Is this Sequences of Complexes of Sheaves Exact?

Question

So in another question of mine there is a sequence of complexes of sheaves which the author asserts is exact.

Let $K^{\bullet} = \underline{\mathbb{C}}^* \ \underrightarrow{d\ log} \ \underline{A}^1_{M, \mathbb{C}}$ and so we have an exact sequence of complexes of sheaves: $$0 \rightarrow {\mathbb{C}}^* \rightarrow K^{\bullet} \rightarrow \underline{A^2}{M, cl}[-1] \rightarrow 0 $$

Where that nastily noted $\underline{A^2}{M, cl}[-1]$ means the two term complex with 0 in the first slot and closed 2 forms on $M$ in the second slot.

The fact that this sequence is exact in itself seems to rely on the fact that the sheafification of the image of the contant $\mathbb{C}^*$ sheaf is isomorphic to the sheaf of smooth functions $\underline{\mathbb{C}^ * }$ right? Well that part is bothersome to me

Answer 1

This is not related to sheafification. The sheaf $\mathbb{C}^{*}$ of locally constant functions on $M$ is already a sheaf, so sheafification will not change it.

This sequence is not an exact sequence of complexes but it is an exact triangle of complexes. That is - it is an exact sequence of complexes, up to quasi-isomorphism. The obvious short exact sequence of complexes is $$ 0 \to \mathbb{C}^{*} \to \left[\begin{array}{c} \underline{\mathbb{C}}^{*} \\ \downarrow \\ A^{1}_{M} \end{array} \right] \to \left[ \begin{array}{c} A^{1}_{M,cl} \\ \downarrow \\ A^{1}_{M} \end{array}\right] \to 0 . $$ Now note that the last complex has an obvious surjective map $$ \left[ \begin{array}{c} A^{1}_{M,cl} \\ \downarrow \\ A^{1}_{M} \end{array}\right] \to \left[ \begin{array}{c} 0 \\ \downarrow \\ A^{2}_{M,cl} \end{array} \right] $$ and that this surjective map is a quasi-isomorphism. So up to a quasi-isomorphism, you can replace the last term in the short exact sequence with the complex you wanted.