Derived tensor product , independence of resolution

Question

Right after Lemma 20.26.13

We have the following paragraph of how to derive the tensor product.

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It claims that the end result is independent of choice of K-flat resolution Suppose we take $L^\bullet \rightarrow F^\bullet$, then we have to show a quasi iso. $$Tot(G^* \otimes K^*) \simeq Tot(G^* \otimes L^*)$$

This would follow if we can exhibit a map $K^* \rightarrow L^*$ (or vice versa) and apply 20.26.13, but this is unclear to me.

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What am I missing?

Some explaination would be appreciated.

Answer 1

Suppose $\alpha_K:K^*\to F^*$ and $\alpha_L:L^*\to F^*$ are two $K$-flat resolutions of $F^*$.

As one typically proves when constructing the derived category (see Prop. 10.4.1 of Weibel's Introduction to Homological Algebra, for example) there is a complex $G^*$ with quasi-isomorphisms $\beta_K:G^*\to K^*$ and $\beta_L:G^*\to L^*$ such that $\alpha_K\beta_K$ and $\alpha_L\beta_L$ are chain homotopic.

Let $\gamma:M^*\to G^*$ be a $K$-flat resolution of $G^*$. Then $\beta_K\gamma:M^*\to K^*$ and $\beta_L\gamma:M^*\to L^*$ are quasi-isomorphisms of $K$-flat complexes, so you can now apply Lemma 20.26.13 of the Stacks Project.