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Suppose $M$ is a smooth 4-manifold, and $U,V \subset M$ are exotic $\mathbb{R}^4$'s, i.e. homeomorphic to standard $\mathbb{R} ^4$. Further more suppose $U$ an $V$ intersect nicely sucht that $U \cup V \subset M$ is again homeomorphic to $\mathbb{R}^4$.

An exotic $\mathbb{R}^4$ is called small if it embeds into $S^4$.

Now my question:

If $U$ and $V$ are both small, is $U \cup V $ then also small? I thought about that for quite a long time, somehow I believe it should be true, but do not see any way to prove it.

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The answer is yes. You can embed U and V disjointly in $S^4$, and then do the connected sum at infinity, or `end-connected sum', using an arc going from one to the other. This construction is, I guess, what you intend by $U \cup V$, and is discussed for example in Gompf's paper, Three exotic R4's and other anomalies, J. Differential Geom. 18 (1983), no. 2, 317–328.

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    $\begingroup$ Thank you for for your answer. This is what you could do if $U$ and $V$ are disjoint. I was rather interested in the case where their intersection is topologically "like an end sum", but not smoothly. $U \cap V$ could also be exotic itself and then I dont see how the argument should work. $\endgroup$ – drunken_monkey Jan 21 '15 at 16:53
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    $\begingroup$ I don't think I understood your question, then. Perhaps you mean something like this: Suppose that there is a manifold $W$, homeomorphic to $R^4$, that is the union of two open submanifolds $U$ and $V$, such that $U \cap V$ is also homeomorphic to $R^4$, and each of $U$ and $V$ embeds in $R^4$. Then are you asking if $W$ embeds. Is that correct? $\endgroup$ – Danny Ruberman Jan 22 '15 at 1:18
  • $\begingroup$ Yes that is how I meant it, only additionally $U$ $V$ are also assumed to be homeomorphic to $\mathbb{R}^4$. I considered them both as submanifolds of a manifold $M$, but if one puts it like you, that's not necessary. $\endgroup$ – drunken_monkey Jan 22 '15 at 12:47

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