For a chart ${\displaystyle (\{q^{i}\}_{i=1}^{n},U)}$ on ${\displaystyle Q}$ (where ${\displaystyle U\subseteq \mathbb {R} ^{n}),}$ let ${\displaystyle \{p_{i},q^{i}\}_{i=1}^{n}}$ be the coordinates on ${\displaystyle T^{*}Q,}$ where the fiber coordinates ${\displaystyle \{p_{i}\}_{i=1}^{n}}$ are associated with the linear basis ${\displaystyle \{dq^{i}\}_{i=1}^{n}.}$ By assumption, for every ${\displaystyle {\mathbf {q} }=(q^{1},\ldots ,q^{n})\in U,}$

$${\displaystyle \beta ({\mathbf {q} })=\sum _{i=1}^{n}\beta _{i}(\mathbf {q} )\,dq^{i},}$$

or

$${\displaystyle \mathbf {q} =(q^{1},\ldots ,q^{n})\ {\stackrel {\beta }{\to }}\ (\underbrace {q^{1},\ldots ,q^{n}} _{\mathbf {q} },\underbrace {\beta _{1}(\mathbf {q} ),\ldots ,\beta _{n}(\mathbf {q} } _{\mathbf {p} })).}$$

It follows that

$${\displaystyle \beta _{*}\left({\frac {\partial }{\partial q^{i}}}{\Biggl |}_{\mathbf {q} }\right)={\frac {\partial }{\partial q^{i}}}{\Biggl |}_{\beta (\mathbf {q} )}+\sum _{j=1}^{n}{\frac {\partial \beta _{j}}{\partial q^{i}}}{\Biggl |}_{\mathbf {q} }\cdot {\frac {\partial }{\partial p_{j}}}{\Biggl |}_{\beta (\mathbf {q} )}}$$

which implies that

$${\displaystyle (\beta ^{*}\,dq^{i})\left({\partial /\partial q^{j}}\right)_{\mathbf {q} }=dq^{i}\left[\beta _{*}\left({\partial /\partial q^{j}}\right)_{\mathbf {q} }\right]=\delta _{ij}.}$$

Step 1. We have

$${\displaystyle {\begin{aligned}(\beta ^{*}\theta )\left(\partial /\partial q^{i}\right)_{\mathbf {q} }&=\theta \left(\beta _{*}\left(\partial /\partial q^{i}\right)_{\mathbf {q} }\right)=\left(\sum _{j=1}^{n}p_{j}dq^{j}\right)\left(\beta _{*}\left(\partial /\partial q^{i}\right)_{\mathbf {q} }\right)\\&=\beta _{i}(\mathbf {q} )=\beta \left(\partial /\partial q^{i}\right)_{\mathbf {q} }.\end{aligned}}}$$

Step 1'. For completeness, we now give a coordinate-free proof that ${\displaystyle \beta ^{*}\theta =\beta ,}$ for any 1-form ${\displaystyle \beta .}$

Observe that, intuitively speaking, for every ${\displaystyle q\in Q}$ and ${\displaystyle p\in T_{q}^{*}Q,}$ the linear map ${\displaystyle d\pi _{(q,p)}}$ in the definition of ${\displaystyle \theta }$ projects the tangent space ${\displaystyle T_{(q,p)}T^{*}Q}$ onto its subspace ${\displaystyle T_{q}Q.}$ As a consequence, for every ${\displaystyle q\in Q}$ and ${\displaystyle v\in T_{q}Q,}$

$${\displaystyle d\pi _{\beta (q)}(\beta _{*q}v)=v,}$$

where ${\displaystyle \beta _{*q}}$ is the instance of ${\displaystyle \beta _{*}}$ at the point ${\displaystyle q\in Q,}$ that is,

$${\displaystyle \beta _{*q}:T_{q}Q\to T_{\beta (q)}T^{*}Q.}$$

Applying the coordinate-free definition of ${\displaystyle \theta }$ to ${\displaystyle \theta _{\beta (q)},}$ obtain

$${\displaystyle (\beta ^{*}\theta )_{q}v=\theta _{\beta (q)}(\beta _{*q}v)=\beta (q)(d\pi _{\beta (q)}(\beta _{*q}v))=\beta (q)v.}$$

