**Proof.**
We first prove this for $r = 1$. Set $f = f_1$. Write $K_ t = R\langle x \rangle $, $K_ n = R\langle y \rangle $, and $K_ N = R\langle z \rangle $ with $x$, $y$, $z$ of degree $1$ and $\text{d}(x) = f^ t$, $\text{d}(y) = f^ n$, and $\text{d}(z) = f^ N$. For all $N > t$ we claim there is a quasi-isomorphism

\[ B_{N, t} = R\langle x, z, u \rangle \longrightarrow K_ t,\quad x \mapsto x,\quad z \mapsto f^{N - t}x,\quad u \mapsto 0 \]

Here the left hand side denotes the divided power polynomial algebra in variables $x$ and $z$ of degree $1$ and $u$ of degree $2$ with $\text{d}(x) = f^ t$, $\text{d}(z) = f^ N$, and $\text{d}(u) = z - f^{N - t}x$. To prove the claim, we observe that the following three submodules of $H_*(R\langle x, z\rangle )$ are the same

the kernel of $H_*(R\langle x, z\rangle ) \to H_*(K_ t)$,

the image of $z - f^{N - t}x : H_*(R\langle x, z\rangle ) \to H_*(R\langle x, z\rangle )$, and

the kernel of $z - f^{N - t}x : H_*(R\langle x, z\rangle ) \to H_*(R\langle x, z\rangle )$.

This observation is proved by a direct computation^{1} which we omit. Then we can apply Lemma 23.12.3 part (2) to see that the claim is true.

Via the homomorphism $K_ N \to B_{N, t}$ of differential graded $R$-algebras sending $z$ to $z$, we may view $B_{N, t} \to K_ t$ as a quasi-isomorphism of left differential graded $K_ N$-modules. To define the arrow in the statement of the lemma we use the homomorphism

\[ B_{N, t} = R\langle x, z, u \rangle \to K_ n \otimes _ R K_ t,\quad x \mapsto 1 \otimes x,\quad z \mapsto f^{N - n}y \otimes 1,\quad u \mapsto - f^{N - n - t}y \otimes x \]

This makes sense as long as we assume $N \geq n + t$. It is a pleasant computation to show that the (pre or post) composition with the multiplication map is the transition map.

For $r > 1$ we proceed by writing each of the Koszul algebras as a tensor product of Koszul algebras in $1$ variable and we apply the previous construction. In other words, we write

\[ K_ t = R\langle x_1, \ldots , x_ r\rangle = R\langle x_1\rangle \otimes _ R \ldots \otimes _ R R\langle x_ r\rangle \]

where $x_ i$ is in degree $1$ and $\text{d}(x_ i) = f_ i^ t$. In the case $r > 1$ we then use

\[ B_{N, t} = R\langle x_1, z_1, u_1 \rangle \otimes _ R \ldots \otimes _ R R\langle x_ r, z_ r, u_ r \rangle \]

where $x_ i, z_ i$ have degree $1$ and $u_ i$ has degree $2$ and we have $\text{d}(x_ i) = f_ i^ t$, $\text{d}(z_ i) = f_ i^ N$, and $\text{d}(u_ i) = z_ i - f_ i^{N - t}x_ i$. The tensor product map $B_{N, t} \to K_ t$ will be a quasi-isomorphism as it is a tensor product of quasi-isomorphisms between bounded above complexes of free $R$-modules. Finally, we define the map

\[ B_{N, t} \to K_ n \otimes _ R K_ t = R\langle y_1, \ldots , y_ r\rangle \otimes _ R R\langle x_1, \ldots , x_ r\rangle \]

as the tensor product of the maps constructed in the case of $r = 1$ or simply by the rules $x_ i \mapsto 1 \otimes x_ i$, $z_ i \mapsto f_ i^{N - n}y_ i \otimes 1$, and $u_ i \mapsto - f_ i^{N - n - t}y_ i \otimes x_ i$ which makes sense as long as $N \geq n + t$. We omit the details.
$\square$

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