Does short exact sequence of vector bundles induce short exact sequence of vector space of sections?

Question

Let $M$ be a smooth manifold.

Let $P\rightarrow M, Q\rightarrow M$ and $R\rightarrow M$ be vector bundles over $M$ which combine to form a short exact sequence of vector bundles

$$0\rightarrow P\rightarrow Q\rightarrow R\rightarrow 0.$$

Question : Does this short exact sequence of vector bundles induce short exact sequence of vector space of sections? Is $$0\rightarrow \Gamma(M,P)\rightarrow \Gamma(M,Q)\rightarrow \Gamma(M,R)\rightarrow 0$$ an exact sequence?

Given a smooth map $s:M\rightarrow R$, we can always define a map $M\rightarrow Q$, but there is no good reason for this to be a smooth map. I am looking for a simple example where this is not true.

Answer 1

Yes. Every short exact sequence of vector bundles splits; for example, if we pick a Riemannian metric on the bundles we can split them by taking an orthogonal complement. So WLOG every short exact sequence has the form

$$0 \to P \to P \oplus R \to R \to 0$$

and then we get a split exact sequence of spaces of sections

$$0 \to \Gamma(M, P) \to \Gamma(M, P) \oplus \Gamma(M, R) \to \Gamma(M, R) \to 0.$$