Step 2. It is enough to show that ${\displaystyle \alpha =0}$ if ${\displaystyle \beta ^{*}\alpha =0,}$ for every one-form ${\displaystyle \beta .}$ Let

$${\displaystyle \alpha =\sum _{i=1}^{n}\alpha _{q^{i}}(\mathbf {p} ,\mathbf {q} )\,dq^{i}+\sum _{i=1}^{n}\alpha _{p_{i}}(\mathbf {p} ,\mathbf {q} )\,dp_{i},}$$

where ${\displaystyle \alpha _{p^{i}},\alpha _{q^{i}}\in C^{\infty }(\mathbb {R} ^{n}\times U,\mathbb {R} ).}$

Substituting ${\displaystyle v=\left(\partial /\partial q_{i}\right)_{\mathbf {q} }}$ into the identity ${\displaystyle \alpha (\beta _{*}v)=0}$ obtain

$${\displaystyle \alpha (\partial /\partial q^{i})_{\beta (\mathbf {q} )}+\sum _{j=1}^{n}(\partial \beta _{j}/\partial q^{i})_{\mathbf {q} }\cdot \alpha (\partial /\partial p_{j})_{\beta (\mathbf {q} )}=0,}$$

or equivalently, for any choice of ${\displaystyle n}$ functions ${\displaystyle p_{i}=\beta _{i}(\mathbf {q} ),}$

$${\displaystyle \alpha _{q^{i}}(\mathbf {p} ,\mathbf {q} )+\sum _{j=1}^{n}\partial p_{j}/\partial q^{i}\cdot \alpha _{p_{j}}(\mathbf {p} ,\mathbf {q} )=0.}$$

Let ${\displaystyle \beta =\sum _{j=1}^{n}c_{j}dq^{j},}$ where ${\displaystyle c_{j}={\text{const}}.}$ In this case, ${\displaystyle \beta _{j}=c_{j}.}$ For every ${\displaystyle \mathbf {q} \in U}$ and ${\displaystyle c_{j}\in \mathbb {R} ,}$

$${\displaystyle \alpha _{q^{i}}(\mathbf {p} ,\mathbf {q} ){\bigl |}_{j=1\ldots n}^{p_{j}=c_{j}}=0.}$$

This shows that ${\displaystyle \alpha _{q^{i}}(\mathbf {p} ,\mathbf {q} )=0}$ on ${\displaystyle \mathbb {R} ^{n}\times U,}$ and the identity

$${\displaystyle \sum _{j=1}^{n}\partial p_{j}/\partial q^{i}\cdot \alpha _{p_{j}}(\mathbf {p} ,\mathbf {q} )=0}$$

must hold for an arbitrary choice of functions ${\displaystyle p_{i}=\beta _{i}(\mathbf {q} ).}$ If ${\displaystyle \beta =\sum _{j=1}^{n}c_{j}q^{j}dq^{j}}$ (with ${\displaystyle {}^{j}}$ indicating superscript) then ${\displaystyle \beta _{j}=c_{j}q^{j},}$ and the identity becomes

$${\displaystyle \alpha _{p_{i}}(\mathbf {p} ,\mathbf {q} ){\bigl |}_{j=1\ldots n}^{p_{j}=c_{j}q^{j}}=0,}$$

for every ${\displaystyle \mathbf {q} \in U}$ and ${\displaystyle c_{j}\in \mathbb {R} .}$ Since ${\displaystyle c_{j}=p^{j}/q^{j},}$ we see that ${\displaystyle \alpha _{p_{i}}(\mathbf {p} ,\mathbf {q} )=0,}$ as long as ${\displaystyle q^{j}\neq 0}$ for all ${\displaystyle j.}$ On the other hand, the function ${\displaystyle \alpha _{p_{i}}}$ is continuous, and hence ${\displaystyle \alpha _{p_{i}}(\mathbf {p} ,\mathbf {q} )=0}$ on ${\displaystyle \mathbb {R} ^{n}\times U.}$