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Let $A$ be a noetherian ring; let $C^{\boldsymbol\cdot}$ be a bounded above complex of flat $A$-modules in positive degrees, let $L^{\boldsymbol\cdot}$ be a bounded above complex of free $A$-modules in positive degrees, and let $g:L^{\boldsymbol\cdot}\rightarrow C^{\boldsymbol\cdot}$ be a morphism of complexes such that the induced map $h^i(L^{\boldsymbol\cdot})\rightarrow h^i(C^{\boldsymbol\cdot})$ is an isomorphism for each $i$. It is then claimed (Hartshorne III.12.3) that the map $$h^i(L^{\boldsymbol\cdot}\otimes M)\rightarrow h^i(C^{\boldsymbol\cdot}\otimes M)$$ is an isomorphism for any $A$-module $M$.

Since $\otimes$ and $h^i$ commute with direct limits we reduce to $M$ finitely generated and write $$0\rightarrow R\rightarrow E\rightarrow M\rightarrow 0$$ for $E$ free f.g., $R$ the finitely-generated kernel. Since $L^{\boldsymbol\cdot}$ is a complex of free $A$-modules and $C^{\boldsymbol\cdot}$ is a complex of flat $A$-modules we get an exact, commutative diagram of complexes $$\require{AMScd} \begin{CD} 0@>>>L^{\boldsymbol\cdot}\otimes R@>>>L^{\boldsymbol\cdot}\otimes E@>>>L^{\boldsymbol\cdot}\otimes M@>>>0\\ @.@VVV@VVV@VVV\\ 0@>>>C^{\boldsymbol\cdot}\otimes R@>>>C^{\boldsymbol\cdot}\otimes E@>>>C^{\boldsymbol\cdot}\otimes M@>>>0 \end{CD} $$ Applying $h^i$, we get a commutative diagram of long-exact sequences $$\require{AMScd} \begin{CD} \cdots@>>>h^i(L^{\boldsymbol\cdot}\otimes R)@>>>h^i(L^{\boldsymbol\cdot}\otimes E)@>>>h^i(L^{\boldsymbol\cdot}\otimes M)@>>>h^{i+1}(L^{\boldsymbol\cdot}\otimes R)@>>>h^{i+1}(L^{\boldsymbol\cdot}\otimes E)@>>>\cdots\\ @.@VVV@VVV@VVV@VVV@VVV\\ \cdots@>>>h^i(C^{\boldsymbol\cdot}\otimes R)@>>>h^i(C^{\boldsymbol\cdot}\otimes E)@>>>h^i(C^{\boldsymbol\cdot}\otimes M)@>>>h^{i+1}(C^{\boldsymbol\cdot}\otimes R)@>>>h^{i+1}(C^{\boldsymbol\cdot}\otimes E)@>>>\cdots \end{CD} $$ Since the result holds for $i+1$ by induction, and for $E$, since $E$ is free and $h^i(L^{\boldsymbol\cdot})\rightarrow h^i(C^{\boldsymbol\cdot})$ is an isomorphism, one deduces that the five vertical arrows in the diagram above are (?,iso,?,iso,iso). It follows that the middle map is an epimorphism from one of the four-lemmas but it is concluded ('from the subtle 5-lemma') that the middle map is in fact an isomorphism. To apply the 5-lemma I know, we also need that the leftmost vertical arrow is an epimorphism. I don't see why this is.

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Let me first give some names to your morphisms: $$\require{AMScd} \begin{CD} \cdots@>>>h^i(L^{\bullet}\otimes R)@>>>h^i(L^{\bullet}\otimes E)@>>>h^i(L^{\bullet}\otimes M)@>>>h^{i+1}(L^{\bullet}\otimes R)@>>>h^{i+1}(L^{\bullet}\otimes E)@>>>\cdots\\ @.@VV{\varphi^i_R}V@V\cong V{\varphi^i_E}V@VV{\varphi^i_M}V@V\cong V{\varphi^{i+1}_R}V@V\cong V{\varphi^{i+1}_E}V\\ \cdots@>>>h^i(C^{\bullet}\otimes R)@>>>h^i(C^{\bullet}\otimes E)@>>>h^i(C^{\bullet}\otimes M)@>>>h^{i+1}(C^{\bullet}\otimes R)@>>>h^{i+1}(C^{\bullet}\otimes E)@>>>\cdots \end{CD} $$ As you say, $\varphi^{i+1}_R$ is an isomorphism by induction, and $\varphi^{i}_E,\varphi^{i+1}_E$ are isomorphisms since $E$ is free.

Now we can say the following:

  1. By the 4-lemma, $\varphi^i_M$ is an epimorphism; that is, we have shown that $$\varphi^i_M\colon h^i(L^\bullet \otimes M) \to h^i(C^\bullet \otimes M)$$ is an epimorphism for any finitely-generated $M$.
  2. Now replacing $M$ with $R$ in Step 1, we also have that $\varphi^i_R$ is an epimorphism.
  3. By the 5-lemma, $\varphi^i_M$ is then an isomorphism.
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  • $\begingroup$ of course! thank you. $\endgroup$ – Tomo Jul 14 '16 at 20:11

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