Proof Verification: A differentiable aplication $\psi : M\to N$ is differentiable if and only if: $\psi^{*}f\in C^{\infty}(M)$ Show that a differentiable aplication $\psi$ over $M$ to a differentiable variety $N$ is differentiable if and only if:
$$\psi^{*}f\in C^{\infty}(M)$$
For: $f\in C^{\infty}(N)$
Where $\psi^{*}f$ is the pull back of $f$ over $\psi$ and $C^{\infty}(N)$ is the set of all differentiable functions of $N$ over $\mathbb R$
What i have done: Knowing that:


*

*Let $(U,\phi)$ and $(V,\chi)$ be two n-charts over $M$. And
$g:M\to \mathbb R$ When the n-charts are $C^{\infty}$-related, the
following identities: $$g\circ  \phi^{-1}=(g\circ
    \chi^{-1})\circ(\chi \circ \phi^{-1})$$
$$g\circ \chi^{-1}=(g\circ \phi^{-1})\circ(\phi \circ \chi^{-1})$$ Then
$g\circ  \phi^{-1}$ is differentiable if and only if $g\circ 
    \chi^{-1}$ is.

*The pull back si defined as: $$\psi^{*}f=f\circ \psi$$

 A: If $\psi$ is smooth, the composition $f\circ \psi$ is obviously smooth for any $f\in C^\infty(N)$.
Conversely, suppose $f\circ\psi\in C^\infty(M)$ for all $f\in C^\infty(N)$.

Claim: if $p\in M$, then for any neighbourhood $V\subset N$ of $\psi(p)$ there is a neighbourhood $U\subset M$ of $p$ with $\psi(U)\subset V$.
proof: Take $\rho\in C^\infty(N)$ a bump function such that $\rho(\psi(p))=1$ and $\text{supp}(\rho)\subset V$. Take any $0<\epsilon<1$ and define $I:=(\epsilon,1+\epsilon)$, so that $p\in(\rho\circ\psi)^{-1}(I)$. Since $\rho\circ \psi$ is smooth by our hypothesis, it is continuous, so $U:=(\rho\circ\psi)^{-1}(I)$ is a neighbourhood of $p$ with $\psi(U)\subset \rho^{-1}(I)\subset V$.

Now take charts $(U,X)$ at $p$ and $(V,Y)$ at $\psi(p)$ with $\psi(U)\subset V$. If $\pi_i:\mathbb{R}^n\to\mathbb{R}$ is the projection of the $i$-th coordinate, then $\pi_i\circ Y\in C^\infty(V)$ for all $i$. Take $V'\subsetneq V$ a neighborhood of $q$ such that $\overline{V'}\subset V$ and a function $\xi_i\in C^\infty(N)$ such that $\xi_i|_{V'}=\pi_i\circ Y|_{V'}$ (this can be done with partitions of unity, see lemma $2.20$ in Lee's Introduction to Smooth Manifolds). Since $\xi_i\circ \psi$ is smooth by hypothesis, so is $\xi_i\circ \psi\circ X^{-1}$. For a neighbourhood $U'$ of $p$ with $\psi(U')\subset V'$, we get:
$$\pi_i\circ Y\circ \psi\circ X^{-1}|_{U\cap U'}=\xi_i\circ \psi\circ X^{-1}|_{U\cap U'}\text{ (smooth }\forall i)$$
$$\Rightarrow Y\circ \psi\circ X^{-1}|_{U\cap U'} \text{ smooth}$$
A: Proof
Let $\varphi : M\to N$ and $f\in C^{\infty}(N)$:
The relation:
$$(\psi^{*}f)\circ  \phi^{-1}=(f\circ
    \chi^{-1})\circ(\chi \circ \psi \circ \phi^{-1})$$
It follows:
$$\psi^{*}f\in C^{\infty}(M)$$
