[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*}
[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]
[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]
