[proofplan] We use the definition of a diffeomorphism: a bijective smooth map whose inverse is smooth. Since smooth maps compose, $G \circ F$ is smooth. We then prove directly that its inverse is $F^{-1} \circ G^{-1}$ by checking both compositions with $G \circ F$. Finally, smoothness of this inverse follows again from closure of smooth maps under composition. [/proofplan] [step:Use the diffeomorphism hypotheses to record the available maps] Since $F: M \to N$ is a diffeomorphism, $F$ is bijective and smooth, and its inverse \begin{align*}F^{-1}: N \to M\end{align*} is smooth. Since $G: N \to P$ is a diffeomorphism, $G$ is bijective and smooth, and its inverse \begin{align*}G^{-1}: P \to N\end{align*} is smooth. [/step] [step:Show that the composition $G \circ F$ is smooth] The codomain of $F$ is $N$, which is the domain of $G$, so the composition \begin{align*}G \circ F: M \to P\end{align*} is well-defined. Since $F$ and $G$ are smooth maps between smooth manifolds, and compositions of smooth maps are smooth, $G \circ F$ is smooth. [guided] The first point is a domain check. The map $F$ has type $F: M \to N$, and $G$ has type $G: N \to P$. Therefore, for every $p \in M$, the point $F(p)$ lies in $N$, so $G(F(p))$ is defined. Thus \begin{align*} G \circ F: M \to P \end{align*} is a well-defined map. We now use the standard closure property of smooth maps: if $H: X \to Y$ and $K: Y \to Z$ are smooth maps between smooth manifolds, then $K \circ H: X \to Z$ is smooth. Applying this with $H = F$, $X = M$, $Y = N$, $K = G$, and $Z = P$, we obtain that $G \circ F$ is smooth. [/guided] [/step] [step:Identify the inverse map as $F^{-1} \circ G^{-1}$] Define \begin{align*} H: P \to M \end{align*} by \begin{align*} H = F^{-1} \circ G^{-1}. \end{align*} This map is well-defined because $G^{-1}: P \to N$ and $F^{-1}: N \to M$. Let $\operatorname{id}_M: M \to M$, $\operatorname{id}_N: N \to N$, and $\operatorname{id}_P: P \to P$ denote the identity maps on $M$, $N$, and $P$, respectively. For every $p \in M$, \begin{align*} (H \circ (G \circ F))(p) = F^{-1}(G^{-1}(G(F(p)))). \end{align*} Since $G^{-1} \circ G = \operatorname{id}_N$ and $F^{-1} \circ F = \operatorname{id}_M$, this gives \begin{align*} (H \circ (G \circ F))(p) = p. \end{align*} Hence \begin{align*} H \circ (G \circ F) = \operatorname{id}_M. \end{align*} For every $q \in P$, \begin{align*} ((G \circ F) \circ H)(q) = G(F(F^{-1}(G^{-1}(q)))). \end{align*} Since $F \circ F^{-1} = \operatorname{id}_N$ and $G \circ G^{-1} = \operatorname{id}_P$, this gives \begin{align*} ((G \circ F) \circ H)(q) = q. \end{align*} Hence \begin{align*} (G \circ F) \circ H = \operatorname{id}_P. \end{align*} Therefore $H$ is the inverse of $G \circ F$, so $G \circ F$ is bijective and \begin{align*} (G \circ F)^{-1} = F^{-1} \circ G^{-1}. \end{align*} [/step] [step:Prove that the inverse is smooth and conclude] The inverse of $G \circ F$ is \begin{align*} (G \circ F)^{-1} = F^{-1} \circ G^{-1}. \end{align*} Both $G^{-1}: P \to N$ and $F^{-1}: N \to M$ are smooth, and compositions of smooth maps are smooth. Therefore \begin{align*} (G \circ F)^{-1}: P \to M \end{align*} is smooth. We have shown that $G \circ F: M \to P$ is bijective, smooth, and has smooth inverse. Hence $G \circ F$ is a diffeomorphism. [/step]