Examples of Same fundamental group but not homeomorphic

Question

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?

Answer 1

If one space is obtained by attaching a free arc to another space at an endpoint of the arc, the two spaces have the same fundamental group (the attached arc can be contracted to the attachment point). If the two spaces are not homeomorphic, you have an example.

So, the circle (like an "O") and the circle with an attached arc (like a "Q") are not homeomorphic but have the same fundamental group (isomorphic to $\mathbb Z$).

Answer 2

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.

Answer 3

Here are a few examples:

The spheres $S^n$ for $n > 1$ all have trivial fundamental group, but are not homeomorphic, or even homotopy equivalent.

This means that $X$ has the same fundamental group as $X \vee S^n$ for all $n > 1$, but $X$ is usually not homeomorphic or homotopy equivalent to $X \vee S^n$. I think the best way to think about this intuitively is that the spheres do in some sense contain "holes", but they are too high-dimensional to be detected by the fundamental group.

Let $I$ be a contractible space other than the point. Then there isn't usually a homeomorphism $X \to X \times I$ (for example (see edit), if $X$ is a subspace of $\mathbb{R}^n$ and $I = \mathbb{R}$, then $X$ and $X \times I$ are never homeomorphic), but $X$ and $X \times I$ always have the same fundamental group, and are in fact homotopy equivalent.

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EDIT: Najib points out in a comment that the cartesian product example is non-obvious. I can't find a reference, so I'll prove it here: Suppose $X \subseteq \mathbb{R}^n$, $x \in X$, and $\varphi : X \times \mathbb{R} \to X$ is a homeomorphism. This can obviously be iterated to obtain a homeomorphism $\varphi_m : X \times \mathbb{R}^m \to X$ for any $m \ge 0$. But then we get a composite map $\mathbb{R}^m \cong \{x\} \times \mathbb{R}^m \hookrightarrow X \times \mathbb{R}^m \cong X \hookrightarrow \mathbb{R}^n$ which shows that $\mathbb{R}^m$ is homeomorphic to a subspace of $\mathbb{R}^n$. It seems obvious that $\mathbb{R}^n$ doesn't contain subspaces homeomorphic to Euclidean spaces of arbitrarily large dimension, but I can't find a reference so I'll prove that too. Suppose $\mathbb{R}^{n+1} \cong S \subseteq \mathbb{R}^n$. The closed ball $D^{n+1} \subset S$ is compact, and thus is closed and bounded in $\mathbb{R}^n$. In particular, $D^{n+1}$ is homeomorphic to a closed subspace $C \subset \mathbb{R}^n$. $\mathbb{R}^n$ has Lebesgue covering dimension $n$ while $C \cong D^{n+1}$ has Lebesgue covering dimension $n+1$. But no space can contain a closed subspace of strictly larger Lebesgue covering dimension, so this is impossible.

Answer 4

Some classic examples from 3-manifold topology are the Lens spaces $L(p, q)$.

Fix $p\geq 3$ and $q \in \mathbb{Z}$ coprime to $p$. If $\zeta$ is a primitive $p^{th}$ root of unity, identify $\mathbb{Z}/ p \mathbb{Z}$ with $\langle \zeta \rangle$. Then $\mathbb{Z}/ p\mathbb{Z}$ acts properly discontinuously on $$S^3= \{ (z, w) \in \mathbb{C}^2 \mid |z|^2 + |w|^2 \} $$ by $\zeta \cdot (z, w) = (\zeta z, \zeta^{q}w)$ since $\mathbb{Z}/ p\mathbb{Z}$ is finite and the action has no fixed points. This means that the quotient space $L(p, q):= S^3 / (\mathbb{Z} / p \mathbb{Z})$ is a 3-manifold and the quotient map $p \colon S^3 \to L(p, q)$ is the universal cover. In particular, $$ \pi_1 ( L( p, q )) \cong \mathbb{Z}/ p \mathbb{Z}$$ for any choice of the $q$'s. In fact, it isn't too hard to show that the Lens spaces all have the same homology and higher homotopy groups (for fixed $p$). This means that something more sensitive is required to detect the difference between the Lens spaces and one of these somethings is Reidemeister torsion. I won't describe it here, but the link does.

The punchline is that $L(p, q_1 )$ is homeomorphic to $L(p, q_2)$ if and only if $q_1 \equiv \pm q_2^{\pm 1} \mod p$.

Answer 5

Sticking to compact manifolds, you have $S^4$ and $CP^2$ that are both simply connected. Taking a connected sum of those with any 4-manifold with fundamental group $G$ (this can be any finitely presented group) you get two nonhomeomorphic spaces with the same fundamental group $G$.