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What follows is, as far as I can tell, totally standard folklore. I have one particular point of confusion, other than that, I wanted to confirm that I am uttering the incantations correctly.

The smooth 4-dimensional Poincare conjecture (SPC4) is the statement that any smooth 4-dimensional manifold $\Sigma$ that is homeomorphic to $S^4$ is diffeomorphic to $S^4$. Is SPC4 equivalent to the conjecture that any smooth 4-manifold $B$ with $\partial B = S^3$ that is homotopy equivalent to $B^4$ is diffeomorphic to $B^4$?

Going one direction: Assume SPC4 is true.

If $B$ is a homotopy 4-sphere with $\partial B = S^3$ then we can fill the boundary with $B^4$ and we obtain a simply connected (by van Kampen) homology $S^4$ (by Mayer-Vietoris), which (by Hurewicz and Whitehead) must be a homotopy $S^4$, which (by Freedman) must be homeomorphic to $S^4$, which (assuming SPC4) is then diffeomorphic to $S^4$.

I imagine that we are about ready to conclude that $B$ must be standard - since it is now sitting inside of $S^4$ with its boundary bounding a standard $B^4$ on the other side. How do we finish up? Can we ambiently isotope $S^4$ so as to have the image of the $S^3$ be just the usual $S^3 \subset S^4$?

Going the other direction: Assume that every smooth homotopy $B^4$ with boundary $S^3$ is diffeomorphic to $B^4$.

Suppose that $\Sigma$ is a smooth homotopy $S^4$. Take a small 4-ball $B^4$ in $\Sigma$ and remove it. What is left (by van Kampen) is a smooth homotopy $B^4$ with boundary $S^3$ which (by assumption) is then diffeomorphic to $B^4$. Now (by Cerf) since $\Sigma$ is just the union of two copies of $B^4$, it is in fact diffeomorphic to $S^4$.

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Yes, you can perform that ambient isotopy: any oriented embedding $i: B^n \to M^n$ is isotopic to any other. (This is a lemma proven independently by Cerf and Palais1, but the idea is quite clear: shrink the image of $i$ until it's contained in the chart, then take the limit that defines the derivative of a map.)

In particular, if $h$ is your diffeomorphism $B \cup_{\partial B} B^4 \to S^4$, you may isotope the embedding of $h(B^4)$ so that it is the standard inclusion of the north hemisphere. Then $h$ restricts to a diffeomorphism $h: B \to B^4$, where this $B^4$ is the southern hemisphere of $S^4$.

1The references given on this Manifold Atlas page are Palais, Theorem 5.5 (the essential content is Lemma 5.2) and somewhere in Cerf's treatise on embedding spaces.

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  • $\begingroup$ Are you sure the fact about balls is is due Smale? This is usually called the Palais (196)-Cerf (1961) lemma. $\endgroup$ – Danny Ruberman Nov 6 '18 at 11:48
  • $\begingroup$ @DannyRuberman No, but I'm unsure what I was thinking of! I will edit the reference. $\endgroup$ – Mike Miller Nov 6 '18 at 14:00
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    $\begingroup$ The specific reference in the Cerf paper is section 5.1.4. (It's easier to find if you search for "boule" rather than "disque" (or "bisque" as auto-correct would have it).) I think of the Palais result as being from Proc. Amer. Math. Soc. 11 1960 274–277, but you are right about the earlier one. $\endgroup$ – Danny Ruberman Nov 6 '18 at 16:25
  • $\begingroup$ @DannyRuberman Thanks for your care in making sure the references were right. $\endgroup$ – Mike Miller Nov 6 '18 at 16:37

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