> [!theorem] Composition of morphisms is a morphism > > Let $X, Y, Z$ be $C^p$ [[Manifold|manifolds]], $f: X \to Y$ and $g: Y \to Z$ be $C^p$-[[Manifold Morphism|morphisms]], then $g \circ f: X \to Z$ is a $C^p$ morphism. > > *Proof*. Let $x \in X$, $(U, \psi) \in X$ and $(V, \varphi) \in Y$ be charts such that $x \in U$ and $f(U) \subset V$. Let $(V', \varphi') \in Y$ and $(W, \pi) \in Z$ be charts such that $f(x) \in V'$ and $g(V') \subset W$. > > Since $g \circ f$ is continuous, $(g \circ f)^{-1}(W)$ is open in $X$. Restrict the domain of the first chart to $U' = (g \circ f)^{-1}(W)$, then $f(U) \subset V'$ and $(g \circ f)(U) \subset W$, then > $ > \begin{align*} > (g \circ f)_{U', W} &= \pi \circ (g \circ f) \circ \psi^{-1} \\ > &= \underbrace{\pi \circ g \circ \varphi^{-1}}_{C^p} \circ \underbrace{\varphi \circ f \circ \psi^{-1}}_{C^p} > \end{align*} > $ > is a $C^p$ morphism. Therefore $g \circ f$ is a $C^p$ morphism.