I am really poor of geometry imagination and struggled to figure this out.
Tag: homeomorphic
Prove that spheres in 2 and 3 dimensions are not homeomorphic.
The spaces $ S^1$ and $ S^2$ , with their topologies inherited from $ \mathbb R^2$ and $ \mathbb R^3$ , respectively, are not homeomorphic. Here $ S^1$ and $ S^2$ are spheres in $ 2$ and $ 3$ dimension.
Topological groups with homeomorphic underlying spaces, isomorphic abstract groups and homotopy equivalent classifying spaces
Define the classifying space $ BG$ of a wellpointed topological group $ G$ as the fat realization of the nerve of $ G$ .
Let $ G$ , $ H$ be wellpointed topological groups. Assume that there is a continuous group homomorphism $ f:G\rightarrow H$ that induces a homotopy equivalence on the underlying topological spaces. It is claimed in an answer to this question that the induced map $ BG\rightarrow BH$ is a homotopy equivalence. It is important here that we have a map between groups that simultaneously respects algebra and topology.
The question is basically what happens if we do not have such a map. Let $ G$ , $ H$ be wellpointed topological groups. Assume that their underlying abstract groups are isomorphic and that their underlying topological spaces are homeomorphic. This by itself does not mean that $ G$ and $ H$ are isomorphic as topological groups (a prochoice example is given by $ p$ adic rationals for different $ p$ ).
But what happens if we assume in addition that the classifying spaces of $ G$ and $ H$ are homotopy equivalent?
Note that a theorem of Notbohm says that two compact Lie groups are isomorphic as Lie groups iff their classifying spaces are homotopy equivalent so at least in some contexts, this question has a positive answer.
Are 0,8 and 9 homeomorphic topological space?
Consider the topological spaces “0”,”8″ and “9” in $ \mathbb{R}^{2}$ . Are they homeomorphic?
I have an approach that doesnt look very rigorous to me. I wanted to know how to formalize this if its correct.

0 and 8 are not homeomorphic since excluding one point of 0 the space is still connected, but excluding the “tangent point” of 8, we have a disconnected space.

Same idea for 8 and 9.

The space 9 is union of one circle and one arc. The arc is homeomorphic to the circle, so we can view 9 as a union of two circles, then 8 and 9 are homeomorphic
PS: the topology of the spaces is induced by topology of $ \mathbb{R}^{2}$ .
Homeomorphic characterization of the real line?
Let $ A$ be a pathconnected subset of $ \mathbb R^2$ such that the removal of any singleton from $ A$ splits $ A$ into two open connected components, each of which is pathconnected.
Is $ A$ necessarily homeomorphic to $ \mathbb{R}$ ?
Connected and homogeneous $T_2$space not homeomorphic to a subset of $\mathbb{R}^n$
What is an example of an connected and homogeneous $ T_2$ space $ (X,\tau)$ with $ X=2^{\aleph_0}$ such that for no $ n\in\mathbb{R}$ the space $ (X,\tau)$ is homeomorphic to a subspace of $ \mathbb{R}^n$ ?