Is the Picard group of a smooth manifold the same as the Picard group of the associated ringed space? $\def\sO{\mathcal{O}}
\def\Pic{\operatorname{Pic}}$Given a ringed space $(X,\sO_X)$, the Picard group of $(X,\sO_X)$ is defined to be the class of isomorphism classes of invertible sheaves of $\sO_X$-modules with the tensor product of sheaves of modules as group operation (defined in 01CX). We denote it by $\Pic(X,\sO_X)$.
On the other hand, if $M$ is a smooth manifold, there is another notion of Picard group for $M$. Namely, the class of isomorphism classes of smooth line bundles over $M$ with the tensor product of bundles as group operation (the inverse is given by the dual bundle, as it is explained here). We denote it by $\Pic(M)$.
However, $M$ is naturally a (locally) ringed space with its sheaf $\sO_M$ of real-valued smooth functions. So my question is: is $\Pic(M)\cong\Pic(M,\sO_M)$?
 A: $\def\VB{\mathsf{VB}}
\def\LFMod{\mathsf{LFMod}}
\def\bbR{\mathbb{R}}
\def\op{\oplus}
\def\sO{\mathcal{O}}
$The answer is yes. First note that over a locally ringed space (as $(M,\sO_M)$ is) invertible modules are the same as locally free of finite rank 1 modules (see 0B8M).
The isomorphism between Picard groups is induced by the functor
\begin{align*}
\Sigma:\VB(M)&\to\LFMod(\sO_M)\\
E&\mapsto\Sigma_E\\
F:E\to E'&\mapsto F_*:\Sigma_E\to\Sigma_{E'},
\end{align*}
from the category of vector bundles over $M$ to the category of locally finite free sheaves of $\sO_M$-modules. Here $\Sigma_E$ is the $\sO_M$-module of sections of the vector bundle $E$, and $F_*$ is the $\sO_M$-linear map induced by postcomposition with $F$. This functor is an equivalence of categories (see this). Moreover, this functor can be seen to preserve the monoidal structures given by the tensor product of vector bundles and of sheaves of $\sO_M$-modules,
$$
\Sigma_E\otimes_{\sO_M}\Sigma_{E'}\cong\Sigma_{E\otimes E'}.
$$
This map is constructed applying the universal property of the tensor product of $\sO_M$-modules to the $\sO_M$-bilinear map $\Sigma_E\oplus\Sigma_{E'}\to\Sigma_{E\otimes E'}$ which maps a section $(\sigma,\sigma')$ over $U$ to $\sigma\otimes\sigma':U\to E\otimes E'$. To see that it is an iso, it suffices to prove it locally, so one can assume $E=M\times \bbR^k$ and $E'=M\times \bbR^{k'}$. Then $E\otimes E' = M\times(\bbR^k\otimes_\bbR\bbR^{k'})\cong M\times\bbR^{kk'}$ and we have a commutative diagram
$$
\require{AMScd}
\begin{CD}
\Sigma_E\otimes_{\sO_M}\Sigma_{E'} @>>> \Sigma_{E\otimes E'} \\
@V{\cong}VV @VV{\cong}V \\
\sO_M^{\op k}\otimes_{\sO_M}\sO_M^{\op k'} @>>> \sO_M^{\op kk'} 
\end{CD}
$$
The bottom map is just the distributivity of the tensor product of $\sO_M$-modules with respect to direct sums, which is an iso; hence, the top map is an iso.
On the other hand, $\Sigma$ restricts to an equivalence of categories between the full subcategories of vector bundles of a fixed rank $k$ and the locally free $\sO_M$-modules of rank $k$. In our case, for $k=1$ it induces an isomorphism between Picard groups, for it also preserves the identity element, $\Sigma_{M\times\mathbb{R}}\cong\mathcal{O}_M$.
