If the suspension of $M$ is homotopy equivalent to a smooth manifold, does $M$ bound a contractible smooth manifold

Question

Let $M$ be a smooth $(n-1)$-manifold. If $M$ bounds a contractible $n$-manifold $C$, then the suspension of $M$ is homotopy equivalent to $$\partial (C\times I) = C\times (\partial I)\cup (\partial C)\times I = (M\times I) \cup C\times\{0,1\}$$ by contracting the two copies of $C$ on the "caps." The space $C\times I$ is a smooth manifold since $C$ is, hence the boundary is smooth, and thus the suspension of $M$ is homotopy equivalent to a smooth manifold.

Is the converse true? That is, if the suspension of (smooth) $M$ is homotopy equivalent to a smooth $n$-manifold, does $M$ necessarily bound a contractible smooth $n$-manifold?

It's clear that $M$ must be a homology sphere by Poincaré duality, but not every homology sphere smoothly bounds a contractible manifold; for example the Poincaré homology sphere does not bound a smooth contractible $4$-manifold, see this question.

Answer 1

Here is a form of Whitehead's theorem that you can find, for instance, in Corollary 6.70 of

Davis, James F.; Kirk, Paul, Lecture notes in algebraic topology, Graduate Studies in Mathematics. 35. Providence, RI: AMS, American Mathematical Society. xvi, 367 p. (2001). ZBL1018.55001.

Suppose that $f: X\to Y$ is a continuous map of simply connected CW complexes which induces an isomorphism of all integral homology groups. Then $f$ is a homotopy-equivalence.

Now, take a 3-dimensional manifold $Y$ which is an integral homology 3-sphere. Let $X$ be the subspension of $Y$. Then $X$ is a simply connected homology 4-sphere. However, if $X$ is an $n$-dimensional simplicial complex which is a pseudo-manifold and a simply-connected $n$-dimensional homology sphere, then there exists a continuous map $f: X\to S^n$ inducing an isomorphism of homology groups: Just take any degree 1 map $X\to S^n$. (Pick an $n$-dimensional open simplex in $X$ and collapse its complement to a point.) Then, by the above version of Whitehead's theorem, $f$ is a homotopy-equivalence. It follows that the suspension of a 3-dimensional manifold which is a homology sphere is homotopy-equivalent to $S^4$.

However, as you correctly noted, a 3-dimensional homology sphere $Y$ need not bound a smooth contractible 4-manifold (e.g., if $Y$ is the Poincare homology sphere). Thus, your question has negative answer.