= Solution
In <cotangent bundle> coordinates $(q^1,\ldots,q^n,p_1,\ldots,p_n)$, the <Liouville one-form> and the canonical <symplectic form> are
$$
\alpha=\sum_i p_i\,dq^i,\qquad
\omega=-d\alpha=\sum_i dq^i\wedge dp_i.
$$
The <twisted cotangent symplectic form> is closed: <pullback of a differential form> commutes with the <exterior derivative>, so
$$
d\omega_\sigma=d\omega+\pi^*(d\sigma)=0.
$$
To prove the form is <nondegenerate>, write a tangent vector in these coordinates as $(u,v)$ and test its pairing with an arbitrary vertical vector $(0,w)$. The base <differential two-form> has zero pairing with vertical vectors, hence
$$
\omega_\sigma((u,v),(0,w))=u^iw_i.
$$
A vector in the <kernel of a linear map> therefore has $u=0$. Pairing it next with every $(z,0)$ gives $-z^iv_i=0$, so $v=0$. Thus the kernel is zero at every point, independently of the size or rank of $\sigma$. Together with closedness, this proves \b[$\omega_\sigma$ is a <symplectic form>]. The argument is local only for convenience; the <nondegenerate> property is intrinsic and the coordinate charts cover the whole <cotangent bundle>.
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