TITLE: Total spaces of $TS^2$ and $S^2 \times R^2$ not homeomorphic 
QUESTION [48 upvotes]: Hello,
I'm looking for an invariant to distinguish the homeomorphism types of homotopy equivalent spaces. Specifically, how does one show that the total spaces of the tangent bundle to $S^2$ and the trivial bundle $S^2 \times R^2$ are not homeomorphic? (I am not asking for a proof that $TS^2$ is not the trivial bundle.)
Also, is there a way to reduce the question, "Are the total spaces of two vector bundles homeomorphic" to "Are the associated sphere bundles homeomorphic"? In the case of $TS^2$ and $S^2\times R^2$ it's not too difficult to show that the sphere bundles are not homeomorphic, and I'm wondering if there's a way to leverage that.
Thanks,
Zygund

REPLY [4 votes]: An argument using only basic topology and fundamental group (no homology or higher $\pi_i$).
Consider the condition $\mathcal{P}$, for a topological space: for every compact subset $K\subset X$ there exists a compact subset $K'$ containing $K$, such that $X-K'$ is path-connected and has a finite fundamental group.
Then $T\mathbf{S}^2$ satisfies $\mathcal{P}$ ($K$ can be enlarged to the set $K'$ of vectors of norm $\le R$ for some $R$, and the complement of $K'$ to the product of $\mathbf{R}$ and the set of vectors of norm $2R$, so is homeomorphic to $\mathbf{R}\times\mathrm{SO}(3)$ and has $\pi_1$ with 2 elements).
While $\mathbf{S}^2\times\mathbf{R}^2$ does not satisfy $\mathcal{P}$: given $K$, for any $K'$ we can retract the complement of $K'$ onto $\mathbf{S}^2\times \mathbf{S}^1(R)$ for large enough $R$, where $\mathbf{S}^1(R)$ is the circle of radius $R$, so the $\pi_1$ complement of $K'$ has an infinite homomorphism onto the group $\mathbf{Z}$ of integers.
Informally, the fundamental groups at infinity of the 2 spaces differ. (This argument is, anyway, related to the previous ones.)