This is a statement in Huybrechts, Fourier Mukai Transform, pg78, where he defines the derived tensor.
If $E^*$ is an acyclic complex bounded with all $E^i$ locally free. $F$ a coherent sheaf , then $F \otimes E^*$ is still acyclic. We are considering sheaves on $X$, a projective scheme over $k$.
Why is this true?
Best Answer
Exactness of $F\otimes_{\mathcal{O}_X}E^*$ is stalk local on $X$. Pick $x$ in $X$. Then $(F\otimes_{\mathcal{O}_X}E^*)_x = F_x\otimes_{\mathcal{O}_{X,x}}E^*_x$. Thus it suffices to show that if $$A^*:0\rightarrow A^n\rightarrow A^{n-1}\rightarrow ...\rightarrow A^1\rightarrow A_0\rightarrow 0$$ is an acyclic complex of free $R$-modules, where $R =\mathcal{O}_{X,x}$, and $M$ is an $R$-module then $M\otimes_RA^*$ is acyclic. Note that homologies of complex $M\otimes_RA^*$ are by definition $$\mathrm{Tor}^R_i(M,A^0)$$ for $1\leq i\leq n$. Indeed, we can view $A^*$ as a free resolution of $R$-module $A^0$. Since $A^0$ is free, we derive $\mathrm{Tor}^R_i(M,A^0) = 0$ for each $i$ and hence $M\otimes_RA^*$ is acyclic.