Thom isomorphism and invertible sheaves
-------------------------------------------

Let $A$ be a commutative ring spectrum; at this point, we remain vague about exactly what we mean by the term. An $A$-module $M$ is free of rank one if there is a weak equivalence of $A$-modules $$A \to M.$$ If $X$ is a space, and $V$ is a virtual vector bundle of rank zero over $X$, then let $X^{V}$ be the associated Thom spectrum. A Thom isomorphism for $V$ is a weak equivalence of $A^{{X_{{+}}}}$-modules $$A^{{X_{{+}}}} {\simeq}A^{X^{V}}.$$ The local triviality of $V$ and the suspension isomorphism together imply that, locally on $X,$ $A^{X^{V}}$ is free of rank one, and suggest that there is a cohomological obstruction to the existence of a Thom isomorphism.

Thus let $${{\underline{A}}}: {(\mbox{spaces})}^{\operatorname{op}} \rightarrow {(\mbox{commutative ring
spectra})}$$ be the presheaf of ring spectra defined by $${{\underline{A}}} (Y) = A^{{Y_{{+}}}},$$ and let $${A_{X}} = {{\underline{A}}} {|_{\operatorname{Open}(X)^{\operatorname{op}}}}$$ be the associated presheaf of ring spectra on $X.$ Similarly, for $U$ open in in $X$, let $U^{V}$ be the Thom space of $V$ restricted to $U$, and let ${A_{V}}$ be the presheaf of ${A_{X}}$-modules whose value on the open set $U$ is given by the formula $${A_{V}} (U) = A^{U^{V}}.$$

An open cover 