Weil Kostant Integrality Result as Stated by Brylisnki

I'm reading through Brylinski's "Loop Spaces, Characteristic Classes and Geometric Quantization" and I am stuck on a piece of Theorem 2.2.15, which asserts that If $K$ is a closed complex-valued 2-form on a paracompact manifold $M$ whose cohomology class is in the image of $H^2(M, \mathbb{Z}(1)) \rightarrow H^2(M, \mathbb{C})$ then there exists a line bundle $L$ with connection on $M$, whose curvature is equal to $K$.

The proof goes like so:

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 is beautifully explained in my other question: here.

Ok well given the above exact sequence, we would thus be able to get the exact sequence: $$0 \rightarrow H^1(M,{\mathbb{C}}^*) \rightarrow H^1(M,K^{\bullet}) \rightarrow \underline{A^2}{M, cl} \rightarrow H^2(M, \mathbb{C}^ * )$$

The above sequence makes sense to me. HOWEVER The claim is now that since our closed 2-form K is in the image of

$H^2(M, \mathbb{Z}(1)) \rightarrow H^2(M, \mathbb{C})$

Then the cohomology class $[K] \in H^2(M, \mathbb{C})$ has zero image in $H^2(M, \mathbb{C}^*)$ and so by the above sequence it must come from $H^1(M, K^{\bullet})$, i.e. it comes from being the curvature of a line bundle with connection.

I hope it is understandable why this is not obvious to me. There seems to be many different identifications going on with all of the possible places you can get a map $H^2(M, \mathbb{Z}(1)) \rightarrow H^2(M, \mathbb{C})$, *not all of which are clearly linked with a map * $H^2(M, \mathbb{C}) \rightarrow H^2(M, \mathbb{C}^*)$

I think that this mysterious map $H^2(M, \mathbb{Z}(1)) \rightarrow H^2(M, \mathbb{C})$ comes from first thinking of the $\mathbb{Z}(1)$-valued Cech cocyle living in $\mathbb{R}(2)$ Deligne Cohomology, obtaining a 2-form and then using the De-Rham Cech isomorphism! phew! That still doesn't explain the mapping properties and identifications made above. I will update this last bit and clarify as soon as possible.

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