Contractible 4-manifold with an exhaustion Let $M^4$ be a noncompact, contractible, 4-dimensional, topological manifold. Suppose there exists an exhaustion $M=\bigcup_{i\ge 1} U_i$ by open sets such that (1) $\bar U_i \subset U_{i+1}$ and (2) each $U_i$ is homeomorphic to $\mathbb R^2 \times \mathbb T^2$.
Can we prove that $M$ is homeomorphic to $\mathbb R^4$?
 A: It seems to me that there is an obvious Whitehead-manifold inspired construction to try here to produce a counterexample. Let $w : S^1 \times D^2 \to S^1 \times \operatorname{int} D^2$ be the familiar Whitehead embedding, which is null-homotopic but not “null-isotopic”, for which the two inclusions $S^1 \times \partial D^2  \hookrightarrow  S^1 \times D^2 \setminus  w(S^1 \times \operatorname{int} D^2)$ and $w(S^1 \times \partial D^2)  \hookrightarrow  S^1 \times D^2 \setminus  w(S^1 \times \operatorname{int} D^2)$ are each $\pi_1$-injective (google Whitehead manifold.) For $i = 1$ and $2$ let $w_i : T^2 \times D^2 \to T^2 \times D^2$ be the two obvious product embeddings engendered by $w$, namely, let $w_2 : = \operatorname{iden}(S^1) \times w : S^1 \times S^1 \times D^2  \to S^1 \times S^1 \times \operatorname{int} D^2$, and similarly let $w_1 =$ “$w \times \operatorname{iden}(S^1)$” (the quotes meaning that you do the appropriate factor permutations here to make it work). It looks as though the contractible direct-limit 4-manifold $M^4$ gotten from the sequence $w^1$, $w^2$, $w^1$, $w^2$, … is not simply-connected at infinity (like the Whitehead 3-manifold). True?
