Smooth atlas and extension of charts in a compact manifold

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Let $M$ be a compact smooth manifold of dimension $m$.

a) Prove that $M$ has a finite smooth atlas $A = \{(U_1, \varphi_1),...,(U_n, \varphi_n)\}$

b) Choose a smooth partition of 1, $\{θ_1,...,θ_n\}$ subordinate to $\{U_1,...,U_n\}$. For each $j$ consider the function $\tilde \varphi_j: M\rightarrow \Bbb R^n $ given by $\tilde \varphi(p) = 0$ for every $p \in M\setminus U_j$ and $\tilde \varphi_j(p) =\theta_j(p)\varphi_j(p) $ for all $p \in U_j$. Prove that $\tilde \varphi_j$ is smooth for every $j$

My try:

a)$A_M = \{(U_i, \varphi_i),i\in I\}$ be the maximal smooth atlas of the manifold $M$ Because M is compact there exist a finite subcover given by $A = \{(U_1, \varphi_1),...,(U_n, \varphi_n)\}$ . This is a finite smooth atlas, because the $\varphi_j, j \in I$ where initially smoothly compatible and this is just a subset of $A_M$, so also the $\varphi_j, j \in \{1,..,n\}$ are smoothly compatible.

b) Let $A$ be as in (a). By definition of smooth partition of unity, $\theta_j: M \to \Bbb R$ is smooth and Supp$(\theta_j)\subseteq U_j$ for every $j$. In particular $\theta_j$ is smooth over $ U_j$.The $\varphi_j:U_j \to \varphi_j(U_j)\subseteq\Bbb R^m $ are also smooth because they are diffeomorphisms. So the product $\tilde \varphi_j(p)=\varphi_j(p)\theta_j(p)$ is smooth over $U_j$. Over $M\setminus U_j$ , $\tilde \varphi_j(p)=0$ so it is smooth over the interior Int$(M\setminus U_j)$. On the boundary $\partial (M\setminus U_j)$ it is also smooth because Supp$(\theta_j)\subseteq U_j$, which means that for every point in that boundary there is an open neighborhood where the $\tilde \varphi_j$ is 0.

Is my solution OK? Am I missing something or is there anything to improve?

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I'd like to make one point about the gluing argument for proving that $\tilde \varphi_j $ is smooth. (Otherwise, I think your proof is fine.)

The standard technique for this kind of gluing argument is to show that there exists an open cover $\{ V_\alpha \}$ of $M$ such that $f|_{V_\alpha}$ is smooth for each $\alpha$. If you can do this, then you'll have proved that $f$ is smooth on $M$. See Proposition 2.6 and Corollary 2.8 in Lee.

For our problem, let's consider the open cover $\{ V_1, V_2 \}$, where $V_1 = U_j$ and $V_2 = M \setminus \text{supp}(\theta_j)$.

  • $\tilde \varphi_j$ is manifestly smooth on $V_1$, as you already observed.
  • $\tilde \varphi_j$ is identically zero on $V_2$, and the zero function is smooth.