Proving that $S^1\times \mathbb{R}^2$ is not homeomorphic to $S^2\times \mathbb{R}$ 
Prove that $S^1\times \mathbb{R}^2$ is not homeomorphic to $S^2\times \mathbb{R}$.

So far our only approach to proving spaces are not homeomorphic is removing points and checking the number of path components.
Latest discussion was on pushouts and pullbacks, so another idea I though of was showing one is a pushout of some diagram that the other isn't, but while we've shown that two spaces satisfying the same pushout are homeomorphic, we have not shown two homeomorphic spaces satisfy the same pushout and I am not sure it is true.
Another approach was to try and prove that $S^1\times\mathbb{R}\not\cong\mathbb{R}^2$, but this has two problems - first is that I don't know how to do it either (without homotopies) and second is that I do not know if the canceling in the factorization is even legal. I know it is not legal in general (Taking any countable set $S$ with the discrete topology and $s\in S$ we have $S\times S \cong S\times\{s\}$), but am unable to prove or disprove the specific case of $\mathbb{R}$.
Are there any more elementary approaches then to prove this?
 A: If you want to solve this with a topological invariant which isn't something like homotopy, you can use the theory of ends. I give the definition here:

Definition: Let $X$ be a topological space and let $\{K_n\}_{n\geq 1}$ be an increasing sequence of compact subsets such that $$X=\bigcup_{n\geq1}\stackrel{\circ}{K_n}.$$
An end of $X$ is a decreasing sequence $\{U_n\}_{n\geq 1}$ such that $U_n$ is a connected component of $X-K_n$.

Then you can prove that the number of ends of a topological space doesn't depend on the sequence $\{K_n\}_{n\geq 1}$ we have chosen (see here), and that it is a topological invariant (this is easy once you proved the first point).
Now using the sequences $K_n=S^1\times B(0,n)$ and ${K_n}^\prime=S^2\times [-n,n]$ for $S^1\times \Bbb R^2$ and $S^2\times \Bbb R$ respectively, you can see that $S^1\times \Bbb R^2$ has one end, whereas $S^2\times \Bbb R$ has two ends, which implies that they can't be homeomorphic.
Here the definition is very formal, you should try to do drawings in $\Bbb R^3$ to see what it represent, it's very intuitive!
