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I am not very conversant with the $3$-manifold or $4$-manifold theory. The literature I found so far related to $3$-manifold tells me that $S^1\times S^2$ is "very special". It is prime but not irreducible. $3$-manifolds with the infinite non-cyclic fundamental group $G$ are $K(G,1)$ spaces. The answer to the following questions may be well-known.

  1. Let $M$ be a compact, oriented manifold of real dimension $4$ with boundary $dM=S^1 \times S^2$. Is there a way to classify all such $M$ up to diffeomorphism?
  1. Let $(M,dM)$ as above. Given an infinite non-cyclic group $G$ so that $dM$ is homotopic to $K(G,1)$, is there a way to classify $M$ (compact, oriented with boundary $dM$)?
  1. Can we say something about the choices of $\pi_1(M)$ in both the above cases?
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    $\begingroup$ For 1: one can't expect to classify oriented closed 4-manifolds (since any finitely presented group occurs as $\pi_1$ of such a manifold). Then in any closed 4-manifold one can remove the tubular neighborhood of a little circle, and that makes a compact oriented manifold with boundary $S^1\times S^2$. This suggests that classifying such manifolds is as intractable as classifying closed 4-manifolds. $\endgroup$
    – YCor
    Commented Oct 11, 2022 at 7:30
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    $\begingroup$ 1) No. 2) No. 3) No. Generally there is no relation between $\pi_1 M$ and $\pi_1 \partial M$. Just take a look at some examples. $\endgroup$
    – Ryan Budney
    Commented Oct 11, 2022 at 7:32
  • $\begingroup$ Here is a less naive question. If a manifold $M$ with $\partial M = S^1 \times S^2$ has $i_*$ an isomorphism of fundamental groups, is $M$ a connected sum of a simply connected closed manifold and $S^1 \times D^3$? I think the answer is plausibly "yes", and shows that the kind of thing the OP is looking for in this and his previous question are more interesting than it seems at first glance. $\endgroup$
    – mme
    Commented Oct 11, 2022 at 13:08
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    $\begingroup$ @mme. In the topological category, I suspect one can give a fairly precise description of such 4-manifolds (in DIFF, there is exotica). For this specific boundary, they are determined by their $\mathbb{Z}[t^{\pm 1}]$-intersection form. This Hermitian form must be nonsingular and "evaluate" to the $\mathbb{Z}$-intersection form at $t=1$. Once this form is fixed, there is a unique manifold in the spin case and two in the odd case (Kirby-Siebenmann;star construction). Could write more, if this is the question the OP is asking. $\endgroup$
    – Anthony Conway
    Commented Oct 11, 2022 at 13:40
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    $\begingroup$ Hambleton-Teichner gave examples of a nonsingular form over $\mathbb{Z}[t^{\pm 1}]$ that is not extended from the integers (which would be the case for a manifold of the form $(S^1 \times D^3) \# X$ with $X$ simply-connected). Using some more recent papers, it is then possible to realise their form by a $4$-manifold with $\pi_1=\mathbb{Z}$ and boundary $S^1 \times S^2$. Don't worry about asking another question :). $\endgroup$
    – Anthony Conway
    Commented Oct 12, 2022 at 11:55

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