[proofplan] We prove equality of one-forms by evaluating both sides at an arbitrary point $q\in Q$ and tangent vector $v\in T_qQ$. The pullback definition converts $(\alpha^*\lambda)_q(v)$ into the value of $\lambda$ on the tangent vector $d\alpha_q(v)$ in $T_{\alpha(q)}(T^*Q)$. The defining formula for the tautological form then projects this tangent vector back to $T_qQ$, and the section identity $\tau\circ\alpha=\operatorname{id}_Q$ makes that projection equal to $v$. [/proofplan]

[step:Evaluate the pullback at an arbitrary tangent vector] Fix $q\in Q$ and $v\in T_qQ$. Since $\alpha:Q\to T^*Q$ is smooth, its differential at $q$ is the [linear map](/page/Linear%20Map) \begin{align*} d\alpha_q:T_qQ\to T_{\alpha(q)}(T^*Q). \end{align*} By the definition of pullback of a one-form, \begin{align*} (\alpha^*\lambda)_q(v)=\lambda_{\alpha(q)}(d\alpha_q(v)). \end{align*} [/step]

[step:Use the tautological form and the section identity]The tautological one-form satisfies \begin{align*} \lambda_\xi(W)=\xi(d\tau_\xi(W)) \end{align*} for every $\xi\in T^*Q$ and $W\in T_\xi(T^*Q)$. Applying this with $\xi=\alpha(q)\in T_q^*Q$ and $W=d\alpha_q(v)\in T_{\alpha(q)}(T^*Q)$ gives \begin{align*} \lambda_{\alpha(q)}(d\alpha_q(v))=\alpha(q)(d\tau_{\alpha(q)}(d\alpha_q(v))). \end{align*} Because $\alpha$ is a section of $\tau$, we have \begin{align*} \tau\circ\alpha=\operatorname{id}_Q. \end{align*} Differentiating this identity at $q$ gives \begin{align*} d\tau_{\alpha(q)}\circ d\alpha_q=d(\operatorname{id}_Q)_q=\operatorname{id}_{T_qQ}. \end{align*} Therefore \begin{align*} d\tau_{\alpha(q)}(d\alpha_q(v))=v. \end{align*} Substituting this into the previous display yields \begin{align*} (\alpha^*\lambda)_q(v)=\alpha(q)(v). \end{align*}[/step]

[guided]We want to compare two one-forms on $Q$, namely $\alpha^*\lambda$ and $\alpha$. Equality of one-forms is pointwise equality on every tangent vector, so fix a point $q\in Q$ and a vector $v\in T_qQ$. The map $\alpha:Q\to T^*Q$ is smooth because $\alpha\in\Omega^1(Q)$ is being viewed as a smooth section of the cotangent bundle. Hence its differential at $q$ is a linear map \begin{align*} d\alpha_q:T_qQ\to T_{\alpha(q)}(T^*Q). \end{align*} The definition of pullback of a one-form says that we first push the tangent vector $v$ forward by $d\alpha_q$, then evaluate the target one-form $\lambda$ at the resulting tangent vector: \begin{align*} (\alpha^*\lambda)_q(v)=\lambda_{\alpha(q)}(d\alpha_q(v)). \end{align*} Now we use the defining property of the tautological one-form. For any covector $\xi\in T^*Q$ and any tangent vector $W\in T_\xi(T^*Q)$, the tautological form is defined by projecting $W$ down to the base manifold through $d\tau_\xi$ and then applying the covector $\xi$: \begin{align*} \lambda_\xi(W)=\xi(d\tau_\xi(W)). \end{align*} In the present situation, the covector is $\xi=\alpha(q)\in T_q^*Q$ and the tangent vector to $T^*Q$ is $W=d\alpha_q(v)\in T_{\alpha(q)}(T^*Q)$. Therefore \begin{align*} \lambda_{\alpha(q)}(d\alpha_q(v))=\alpha(q)(d\tau_{\alpha(q)}(d\alpha_q(v))). \end{align*} It remains to identify the vector inside $\alpha(q)$. Since $\alpha$ is a section of the bundle projection $\tau:T^*Q\to Q$, it satisfies \begin{align*} \tau\circ\alpha=\operatorname{id}_Q. \end{align*} Differentiating this equality at the point $q$ gives the equality of linear maps from $T_qQ$ to $T_qQ$: \begin{align*} d\tau_{\alpha(q)}\circ d\alpha_q=d(\operatorname{id}_Q)_q=\operatorname{id}_{T_qQ}. \end{align*} Applying this linear-map identity to $v$ gives \begin{align*} d\tau_{\alpha(q)}(d\alpha_q(v))=v. \end{align*} Substituting into the tautological-form computation gives \begin{align*} (\alpha^*\lambda)_q(v)=\alpha(q)(v). \end{align*}[/guided]

[step:Conclude equality of the two one-forms] For every $q\in Q$ and every $v\in T_qQ$, we have shown \begin{align*} (\alpha^*\lambda)_q(v)=\alpha(q)(v). \end{align*} Since $\alpha(q)$ is exactly the value $\alpha_q$ of the one-form $\alpha$ at $q$, this proves \begin{align*} \alpha^*\lambda=\alpha \end{align*} as one-forms on $Q$. [/step]