Comparison between pushforward-pullback and quasi-coherent pushforward-pullback

Question

In the following, an algebraic stack means a stack over the big fppf site of a scheme, admitting a smooth, representable, surjective morphism from a scheme (no separation hypotheses), and let $\mathsf{D}(X)$ (resp. $\mathsf{D}_{\textrm{qc}}(X)$) denote the unbounded derived category of the abelian category of $\mathcal{O}_X$-modules on the lisse-'etale topos of $X$ (resp. the full subcategory of $\mathsf{D}(X)$ consisting of objects whose cohomology sheaves are quasi-coherent $\mathcal{O}_X$-modules).

For a representable morphism of algebraic stacks $f : X\to Y,$ there exists a functor $$\mathbf{L}(f_{\textrm{qc}})^* : \mathsf{D}_{\textrm{qc}}(Y)\to\mathsf{D}_{\textrm{qc}}(X)$$ such that $\mathscr{H}^0(\mathbf{L}f_{\textrm{qc}}^*\mathcal{F}[0])\cong f^*\mathcal{F},$ for any $\mathcal{F}\in\mathsf{QCoh}(Y)$. Moreover, $\mathbf{L}f_{\textrm{qc}}^*$ admits a right adjoint $$\mathbf{R}(f_{\textrm{qc}})_* : \mathsf{D}_{\textrm{qc}}(X)\to\mathsf{D}_{\textrm{qc}}(Y),$$ which does not in general agree with the ordinary derived pushforward $\mathbf{R}f_*.$ However, there is a comparison morphism $$\mathbf{R}(f_{\textrm{qc}})_*\mathcal{F}\to\mathbf{R}f_*\mathcal{F}$$ for any $\mathcal{F}\in\mathsf{D}_{\textrm{qc}}(X).$ (See section 1 of "Perfect Complexes on Algebraic Stacks" by Hall and Rydh for details.)

Let $f : X\to\mathcal{Y}$ be a flat morphism from an algebraic space $X$ to an algebraic stack $\mathcal{Y}$ over some base scheme $S$ and $\Delta : X\to X\times_{\mathcal{Y}} X$ be the relative diagonal of $f.$ Suppose also that $\mathcal{F}$ is a complex of $\mathcal{O}_X$-modules with quasi-coherent cohomology, and that $\mathbf{L}\Delta^*\mathbf{R}\Delta_*\mathcal{F}\in\mathsf{D}_{\textrm{qc}}(X).$

Does this guarantee that the natural comparison map $$\mathbf{L}\Delta^*\mathbf{R}(\Delta_{\textrm{qc}})_*\mathcal{F}\to\mathbf{L}\Delta^*\mathbf{R}\Delta_*\mathcal{F}$$ is an isomorphism? If it does not guarantee this in general, are there conditions on $\mathcal{F}$ which would make this true? I would be happy to assume that $\mathcal{F}$ is a genuine quasi-coherent sheaf on $X,$ and possibly more as well.

I'm interested in the question above, and potentially generalizations of it, but I would also be interested in an answer in the case when $X$ is a scheme. I know that this will be true when $\Delta$ is qcqs, but I want to avoid restricting my attention to this situation if possible.

Edit with further thoughts: As discussed in the comments, $\Delta$ is separated, but not necessarily quasi-compact. I haven't seen much in the literature dealing with the quasi-coherent pushforward for a separated but not quasi-compact morphism, so I'm not sure to what extent this helps.

One thing which could possibly be valuable is the construction of a natural transformation $Q_{X}\mathbf{L}\Delta^*\to \mathbf{L}\Delta^* Q_{X\times_{\mathcal{Y}} X},$ where $Q_S$ denotes the derived coherator functor $\mathsf{D}(S)\to\mathsf{D}_{\textrm{qc}}(S).$ This would construct a morphism $$\mathbf{L}\Delta^*\mathbf{R}\Delta_*\mathcal{F} = Q_{X}\mathbf{L}\Delta^*\mathbf{R}\Delta_*\mathcal{F}\to \mathbf{L}\Delta^*Q_{X\times_{\mathcal{Y}} X}\mathbf{R}\Delta_*\mathcal{F}=\mathbf{L}\Delta^*\mathbf{R}(\Delta_{\textrm{qc}})_*\mathcal{F},$$ which would hopefully provide the desired inverse. Of course, it could be too much to hope for that such a transformation exists in general.

Of course, one could find an explicit $\mathcal{G}$ completing a distinguished triangle $[\mathbf{R}(\Delta_{\textrm{qc}})_*\mathcal{F}\to\mathbf{R}\Delta_*\mathcal{F}\to\mathcal{G}\to]$ and then analyze (the cohomology of) $\mathbf{L}\Delta^*\mathcal{G}.$ I was trying to avoid an argument as explicit as this, but it could be the most fruitful approach. Does this strategy become any more reasonable if $\mathcal{F} = \mathcal{O}_X$?

Answer 1

Thanks to David Benjamin Lim's comment above, I have found an answer in the main case of interest to me, detailed below. However, I will leave this question open, since I am also interested in the answer when the diagonal of the stack in question is not qcqs, and when the algebraic space $X$ is not qcqs.

Proposition: Let $(S,\mathcal{M})$ be a fine log scheme, and let $\mathcal{L}$ be Olsson's log stack classifying fine log structures over $(S,\mathcal{M}).$ Let $f : X\to\mathcal{L}$ be a morphism from a qcqs algebraic space over $S,$ and let $\Delta : X\to X\times_{\mathcal{L}} X$ be the relative diagonal of $f.$ Then $\mathbf{R}(\Delta_{\textrm{qc}})_*$ and $\mathbf{R}\Delta_*$ coincide.

Proof: By Theorem 3.2 of "Logarithmic Geometry and Algebraic Stacks," the diagonal of $\mathcal{L}\to S$ is representable, locally separated, and of finite presentation. In particular, $\Delta_{\mathcal{L}} : \mathcal{L}\to\mathcal{L}\times_S\mathcal{L}$ is qcqs. Since $X\times_{\mathcal{L}} X\cong \mathcal{L}\times_{\mathcal{L}\times_S \mathcal{L}}(X\times_S X),$ it follows that the base change $X\times_{\mathcal{L}} X\to X\times_S X$ of $\Delta_{\mathcal{L}}$ is also qcqs.

By assumption, the diagonal $X\to X\times_S X$ is quasi-compact. Since this morphism factors as $X\to X\times_{\mathcal{L}} X\to X\times_S X,$ and the morphism $X\times_{\mathcal{L}} X\to X\times_S X$ is quasi-separated as shown above, it follows from Lemma 67.8.9 of the Stacks Project that $X\to X\times_{\mathcal{L}} X$ is quasi-compact.

Because $X\to\mathcal{L}$ is representable, $\Delta$ is separated. Then $\Delta : X\to X\times_{\mathcal{L}} X$ is qcqs, and therefore $\mathbf{R}(\Delta_{\textrm{qc}})_*$ and $\mathbf{R}\Delta_*$ agree. $\square$