$\DeclareMathOperator{\Id}{Id} \require{cancel}$ Jet Nestruev's "Smooth Manifolds and Observables" contains following exercise:

Exercise.Show that $P$ is geometric if and only if the two modules $P$ and $\Gamma(P)$ are isomorphic.

It is clear that there is an injective map from $P$ to $\Gamma(P)$ iff $P$ is geometric. However, I do not see why this map needs to be surjective. I doubt that it holds, cause $\Gamma(P)$ is "sort of bidual" of $P.$

**Notions and the question.**

Let $P$ be a $C^\infty(M)$-module. We define following objects. For any $x\in M$ $$\mu_x:=\lbrace f\in C^\infty(M): f(x)=0\rbrace \hspace{5pt}\text{and}\hspace{5pt}P_x:=P/\mu_xP \hspace{5pt}\text{with}\hspace{5pt}\pi_x:P\to P_x.$$ Next we define $$|P|:=\bigcup_{x\in M}P_x\hspace{5pt}\text{with}\hspace{5pt}\pi_P:|P|\to M\hspace{5pt}\text{and}\hspace{5pt}\Gamma(P):=\overbrace{\cancel{\lbrace s:M\to|P|: \pi_P\circ s=\Id_M\rbrace}}^{\color{red}{\text{MY MISREAD! IT IS DEFINED AS }\phi(P)}}.$$

We will say that $P$ is **geometric** if
$$\bigcap_{x\in M}\mu_xP=0.$$

For convenience consider natural $C^\infty(M)$-module homomorphism $\phi:P\to\Gamma(P)$ $$\phi(p)(x):=\pi_x(p)$$

Question.Why is $\phi$ surjective? Is geometricity needed for $\phi$ to be surjective?

**Some note after Wille Liou's comment.**

It is suspicious that they made no constraints on sections (elements of $\Gamma(P))$. If we consider $P=C^\infty(M)$, then $\Gamma(P)$ contains all functions from $M$ to $\mathbb{R}.$