Local morse sublevel pair is a handle pair

Use the Morse chart and choose $\epsilon$ small enough that the ball of squared radius $6\epsilon$ is contained in it. Apply the lowering construction, put $x=\mathrm{\mid }u{\mathrm{\mid }}^2$, $y=\mathrm{\mid }v{\mathrm{\mid }}^2$, and subtract $c$ from both functions. Thus $F=-x+y-\mu (x+2y)$. Put $r=\sup \{t:\mu (t)>0\}<2\epsilon$. Since $\mu (0)>\epsilon$ and ${\mu }^{\mathrm{\prime }}>-1$, $r>\epsilon$.

First let $0<k<n$. For each $x\ge 0$ the equation $-x+s-\mu (x+2s)=-\epsilon$ has a unique positive solution $s=s(x)$: its left side is strictly increasing in $s$, is below $-\epsilon$ at zero because $x+\mu (x)\ge \mu (0)>\epsilon$, and tends to infinity. Its derivative in $s$ is $1-2{\mu }^{\mathrm{\prime }}>0$, so $s$ is smooth and $s^{\mathrm{\prime }}=(1+{\mu }^{\mathrm{\prime }})\mathrm{/}(1-2{\mu }^{\mathrm{\prime }})>0$. Also $s(x)\ge x-\epsilon$, with equality for $x\ge r$ (in fact it holds earlier). Choose $d>0$ with $d<s(0)$ and $\epsilon +3d<r$. The compact region $H_0=\{y\le d,x\le \epsilon +y\}$ is parametrized by $(U,V)\mapsto (\sqrt{\epsilon +d\mathrm{\mid }V{\mathrm{\mid }}^2}U,\sqrt{d}V)$ on $D^k\times D^{n-k}$. Its inverse is $(u,v)\mapsto (u\mathrm{/}\sqrt{\epsilon +\mathrm{\mid }v{\mathrm{\mid }}^2},v\mathrm{/}\sqrt{d})$. These formulas are smooth on the axes. Its attaching face is $\mathrm{\mid }U\mathrm{\mid }=1$, on $f=-\epsilon$, and its core is $v=0$.

The union of the lower sublevel with $H_0$ has local boundary $y=\max (d,x-\epsilon )$. Round this single corner by a smooth nondecreasing function $j(x)$ equal to that maximum off a small neighborhood of $x=\epsilon +d$. Choose the rounding above the maximum and below $s(x)$; the strict gap at the corner permits this, for example by smoothing the absolute-value formula for the maximum on a sufficiently short interval. Then $j>0$ and $j=s=x-\epsilon$ for $x\ge r$. This is precisely the compatible rounding of the attached product handle.

The graphs $y=j_t(x)=(1-t)j(x)+ts(x)$ stay strictly positive. Near them use the smooth field $V_t=(s(x)-j(x))v\mathrm{/}(2j_t(x))$ in the $v$ coordinates and zero in the $u$ coordinates. Then $dy(V_t)=s-j$ on the graph. Multiply this field by a bump that is one on the moving graphs where they differ and zero near $v=0$ and outside the chart. The graph difference has compact support in $x$, and all these graph points lie in the chosen chart. Integrate ${\mathrm{\partial }}_t+V_t$ with this cutoff on time times the chart. Smooth dependence and uniqueness give inverse evolution; compact spatial support gives continuation throughout $0\le t\le 1$. Since $dy\mathrm{/}dt=s-j$ on the moving graph, this isotopy carries $\{y\le j(x)\}$ onto $\{y\le s(x)\}$, is identity away from the chart, and fixes the core. Hence the rounded attachment is diffeomorphic to $\{F\le -\epsilon \}$. This uses positive smooth radii, never an inverse of a flat cutoff at its endpoint.

The lowering lemma gives $\{F\le \epsilon \}=\{f\le \epsilon \}$ and a compact regular $F$-band from $-\epsilon$ to $\epsilon$. Its product description supplies the complementary collar. An extra finite collar does not change the diffeomorphism type: join it to an inner collar and reparametrize the collar interval by a smooth increasing map fixed near its inner end. Thus the smooth handle change reaches the upper sublevel.

For $k=0<n$, the lower local sublevel is empty and $\{F\le -\epsilon \}$ is the disk $\mathrm{\mid }v{\mathrm{\mid }}^2\le s(0)$, so it is a disjoint zero-handle; the same regular collar finishes. For $k=n>0$, there is no $v$ variable: $x+\mu (x)>\epsilon$ for every $x\ge 0$, so the missing disk $x\le \epsilon$ is filled along its whole sphere; outside that disk the lower sublevel was already present locally. For $n=0$, the chart is a single point and crossing its value adds that point. These descriptions require no corner or angular coordinate.