[proofplan] We verify the category axioms using the local coordinate definition of smoothness. The identity map is smooth because, in overlapping charts, its coordinate representation is exactly a transition map of the smooth atlas. The composition of two smooth maps is smooth because, after choosing charts around $p$, $F(p)$, and $G(F(p))$, the coordinate representation of $G \circ F$ is the ordinary Euclidean composition of the coordinate representations of $F$ and $G$. The identity and associativity laws then follow from the corresponding laws for ordinary functions. [/proofplan] [step:Show that identity maps are smooth by identifying their coordinate representations with transition maps] Let $M$ be a smooth manifold with smooth atlas $\mathcal{A}_M$, and let \begin{align*} \operatorname{id}_M: M &\to M \\ p &\mapsto p \end{align*} be the identity map. To prove that $\operatorname{id}_M$ is smooth, let $p \in M$. Choose two charts $(U,\varphi)$ and $(V,\psi)$ in $\mathcal{A}_M$ with $p \in U \cap V$. The coordinate representation of $\operatorname{id}_M$ on the overlap $U \cap V$ is the map \begin{align*} \psi \circ \operatorname{id}_M \circ \varphi^{-1}: \varphi(U \cap V) &\to \psi(U \cap V) \\ x &\mapsto \psi(\varphi^{-1}(x)). \end{align*} Since $\operatorname{id}_M(\varphi^{-1}(x))=\varphi^{-1}(x)$ for every $x \in \varphi(U \cap V)$, this coordinate representation is precisely the transition map \begin{align*} \psi \circ \varphi^{-1}: \varphi(U \cap V) &\to \psi(U \cap V). \end{align*} The charts $(U,\varphi)$ and $(V,\psi)$ belong to the same smooth atlas, so their transition map $\psi \circ \varphi^{-1}$ is smooth as a map between open subsets of Euclidean spaces. Therefore $\operatorname{id}_M$ is smooth at $p$. Since $p \in M$ was arbitrary, $\operatorname{id}_M$ is smooth on all of $M$. [guided] We need to check the smoothness of the identity map using the definition of smoothness between manifolds. That definition says: around each point, and after passing to coordinates in the domain and target, the resulting map between open subsets of Euclidean space must be smooth. Let $p \in M$. Because $M$ is a smooth manifold, there are charts $(U,\varphi)$ and $(V,\psi)$ in its smooth atlas with $p \in U \cap V$. We use $(U,\varphi)$ as the domain chart and $(V,\psi)$ as the target chart. On the common region $U \cap V$, the coordinate representation of the identity map is \begin{align*} \psi \circ \operatorname{id}_M \circ \varphi^{-1}: \varphi(U \cap V) &\to \psi(U \cap V) \\ x &\mapsto \psi(\operatorname{id}_M(\varphi^{-1}(x))). \end{align*} Since $\operatorname{id}_M(q)=q$ for every $q \in M$, this becomes \begin{align*} \psi \circ \operatorname{id}_M \circ \varphi^{-1}(x) = \psi(\varphi^{-1}(x)). \end{align*} Thus the coordinate representative is exactly \begin{align*} \psi \circ \varphi^{-1}: \varphi(U \cap V) &\to \psi(U \cap V), \end{align*} which is a transition map between two charts in the same smooth atlas. The defining compatibility condition for a smooth atlas is precisely that every such transition map is smooth as a map between open subsets of Euclidean spaces. Therefore the coordinate representation of $\operatorname{id}_M$ is smooth near $p$. Since $p$ was arbitrary, $\operatorname{id}_M: M \to M$ is smooth. [/guided] [/step] [step:Express the coordinate representative of a composite as a Euclidean composite] Let $M,N,P$ be smooth