= Solution
The smooth <h-cobordism theorem> states: if $(W;M_0,M_1)$ is a compact smooth <h-cobordism>, $\dim W\ge6$, and its connected boundary <manifolds> are simply connected, then there is a <diffeomorphism> $W\cong M_0\times[0,1]$ restricting to the prescribed identification on $M_0$. In particular $M_0$ and $M_1$ are <diffeomorphic>. Being an <h-cobordism> means that both inclusions $M_i\hookrightarrow W$ are <homotopy equivalences>; their fundamental groups then also identify with that of $W$.
The homotopy-equivalence condition is necessary for a <trivial cobordism>: each end of a cylinder is a deformation retract, by $(x,t)\mapsto(x,(1-s)t)$ for the incoming end and $(x,t)\mapsto(x,(1-s)t+s)$ for the outgoing end. Its <relative homology> with respect to either end is consequently zero. Compactness is also necessary for a product with a closed compact end. The dimensional bound and simple connectivity are not necessary for an individual <cobordism> to be a product. The cylinder $S^2\times[0,1]$ is a simply connected <trivial cobordism> of dimension three; and, in the theorem's dimension range, $(S^1\times S^4)\times[0,1]$ is a six-dimensional <trivial cobordism> with <fundamental group> $\mathbb Z$. These prove the two non-necessity assertions explicitly, while the inclusion argument proves necessity of the h-condition.
For the remaining claims, use a <cobordism Morse function>: its two regular boundary levels are zero and one, and all <critical points> are interior. An arbitrary function with unrelated boundary values would not have the relative handle-count property. The relative <Morse chain complex> for $(W,M_0)$ has one generator per <critical point>, in its <Morse index>. Thus
$$
\chi(W,M_0)=\sum_{r\in\operatorname{Crit}(f)}(-1)^{\operatorname{ind}(r)}.
$$
A <trivial cobordism> has zero <relative homology>, hence zero relative <Euler characteristic>. Modulo two the displayed sum is the number of <critical points>. An odd number makes it odd, and therefore nonzero. This proves \b[an odd-critical-point <cobordism> is not trivial].
With precisely two <critical points> of indices $\lambda\le\mu$, a <trivial cobordism> requires the two-generator relative complex to be acyclic over $\mathbb Z$. Equal indices leave a rank-two <homology> group; a gap of more than one leaves both generators with zero possible differential. Therefore \b[$\mu=\lambda+1$ is necessary]. The remaining two-term complex is
$$
0\longrightarrow\mathbb Z q\xrightarrow{\times a}\mathbb Z p\longrightarrow0.
$$
Choose a transverse <gradient-like vector field>. A nonzero differential requires the index-$\mu$ point to be above the index-$\lambda$ point; otherwise no descending connecting trajectories exist. In a common regular level between them, the <belt sphere> $S_U(p)$ has dimension $d-\lambda-1$, and the <attaching sphere> $S_L(q)$ has dimension $\mu-1=\lambda$, where $d=\dim W$. Their dimensions add to $d-1$, the level's dimension, so their transverse intersection consists of points. Each intersection is one descending trajectory from $q$ to $p$, modulo time translation. Orient the <attaching sphere> and use an <orientation of a vector space> on the <handle> core to coorient the <belt sphere>; the resulting signed count is the cellular boundary coefficient
$$
a=S_U(p)\cdot S_L(q).
$$
The two <homology> groups are the kernel and cokernel of multiplication by $a$. They vanish precisely when $a$ is a unit of $\mathbb Z$. Consequently
$$
\boxed{\mu=\lambda+1,\qquad S_U(p)\cdot S_L(q)=\pm1}
$$
are necessary. This proves the <two-critical-point obstruction to a trivial cobordism>; it makes an algebraic incidence statement and does not assume that an algebraic count of one automatically supplies geometric <handle> cancellation.
For an explicit nontrivial <cobordism> from $S^3$ to itself, attach two zero-framed <two-handles> to the outgoing end of $S^3\times[0,1]$ along the standard <Hopf link>. Its two components are unknots with linking number one. This realizes two successive $(2,2)$ surgeries, as suggested by the hint. The outgoing boundary is again $S^3$, and this can be checked geometrically: the Hopf-link exterior is $T^2\times[0,1]$, and each component's zero-framing longitude is the other's meridian. Filling along these two slopes glues two solid <tori> with meridians intersecting once, the standard genus-one Heegaard decomposition of $S^3$. Thus <zero surgery on the Hopf link> does return the three-sphere, not just an unspecified <homology> <sphere>.
The trace has only the two added index-two <handles>. Its relative cellular complex has $C_2(W,S^3)=\mathbb Z^2$ and all other groups zero, so
$$
\boxed{H_2(W,S^3;\mathbb Z)=\mathbb Z^2\ne0.}
$$
It cannot be a product cylinder. Equivalently this trace is $S^2\times S^2$ with two disjoint open four-balls removed: the usual <handle> presentation of $S^2\times S^2$ has one <zero-handle>, the two zero-framed Hopf-linked <two-handles>, and one four-handle. Removing the bottom and top balls exposes the two three-sphere ends. The relative-homology calculation alone already proves \b[the requested <cobordism> is nontrivial].
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