[proofplan] We prove equality of cotangent vectors by evaluating both sides on an arbitrary tangent vector $v \in T_pM$. The left-hand side expands by the definition of cotangent pullback as $dg_{F(p)}(dF_p(v))$. The pointwise chain rule for smooth maps identifies this expression with $d(g \circ F)_p(v)$, and equality on every tangent vector gives equality in $T_p^*M$. [/proofplan] [step:Expand the cotangent pullback on an arbitrary tangent vector] Let $v \in T_pM$ be arbitrary. By definition of the cotangent pullback induced by \begin{align*} dF_p: T_pM \to T_{F(p)}N, \end{align*} the covector $F_p^*(dg_{F(p)}) \in T_p^*M$ is the composite [linear map](/page/Linear%20Map) $dg_{F(p)} \circ dF_p: T_pM \to \mathbb{R}$. Therefore \begin{align*} (F_p^*(dg_{F(p)}))(v) = dg_{F(p)}(dF_p(v)). \end{align*} [/step] [step:Apply the pointwise chain rule to the composite function] The composite \begin{align*} g \circ F: M \to \mathbb{R} \end{align*} is smooth because $F$ and $g$ are smooth. By the pointwise chain rule for smooth maps, applied to $F: M \to N$ and $g: N \to \mathbb{R}$ at the point $p$, \begin{align*} d(g \circ F)_p = dg_{F(p)} \circ dF_p \end{align*} as linear maps from $T_pM$ to $\mathbb{R}$. Hence, for the arbitrary vector $v \in T_pM$ fixed above, \begin{align*} d(g \circ F)_p(v) = dg_{F(p)}(dF_p(v)). \end{align*} [guided] We now identify the expression obtained from the pullback with the differential of the composite function. The relevant result is the pointwise chain rule for smooth maps: if $F: M \to N$ and $g: N \to \mathbb{R}$ are smooth, then the differential of their composite at $p \in M$ is the composite of the differentials at the corresponding points. In the present notation, this says \begin{align*} d(g \circ F)_p = dg_{F(p)} \circ dF_p \end{align*} as maps $T_pM \to \mathbb{R}$. The hypotheses of this chain rule are exactly the hypotheses in the theorem: $F$ is smooth, $g$ is smooth, and $p$ is a point of $M$. The target point for the differential of $g$ is $F(p) \in N$, so $dg_{F(p)}$ has domain $T_{F(p)}N$. The differential $dF_p$ has codomain $T_{F(p)}N$, so the composition $dg_{F(p)} \circ dF_p$ is well-defined as a linear map \begin{align*} dg_{F(p)} \circ dF_p: T_pM \to \mathbb{R}. \end{align*} Evaluating the chain-rule identity on the tangent vector $v \in T_pM$ gives \begin{align*} d(g \circ F)_p(v) = dg_{F(p)}(dF_p(v)). \end{align*} This is the same scalar obtained by expanding the cotangent pullback in the previous step. [/guided] [/step] [step:Conclude equality of the cotangent vectors] Combining the two identities, for every $v \in T_pM$ we have \begin{align*} (F_p^*(dg_{F(p)}))(v) = d(g \circ F)_p(v). \end{align*} Both $F_p^*(dg_{F(p)})$ and $d(g \circ F)_p$ are elements of the [cotangent space](/page/Cotangent%20Space) $T_p^*M$, hence linear functionals on $T_pM$. Since they agree on every vector $v \in T_pM$, they are equal: \begin{align*} F_p^*(dg_{F(p)}) = d(g \circ F)_p. \end{align*} This proves the asserted identity. [/step]