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I ran into an oriented smooth $h$-cobordism from $S^1\times S^2$ to itself in my project. I wish to argue that it is diffeomorphic/homeomorphic to the product.

From this question 4-dimensional h-cobordisms, it seems that in dimension $4$, $h$-cobordism is automatically $s$-cobordism. But $s$-cobordism is not necessarily trivial in general.

I wonder whether one can tackle the question, with the further constraint that the boundaries are both $S^1\times S^2$.

My cobordism is also symplectic, where the boundaries are the contact boundary of the subflexible domain $T^*S^1\times \mathbb{C}$. It seems from the mentioned post symplectic property also plays a role (at least for elliptic boundaries).

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    $\begingroup$ You don't need any special argument to conclude that your h-cobordism is an s-cobordism: the Whitehead group of the integers is trivial. (This is an old theorem of Higman.) $\endgroup$
    – Danny Ruberman
    Commented Jun 2, 2020 at 0:34

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