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I am reading through the first couple of chapters of Milnor's Morse theory, and I've gotten to Reeb's sphere theorem (theorem 4.1),

If $M$ is a compact manifold and $f$ is a differentiable function on $M$ with only two critical points, both of which are nondegenerate, then $M$ is homeomorphic to a sphere.

Milnor states that for some small enough $\varepsilon > 0$, the sublevel sets $f^{-1}[0,\varepsilon]$ and $f^{-1}[1-\varepsilon,1]$ are closed $n$-cells, which follows by the Morse Lemma. But I don't quite follow this. I've read other related posts, but none of them quite flesh out why we need a "small enough" $\varepsilon$, or why exactly the Morse lemma implies that the aforementioned preimages are closed $n$-cells.

Now, I understand that the two critical points will have index $0$ and $1$, which correspond to the minimum and maximum, but why does applying the coordinate maps given to us my the Morse lemma tell us that we get $n$-cells, and not only containments? If anyone could clarify, that would be great.

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The neighborhood of a critical point where Morse Lemma applies is not necessarily a disk, but it contains a, (sufficiently small radius) closed disk.

Suppose that $p$ is a minimum of $f$ with $f(p)=0$. By Morse Lemma, there is a coordinate system $(U,y^i)$ of $p$ such that $y^i(p) = 0$ for all $ i=1,\dots,n$ and $f$ has representation $$ f = (y^1)^2 + \dots + (y^n)^2, \quad \text{on }U. $$ So, for sufficiently small $\varepsilon>0$ we have $D \equiv f^{-1}[0,\varepsilon] \subseteq U$. That is $$ D = \{q \in U \, \mid \, f(y_q) = (y^1_q)^2+ \dots + (y^n_q)^2 \leq \varepsilon \}, $$ which is clearly diffeomorphic to a closed $n$-disk in $\mathbb{R}^n$, via coordinate map $y$.

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