In the proof of theorem 3.1 they put : $\langle X,\nabla f \rangle =X(f)$ after that they say that: for a curve $c$ on $M$ then $\left\langle\dfrac{\mathrm{d}c}{\mathrm{d}t},\nabla f\right\rangle=\dfrac{\mathrm{d}(f\circ c)}{\mathrm{d}t}$
I have three question please:
1) What is the purpose of seeing that $t\rightarrow f(\varphi_t(q))$ is linear with derivative +1 ?
2)How to see that $\varphi_{b−a}$ is a difeomorphisme and it's carrie $M^a$ homeomorphicly onto $M^b$
3)Why $r_1$ is a retraction from $M^a$ to $M^b$ ?
Please ,help me
Thank you
The fact that $t \mapsto f(\phi_t(q))$ is linear with derivative $1$ implies that $f(\phi_t(q)) = t + c$ for some constant $c$ (as long as $\phi_t(q) \in f^{-1}[a,b]$). Take a point $q$ such that $f(q) = a$. Since $\phi_0 = id$, we have $a = f(q) = f(\phi_0(q)) = c$; thus $f(\phi_t(q)) = t + a$. In particular $f(\phi_{b-a}(q)) = b - a + a = b$, so $\phi_{b-a}(q)$ maps $M^a$ into $M^b$. $\phi_t$ is automatically a diffeomorphism onto its image for every $t$, and repeating the argument but reversing the direction of the gradient flow for $f$ shows that $\phi_{b-a} \colon M^a \to M^b$ is onto. This should answer (1) and (2).
As for (3), there is nothing to prove: just look up the definition of "retraction" and stare at the formula for $r_1$.