The two derived Hall correspondence squares are homotopy cartesian

For chain maps $x\stackrel{f}{\to }y\stackrel{g}{\to }z$, put $h=gf$. With the mapping-cone convention $\mathrm{Cone}(f{)}^n=y^n\oplus x^{n+1}$ and $d(y,x)=(d_{y}y+fx,-d_{x}x)$, the induced map is $k(y,x)=(gy,x)$. A point of the first homotopy pullback consists of arrows $f$ and $g$ with the target of $f$ identified with the source of $g$, together with an object identified with the cofiber of $g$. To compare this cofiber coordinate with the image of a composable pair, write a degree-$n$ element of $\mathrm{Cone}(k)$ as $(z,u,y^{\mathrm{\prime }},x^{\mathrm{\prime }})\in z^n\oplus x^{n+1}\oplus y^{n+1}\oplus x^{n+2}$. Its differential is $(d_{z}z+gfu+g{y}^{\mathrm{\prime }},-d_{x}u+x^{\mathrm{\prime }},-d_y{y}^{\mathrm{\prime }}-f{x}^{\mathrm{\prime }},d_x{x}^{\mathrm{\prime }})$. Under the invertible coordinate change $\tilde{y}=y^{\mathrm{\prime }}+fu$, $\tilde{x}=x^{\mathrm{\prime }}-d_{x}u$, this becomes $(d_{z}z+g\tilde{y},\tilde{x},-d_y\tilde{y},0)$, the direct sum of $\mathrm{Cone}(g)$ and a contractible identity pair. Thus $\mathrm{Cone}(k)\simeq \mathrm{Cone}(g)$. The homotopy pullback of the source and target maps on arrow cores is the core of composable pairs: its data are two arrows and an equivalence between their middle objects; transporting one arrow along that equivalence gives a composable pair, and the same transport identifies equivalences of such data with objectwise equivalences of pairs. The additional cofiber coordinate is an object equipped with an equivalence to $\mathrm{Cone}(g)$, whose space of choices is contractible over each $g$. The coordinate change supplies the equivalence from $\mathrm{Cone}(k)$, so this coordinate adds no further homotopy data. Hence the first square is homotopy cartesian.

A point of the homotopy pullback in the second square consists of an object $x$, an arrow $h:x\to z$, an arrow $k:a\to b$, and an equivalence $b\simeq \mathrm{Cone}(h)$. Form $y:=z{\times }_{\mathrm{Cone}(h)}^{h}a$, the homotopy pullback of the cofiber map $z\to \mathrm{Cone}(h)$ along $k$. Pulling the fiber sequence $x\to z\to \mathrm{Cone}(h)$ back along $k$ gives a fiber sequence $x\to y\to a$, so $\mathrm{Cone}(x\to y)\simeq a$. The pullback square also identifies $\mathrm{Cone}(y\to z)$ with $\mathrm{Cone}(k)$. The composite $x\to y\to z$ is $h$, and the induced cofiber map is $k$ by the defining pullback square. The object $y$ is perfect because the perfect modules form a stable subcategory closed under finite homotopy limits. These constructions are inverse, functorially on equivalences, to the map from composable pairs to $(h,k)$; hence the second square is homotopy cartesian.

The constructions in steps 1.1 and 1.2 are functorial in maps and equivalences of the input diagrams, and their displayed inverse maps recover every object, arrow, and comparison path. They therefore give equivalences of the full homotopy pullbacks with $X^{(2)}$, proving both squares homotopy cartesian. The argument uses the enhancement's functorial cones; an unenhanced triangulated category alone would not supply these maps.