Can you give me some example that their fundamental group (which is non-trivial) is same but their topological spaces are not homeomorphic?
$i.e$,
\begin{align}
\pi_1(X) = \pi_1 (Y), \qquad X \ncong Y
\end{align}
For simply connected space, i know some examples which their fundamental group is same but their topological spaces are not homeomorphic. $i.e$,
one such a example is convex set. $e,g$, $R^1$, $R^2$ is simply connected but they are not homeomorphic.
Can you give me some other examples?
Best Answer
It is easy to construct examples if you remember that homotopy equivalent spaces which are path connected have isomorphic fundamental groups. Therefore, you need only construct $X,Y$ to be homotopy equivalent, path connected spaces that are not homeomorphic.
For an example, take $X = S^1$ and $Y = \mathbb{R}^2 - \{(0,0)\}$. The inclusion map $S^1 \to \mathbb{R}^2 - \{(0,0)\}$ has, for a homotopy inverse, the deformation retraction $\mathbb{R}^2 - \{(0,0)\} \to S^1$ which is defined by the formula $p \mapsto \frac{p}{|p|}$.
It is easy to see that $S^1$ and $\mathbb{R}^2 - \{(0,0)\}$ are not homeomorphic, and there are several ways to do it. Perhaps the most elementary way is to show that each $x \in S^1$ has a neighborhood basis of connected open sets each of which is disconnected by $x$, but no point of $\mathbb{R}^2-\{(0,0)\}$ has such a neighborhood basis.