## What is the one dimensional image that the torus with one circular hole on its surface homeomorphic to?

I am really poor of geometry imagination and struggled to figure this out.

## 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 well-pointed topological group $$G$$ as the fat realization of the nerve of $$G$$.

Let $$G$$, $$H$$ be well-pointed 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 well-pointed 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 pro-choice 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 path-connected 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 path-connected.

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 $$\mathbb{R}^n$$?