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Corollary 2.3 of these notes on Morse theory seem to suggest the "easy" part of the Morse lemma is a corollary of Ehresmann's fibration theorem.

That is, if $f:X\to \mathbb R$ is a proper smooth map without critical values in $[a,a+\varepsilon]$ then there's a diffeomorphism $f^{-1}(-\infty,a]\cong f^{-1}(-\infty,a+\varepsilon]$.

I don't understand how to deduce this from Ehresmann's fibration theorem.

The proof I know constructs a normalized gradient field on $X$ which is supported on a compact neighborhood of $f^{-1}[a,a+\varepsilon]$ and flows along it.

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As you say I am not sure the other answer covers the detail you want. If $f$ is proper, then the set of critical values is closed (exercise) so if there are no critical values in $[a, a+\epsilon]$, then nor are there in $[a-\delta, a+\epsilon]$ for $\delta$ small. Now you know from Ehressman that there is a diffeomorphism

$$g: X_{[a-\delta, a+\epsilon]} \cong [a-\delta, a+\epsilon] \times X_a,$$ so that under this diffeomorphism $f(g^{-1}(t,x)) = t$; that is, $f$ is taken to projection onto the first factor. Here I write $X_S = f^{-1}(S)$ as notation I prefer.

We will use this to construct the desired diffeomorphism. To do so, pick a diffeomorphism $$h: [a-\delta, a] \to [a-\delta, a+\epsilon]$$ which is the identity near $a-\delta$. Of course, this induces a diffeomorphism $$H: X_{[a-\delta, a] \to [a-\delta, a+\epsilon]}$$ which is the identity near $X_{a-\delta}$; this induced diffeo comes from the diffeo $g$ above (and its restriction to $X_{[a-\delta, a]}$ as well).

Then the desired diffeomorphism $$F: X_{(-\infty, a]} \to X_{(-\infty, a+\epsilon]}$$ by saying that $F$ is the identity on $X_{(-\infty, a-\delta]}$, and on $X_{[a-\delta, a]}$, $F = H$, the function defined above.

The idea is that we are flowing backwards from $a+\epsilon$ to $a$, but we use $h$ to "slow down the flow" past $a$ so that by $a-\delta$ we have stopped flowing completely.

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  • $\begingroup$ This is a very beautiful argument! A bit of a nitpick that I should have mentioned in my question but this really uses Ehresmann's theorem with boundaries. Your argument also works for the right reformulation with open intervals though. $\endgroup$ – Arrow Jan 5 at 16:28
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If there is no critical points in $[a,a+\epsilon], f:f^{-1}([a,a+\epsilon])\rightarrow [a,a+\epsilon]$ is a submersion. If follows from Ehresmann that it is a fibration. Since $[a,a+\epsilon]$ is contractible this fibration is trivial, we deduce that $f^{-1}([a,a+\epsilon])=f^{-1}(a)\times [a,+\epsilon]$which enables to construct an homotopy (via a deformation retract) between $f^{-1}((-\infty,a])$ and $f^{1}(a,a+\epsilon])$.

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  • $\begingroup$ I assume you meant a deformation retract from $f^{-1}(-\infty,a+\epsilon]$ to $f^{-1}(-\infty,a]$, but I don't see how Ehremann furnishes such a homotopy, and I also don't see how such a homotopy might give a the desired diffeomorphism. Could you please add details? $\endgroup$ – Arrow Jan 2 at 14:38
  • $\begingroup$ Consider $f_t$ defined on $f^{-1}(-\infty,a+\epsilon])$ by $f_t(x)=x$ if $x\in f^{-1}(-\infty,a])$, if $x\in f^{-1}((a,a+\epsilon])$ identify it (via Ehresmann) to $(u(x),v(x)), u(x)\in f^{-1}(a), v(x)\in [a,a+\epsilon]$ and $f_t(x)=(u(x),tv_x)$. $\endgroup$ – Tsemo Aristide Jan 2 at 14:43
  • $\begingroup$ Dear Tsemo, how does your comment address the desired diffeomorphism? $\endgroup$ – Arrow Jan 2 at 22:18

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