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A well-known theorem in topology says that for a smooth manifold $M$ of dimension $n$ the map $f: M \rightarrow point$ satisfies $$f^! \mathbf R = \mathbf R[n]$$ Here $\mathbf R$ is the constant sheaf.

Here is my question: is there any kind of converse statement to this? (I.e. if $f$ is such that the above equation holds, is $M$ smooth? I don't expect this implication to hold literally, any partial or weakened statement is OK, too.)

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    $\begingroup$ May I suggest you to edit the question to include some conjectural "converse statement"? It took me some time to figure out you probably want something like "if the exceptional inverse image along the constant map is just a shifted-by-dimension constant sheaf, the manifold is smooth". I would guess this is wrong, but I'm not an expert. $\endgroup$
    – Konrad Voelkel
    Commented Jul 29, 2012 at 17:05
  • $\begingroup$ Yes, I hope the parenthetical addition makes it clearer. Indeed, I also don't expect the converse to be true, but I'm looking for a statement that is (true and) somehow close, in spirit, to the converse statement. I.e. I am looking for "Assuming $f^! R = R[n]$, then $M$ is "nice" in some sense." $\endgroup$
    – Jakob
    Commented Jul 30, 2012 at 9:14

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Let $X$ be a homology sphere which is not homotopic to a sphere. (For example, the Poincare $3$-sphere.) Denote by $M$ the suspension of $X$. Then I believe that $f^!\mathbb{R}=\mathbb{R}[4]$, yet $M$ is not even a topological manifold. (Homologically one cannot distinguish $M$ from a $4$-manifold.)

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