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



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




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**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 $\phi:P\to\Gamma(P)$ 
$$\phi(p)(x):=\pi\_x(p)$$



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> **Question.** Why is $\phi$ surjective? Is geometricity needed for $\phi$ to be surjective?
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**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 $\mathbb{R}.$



Why is $C^\infty(M)$-module homomorphism $P\mapsto\Gamma(P)$ surjective?
The exercise is fine. I misread the definition of $\Gamma(P).$ Jet Nestruev defines $\Gamma(P)$ as an image $\phi(P).$ Hence surjectivity follows obviously from the definition of $\Gamma(P)$.


