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This came up in a conversation:

Question: Is there an exotic $\mathbb{R}^4$ that smoothly splits off an $\mathbb{R}$ factor? More precisely, suppose that $\mathcal R$ is a smooth 4-manifold which is homeomorphic to $\mathbb{R}^4$ and has the form of a smooth product $\mathcal R = N\times\mathbb R$ where $N$ is a smooth 3-manifold (necessarily contractible and open). Is such an $\mathcal R$ always diffeomorphic to $\mathbb{R}^4$?

I'd like to think that exotic $\mathbb R^4$'s are inherently 4-dimensional things, and that all four dimensions would have to be involved in any construction. But then again, I wouldn't really know how to prove this. It's clear that an exotic $\mathbb R^4$ cannot smoothly split off $\mathbb R^2$ or $\mathbb R^3$, but with $\mathbb R$ the situation is more complicated because there are contractible open manifolds that are not diffeomorphic to $\mathbb R^3$.

In fact, I don't even know whether the most famous example, namely $\mathcal R=Wh\times\mathbb R$ where $Wh$ is the Whitehead manifold, is actually exotic. The proof that $Wh\times\mathbb R$ is homeomorphic to $\mathbb R^4$ via Bing shrinking certainly doesn't look like it would produce a diffeomorphism. But that doesn't mean there can't be one.

Weaker question: Is it known whether $Wh\times\mathbb{R}$ is diffeomorphic to $\mathbb R^4$?

I would be surprised if this weaker question was still open, but I couldn't find an answer in the limited literature on the subject that I am aware of. Any sort of information in the form of a reference or an explanation would be much appreciated.

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    $\begingroup$ This is a duplicate question, answered affirmatively: mathoverflow.net/questions/33841/… In particular, the answer is cleverly explained by Andy Putman in a comment here: mathoverflow.net/questions/24970/… $\endgroup$ Jul 14, 2017 at 2:39
  • $\begingroup$ Thanks for the links, Chris. And sorry for the duplicate, everyone else! In retrospect, I probably should have found the other questions, but web searches can be tricky. In any case, I'm very happy to have my suspicion confirmed. I also find it interesting that the proof based on McMillan and Munkres apparently depends on the Poincaré conjecture, although this is not too surprising. $\endgroup$ Jul 14, 2017 at 3:44

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