manifolds, and let \begin{align*} F: M &\to N,\\ G: N &\to P \end{align*} be smooth maps. Define their composite \begin{align*} G \circ F: M &\to P \\ p &\mapsto G(F(p)). \end{align*} Fix $p \in M$. Choose a chart $(U,\varphi)$ on $M$ with $p \in U$, a chart $(V,\psi)$ on $N$ with $F(p) \in V$, and a chart $(W,\chi)$ on $P$ with $G(F(p)) \in W$. Since $G(F(p)) \in W$, the set $G^{-1}(W)$ is a neighbourhood of $F(p)$ in the domain on which the coordinate expression of $G$ into the chart $(W,\chi)$ is considered. Replace $V$ by the smaller chart domain \begin{align*} V_0 := V \cap G^{-1}(W), \end{align*} and restrict $\psi$ to $V_0$. Then $F(p) \in V_0$ and $G(V_0) \subset W$. Next define \begin{align*} U_0 := U \cap F^{-1}(V_0). \end{align*} Then $p \in U_0$ and $F(U_0) \subset V_0$. On the coordinate domain $\varphi(U_0)$, define \begin{align*} f: \varphi(U_0) &\to \psi(V_0) \\ x &\mapsto \psi(F(\varphi^{-1}(x))), \end{align*} and on the coordinate domain $\psi(V_0)$ define \begin{align*} g: \psi(V_0) &\to \chi(W) \\ y &\mapsto \chi(G(\psi^{-1}(y))). \end{align*} These are the coordinate representatives of $F$ and $G$ on the chosen chart domains. Because $F$ and $G$ are smooth maps of manifolds, the maps $f$ and $g$ are smooth maps between open subsets of Euclidean spaces. For every $x \in \varphi(U_0)$, we have $f(x) \in \psi(V_0)$, and hence \begin{align*} (g \circ f)(x) &= g(\psi(F(\varphi^{-1}(x))))\\ &= \chi(G(\psi^{-1}(\psi(F(\varphi^{-1}(x))))))\\ &= \chi(G(F(\varphi^{-1}(x))))\\ &= \bigl(\chi \circ (G \circ F) \circ \varphi^{-1}\bigr)(x). \end{align*} Thus the coordinate representative of $G \circ F$ near $p$ is the Euclidean composite $g \circ f$. [guided] The goal is to prove that $G \circ F$ is smooth at an arbitrary point $p \in M$. The definition of smoothness between manifolds asks us to choose charts near $p$ in $M$ and near $(G \circ F)(p)$ in $P$, and then prove that the resulting coordinate map is smooth. Choose a chart $(U,\varphi)$ on $M$ with $p \in U$, a chart $(V,\psi)$ on $N$ with $F(p) \in V$, and a chart $(W,\chi)$ on $P$ with $G(F(p)) \in W$. We need the image of the middle chart under $G$ to land inside the target chart domain $W$, so we shrink the middle chart domain to \begin{align*} V_0 := V \cap G^{-1}(W). \end{align*} This is an open neighbourhood of $F(p)$ inside $N$, and the restricted chart $\psi|_{V_0}: V_0 \to \psi(V_0)$ is still a chart. By construction, $G(V_0) \subset W$. We also need $F$ to land inside this refined middle chart domain. Therefore we shrink the domain chart around $p$ to \begin{align*} U_0 := U \cap F^{-1}(V_0). \end{align*} Then $p \in U_0$ and $F(U_0) \subset V_0$. The restricted map $\varphi|_{U_0}: U_0 \to \varphi(U_0)$ is again a chart. Now define the coordinate representative of $F$ by \begin{align*} f: \varphi(U_0) &\to \psi(V_0) \\ x &\mapsto \psi(F(\varphi^{-1}(x))). \end{align*} Because $F: M \to N$ is smooth, this map $f$ is smooth as a map between open subsets of Euclidean spaces. Similarly, define the coordinate representative of $G$ by \begin{align*} g: \psi(V_0) &\to \chi(W) \\ y &\mapsto \chi(G(\psi^{-1}(y))). \end{align*} Because $G: N \to P$ is smooth, this map $g$ is also smooth as a map between open subsets of Euclidean spaces. The reason for this careful shrinking is that it makes the composition $g \circ f$ well-defined: for every $x \in \varphi(U_0)$, the point $f(x)$ lies