Background
As is well known cotangent bundles are (`the') examples of symplectic manifolds. In particular they are examples of almost symplectic manifolds (there one merely requires the symplectic form to be non-degenerate, not necessarily closed).
On the other hand -if we phrase things in the language of classifying spaces- the existence of an almost symplectic structure on $T^*M$ means we can reduce the structure group from $GL_{2n}$ to $Sp_{2n}$. Here dim$(T^*M) = 2n$. Differently said, there exists a lift $\tilde{\tau}$ of the classifying map $\tau$ as follows:
The Question
Is there a simple (topological) argument, why such lifts always exist for cotangent bundles?
Further Comment
I would also be interested in other non-constructive proofs of cotangent bundles being symplectic manifolds, so feel free to interpret the question slightly differently and answer that.
I like this question. The following is a bit long for a comment, and not fully fleshed out. You probably want to use something like this: Let $\pi:T^*M\rightarrow M$ be the projection. Then $d\pi:TT^*M\rightarrow TM$ is surjective, so we have an exact sequence
$$ 0\rightarrow T_vT^*M\rightarrow TT^*M\rightarrow \pi^*TM\rightarrow 0 $$ of vector bundles over $T^*M$. Here $T_vT^*M$ is defined as the kernel of $d\pi$. Now $T_vT^*M=\pi^*(T^*M)$. Hence we get an isomorphism $TT^*M\cong \pi^*(T^*M)\oplus (\pi^*TM)\cong (\pi^*(TM))^*\oplus \pi^*TM$.
If $E$ is a vector space, then each vector space $E^*\oplus E$ can be endowed with a symplectic structure via $\omega((\eta,v),(\eta',v'))=\eta(v')-\eta'(v)$. But this reduced the structure group of $TT^*M$ if applied fiberwise.