example $Z \in \mathcal{S}p(P)$ is $F$-local iff for every $F$-acyclic $U$ we have $\text{Hom}_{\mathcal{S}p(P)}(U,Z)=0$.

I define a cocontinuous morphism of derivators as the composition \begin{equation}s \colon \mathscr{E}\subset \mathcal{S}p^{2} \xrightarrow{0^*} \mathcal{S}p \end{equation} which is sends a $F$-localization $f\colon X \rightarrow Y$ to the source $X$.

My aim is to prove that this is an equivalence.

I tried to show directly that for a general finite poset $P$ the functor $s_P$ is an equivalence. Using that $\mathcal{S}p$ is a strong derivator and the universal properties I mentioned it is easy to show essential surjectivity and fullness of the functor. I am stuck on faithfulness: since $s_P$ is additive wlog we can show $s(\phi)=0$ implies $\phi=0$ where $\phi \colon f \rightarrow g$ is a morphism between the objects $f\colon X \rightarrow Y$ and $g \colon A \rightarrow B$. By applying the partial diagram functor \begin{equation} \text{dia}_2 \colon \mathcal{S}p(2\times P)\rightarrow \mathcal{S}p(P)^2 \end{equation} we can consider $f$ and $g$ as morphisms of the category $\mathcal{S}p(P)$. Using that $\text{dia}_{2}f$ is an $F$-equivalence and $B$ a $F$-local diagram we get \begin{equation} (\text{dia}_2 f)^*\colon \text{Hom}_{\mathcal{S}p(P)}(Y, B)\rightarrow \text{Hom}_{\mathcal{S}p(P)}(X, B) \end{equation} is a bijection, thus the composition $\text{dia}_2 g\circ 0^*\phi$ admits a unique preimage. Since $0^*\phi=0$ by assumption this preimage must be $0$ and we deduce $1^*\phi=0$. From this I would like to conclude that $\phi=0$ but what I am basically asking is the map \begin{equation} \text{dia}_2 \colon \text{Hom}_{\mathcal{S}p(2\times P)}(f,g) \rightarrow \text{Hom}_{\mathcal{S}p(P)^2}(\text{dia}_2 f, \text{dia}_2 g) \end{equation} to be injective. As it is known this is generally not the case: strength of the derivator guarantees that this is a surjection but injectivity is what usually fails even for nice derivators. My hope is to use the fact that $f$ and $g$ are not any coherent morphisms but $F$-localizations to show that the map I wrote is actually a bijection. Unfortunately I could not prove this and I have no idea how to produce an argument.

The alternatives I see are either to present an adjoint to $s$ and show that this is an equivalence or to consider another class of maps $X \rightarrow Y$ for which the above morphism is a bijection and complete the first argument I gave.

Answer 1

I think I found a proof which uses only derivator theory. For sake of simplicity we deal the case $P=1$, but the proof clearly will be the same for any $P$.

Associated to the coherent morphism $f \colon X \rightarrow Y$ there is a distinguished triangle \begin{equation} 0_!0^*f \xrightarrow{\epsilon_f} f \rightarrow K \end{equation} Where $\epsilon_f$ is the counit of the adjucntion $0_! \dashv 0^*$ evaluated at $f$.It is easy to see that $0_!0^*f$ is a constant diagram in the form $X \xrightarrow{\cong} X$ and the inchoerent diagram associated to the above triangle is $\require{AMScd}$ \begin{CD} X @>{=}>> X @>>>0 \\ @V {\cong}VV @V{f}VV @VVV\\ X @>{f}>> Y @>>>C(f) \end{CD} so we can identify $K$ with $1_!C(f)$.

Now apply $[-,g]$ (for $g \colon A \rightarrow B$ as in my first post) to this triangle to get a long exact sequence \begin{equation} \dots \leftarrow [0_!0^*f,g] \leftarrow [f,g] \leftarrow [1_!C(f),g] \leftarrow \dots \end{equation} now $[1_!C(f),g]\cong [C(f),1^*g]$ and this is zero since $C(f)$ is $F$-acyclic while $B$ is $F$-local. So we get that $(\epsilon_f)^*\colon [f,g] \xrightarrow{\cong}[0_!0^*f,g]$ and this last group is $[0^*f, 0^*g]$. So we have that $0^*\colon [f,g] \rightarrow [0^*f, 0^*g]$ is an isomorphism.

My mistake is that I was seeing the question as a problem of lifting incoherent diagrams to coherent ones which is not an easy task. This new approach is more elegant and simple.