Why is the map $(p,t) \mapsto \gamma_p(t)$, from the flow of the gradient, an immersion

Question

In Morse theory, we have the following (a variation of this theorem): if $M$ be a smooth compact manifold with boundary and $f:M \to [a,b]$ is a smooth function with no critical points such that $f(\partial M) = \{a,b \}$, then $f^{-1}(\{a\}) \times [a,b]$ is diffeomorphic to $M$.

To prove this, let $g$ be a Riemannian metric on $M$. Then, define the smooth vector field $V = \displaystyle \frac{\text{grad}(f)}{||\text{grad}(f)||^2_g}$ and consider it's flow, $(p,t) \mapsto \gamma_p(t)$, where $\gamma_p$ is the maximal integral curve of $V$ starting at $p$, i.e. with $\gamma_p(0) = p$.

Then, the function $F:f^{-1}(\{a\}) \times [a,b] \to M, F(p,t) = \gamma_p(t-a)$ is the diffeomorphism we are looking for. However, I don't see why this is the case.

Of course $F$ is smooth, since it's definition is using the integral curves of $V$, which are smooth. It is also injective, since $f$ is increasing along the integral curves of its gradient vector field. But why is $F$ an immersion?

The reason I am given (from Hirsch's Differential Topology book, for instance), is the following: "$F$ is an immersion because the gradient vector field is orthogonal to the level sets of $f$". Of course the gradient vector field is orthogonal to the level sets of $f$, but why does this make $F$ into an immersion? What is the actual differential of $F$ (because, we should be differentiating the integral curves with respect to the points they start at, not with respect to time)?

I would appreciate a hint for the above question, instead of a complete answer.

Answer 1

An easy way out is to use so called Flowout Theorem in Lee’s Smooth manifold for example (Theorem 9.20. 2nd ed). I think your problem is a special case of this theorem.

Theorem 9.20: Suppose $M$ is a smooth manifold $S \subseteq M$ is an embedded $k$-dimensional submanifold and $V \in \mathfrak{X}(M)$ is a smooth vector field that is nowhere tangent to $S$. Let $\theta: \mathfrak{D} \rightarrow M$ be the flow of $V$, let $\mathfrak{O} = (\Bbb R \times S) \cap \mathfrak{D}. $ Let $\Phi = \theta|_{\mathfrak O}$.

1. $\Phi: \mathfrak{O} \rightarrow M$ is an immersion.
2. $\frac{\partial}{\partial t } \in \mathfrak{X}(\mathfrak{O})$ is $\Phi$-related to $V$.

You may want to look at the general idea of the proof of that theorem. The proof is similar to the Theorem and I would like to write the details of this but i don’t have that luxury right now. Since you just want a hint i think this comment appropriate as an answer.