in $\psi(V_0)$, which is exactly the domain of $g$. We compute the composite: \begin{align*} (g \circ f)(x) &= g(\psi(F(\varphi^{-1}(x))))\\ &= \chi(G(\psi^{-1}(\psi(F(\varphi^{-1}(x))))))\\ &= \chi(G(F(\varphi^{-1}(x))))\\ &= \bigl(\chi \circ (G \circ F) \circ \varphi^{-1}\bigr)(x). \end{align*} Therefore the coordinate representative of $G \circ F$ in the charts $(U_0,\varphi|_{U_0})$ and $(W,\chi)$ is exactly $g \circ f$. [/guided] [/step] [step:Apply the Euclidean chain rule to prove smoothness of the composite] The maps \begin{align*} f: \varphi(U_0) &\to \psi(V_0),\\ g: \psi(V_0) &\to \chi(W) \end{align*} are smooth maps between open subsets of Euclidean spaces. By the Euclidean chain rule for smooth maps between open subsets of Euclidean spaces (citing a result not yet in the wiki: Euclidean chain rule), the composite \begin{align*} g \circ f: \varphi(U_0) &\to \chi(W) \end{align*} is smooth. Since \begin{align*} g \circ f = \chi \circ (G \circ F) \circ \varphi^{-1} \end{align*} on $\varphi(U_0)$, the coordinate representative of $G \circ F$ is smooth near $p$. Since $p \in M$ was arbitrary, $G \circ F: M \to P$ is smooth. [guided] We have reduced the manifold statement to an ordinary statement about maps between open subsets of Euclidean spaces. The two maps in coordinates are \begin{align*} f: \varphi(U_0) &\to \psi(V_0),\\ g: \psi(V_0) &\to \chi(W). \end{align*} The smoothness of $F$ gives the smoothness of $f$, and the smoothness of $G$ gives the smoothness of $g$. We now use the Euclidean chain rule for smooth maps between open subsets of Euclidean spaces (citing a result not yet in the wiki: Euclidean chain rule). Its hypotheses are satisfied because $\varphi(U_0)$, $\psi(V_0)$, and $\chi(W)$ are open subsets of Euclidean spaces, the map $f$ has codomain $\psi(V_0)$, the map $g$ has domain $\psi(V_0)$, and both $f$ and $g$ are smooth. Therefore \begin{align*} g \circ f: \varphi(U_0) &\to \chi(W) \end{align*} is smooth. From the previous computation, \begin{align*} g \circ f = \chi \circ (G \circ F) \circ \varphi^{-1} \end{align*} on $\varphi(U_0)$. Thus the coordinate representative of $G \circ F$ is smooth near $p$. Since the point $p \in M$ was arbitrary, the composite $G \circ F: M \to P$ is smooth on all of $M$. [/guided] [/step] [step:Verify the category laws from the corresponding laws for functions] For each smooth manifold $M$, the identity morphism is the smooth map $\operatorname{id}_M: M \to M$ proved smooth above. For smooth maps \begin{align*} F: M &\to N,\\ G: N &\to P,\\ H: P &\to Q, \end{align*} the composites $G \circ F$, $H \circ G$, and $H \circ (G \circ F)=(H \circ G)\circ F$ are smooth by the composition result. For every $p \in M$, \begin{align*} (H \circ (G \circ F))(p) &= H(G(F(p)))\\ &= ((H \circ G)\circ F)(p), \end{align*} so composition is associative. Also, for every smooth map $F: M \to N$ and every $p \in M$, \begin{align*} (F \circ \operatorname{id}_M)(p) &= F(p), \end{align*} and for every $q \in N$, \begin{align*} (\operatorname{id}_N \circ F)(q) \end{align*} is interpreted after writing $q=F(p)$ when evaluating on the image; equivalently, for every $p \in M$, \begin{align*} (\operatorname{id}_N \circ F)(p) &= F(p). \end{align*} Thus $F \circ \operatorname{id}_M=F$ and $\operatorname{id}_N \circ F=F$. Therefore smooth manifolds with smooth maps, ordinary composition, and identity maps form a category. [/step]