algebraic topology – the boundary of Mobius band is homeomorphic to a circle

It is know that $partial Mapprox S^1$, where $partial M$ is the boundary of the mobius band. But how do we prove that? I am trying to define the homeomorphism explicitly, but it is messy.

The Mobius band $M$ is defined by the quotient $q:Itimes Irightarrow M$ with the equivalence relation $(0,x)sim(1,1-x)$.

$partial M=q(Itimes{0,1})$.

continuity – Example 4, Sec. 29, in Munkres’ TOPOLOGY, 2nd ed: The one-point compactification of $mathbb{R}^2$ is homeomorphic with the unit sphere $S^2$

Here is Theorem 29.1 in the book Topology by James R. Munkres, 2nd edition:

Let $X$ be a (topological) space. Then $X$ is locally compact Hausdorff if and only if there exists a (topological) space $Y$ satisfying the following conditions:

(1) $X$ is a subspace of $Y$.

(2) The set $Y – X$ consists of a single point.

(3) $Y$ is a compact Hausdorff space.

If $Y$ and $Y^prime$ are two spaces satisfying these conditions, then there is a homeomorphism of $Y$ with $Y^prime$ that equals the identity map on $X$.

Immediately following the proof of Theorem 29.1, Munkres states

If $X$ itself should happen to be compact, then the space $Y$ of the preceding theorem is not very interesting, for it is obtained from $X$ by adjoining a single isolated point. However, if $X$ is not compact, then the point of $Y – X$ is a limit point of $X$, so that $overline{X} = Y$.

Definition If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a compactification of $X$. If $Y – X$ equals a single point, then $Y$ is called the one-point compactification of $X$.

We have shown that $X$ has a one-point compactification $Y$ if and only if $X$ is a locally compact Hausdorff space that is not itself compact. We speak of $Y$ as “the” one-point compactification because $Y$ is uniquely determined up to a homeomorphism.

Here is my Math Stack Exchange post on the fact that the one-point compactification of the real line $mathbb{R}$ is the unit circle $S^1$ (regarded as a subspace of $mathbb{R}^2$).

Now my question is as follows:

Let $mathbb{R}$ have the standard (or usual) topology, and let $mathbb{R}^2 = mathbb{R} times mathbb{R}$ have the product topology. Then how to show that the one-point compactification of $mathbb{R}^2$ is (homeomorphic with) the unit sphere $S^2$ given by
$$
S^2 = left{ (x, y, z) in mathbb{R}^3 colon x^2 + y^2 + z^2 = 1 right},
$$

where $S^2$ is regarded as a subspace of $mathbb{R}^3$? That is, can we find a point $P(a, b, c)$ on $S^2$ and a homeomorphism $f colon mathbb{R}^2 longrightarrow S^2 setminus { (a, b, c) }$?

A supplementary question:

More generally, for each $n = 3, 4, 5, ldots$, can we show that the one-point compactification of the euclidean space $mathbb{R}^n$ is (homeomorphic with) the unit sphere $S^n subset mathbb{R}^{n+1}$ given by
$$
S^n := left{ left( x_1, ldots, x_n, x_{n+1} right) in mathbb{R}^n colon x_1^2 + cdots + x_n^2 + x_{n+1}^2 = 1 right}?
$$

That is, can we find a point $Pleft(a_1, ldots, a_n, a_{n+1} right) in S^n$ and a homeomorphism $f colon mathbb{R}^n longrightarrow S^n setminus left{ left( a_1, ldots, a_n, a_{n+1} right) right}$?

algebraic topology – the n-dimensional cube module at its boundary is homeomorphic to an n-dimensional sphere

To let $ {I} ^ {n} $ denote the n-dimensional cube, $ partially {I} ^ {n} $ be his limit and $ {S} ^ {n} $ denote the n-dimensional unit sphere.

Now for a great room $ (X, x_ {0}) $ the $ n ^ {th} $$ homotopy $ $ group $ of $ X $ is

  1. The set of homotopy classes of cards $ f: ({I} ^ {n}, partially {I} ^ {n}) right arrow (X, x_ {0}) $ so the limit $ partially {I} ^ {n} $ goes to the base point $ x_ {0} $ where a homotopy $ f_ {t} $ satisfied $ f_ {t} ( partially {I} ^ {n}) = x_ {0} forall t in (0,1) $.

Alternative;

  1. The set of homotopy classes of cards $ g: ({S} ^ {n}, s_ {0}) rightarrow (X, x_ {0}) $.

Now this is considered $ I ^ {n} Backslash Partial {I} ^ {n} $ is homeomorphic to $ S ^ {n} $.

I fight here. Is it true to say that we would send out the second definition $ s_ {0} $ to $ x_ {0} $ (hence a point to a point), but in the first we would send the set of boundary points to a unique point $ x_ {0} $. This difference of many to one and one to one in the mapping defined in the first and second definitions, respectively, makes it difficult for me to see that the two alternatives are mutually exclusive, even if I understand that they are one obvious consequence of the homeomorphism between the $ I ^ {n} Backslash Partial {I} ^ {n} $ and $ S ^ {n} $.

gt.geometric topology – is a gluing of homeomorphic Mazur manifolds diffeomorphic to $ S ^ 4 $?

A recently published work confirms the existence of homeomorphic but not diffeomorphic Mazur manifolds (see also examples of exotic pairs of contractible stone manifolds).

Let's call them $ M_1 $ and $ M_2 $. If we stick $ W = M_1 cup _ { partially M_1 = partially M_2} ​​M_2 $, then we get a diverse homeomorphic $ S ^ 4 $ according to Freedman's theorem. doubling $ M_1 $ or $ M_2 $ gives the standard smooth $ S ^ 4 $ after a sentence by Mazur. But I wonder if $ W $ is diffeomorphic too $ S ^ 4 $?

Actually, I don't know if twisting the double of an Akbulut cork can lead to an exotic one $ S ^ 4 $? Each exotic pair of manifolds is related by rotating along a cork. So I wonder if something more is known if the complementary contractible distributors are homeomorphic.

general topology – line and level not homeomorphic

I studied Kahn's topology and came across this exercise.
All of the solutions I've looked up mention the connection, but the connection comes in the next chapter, so I figured there had to be another way.
The question asks the fact that if X and Y are homeomorphic and X-x is a disjoint union of two open sets, y s exists. t Y-y is a disjoint union of two open sets.
Intuitively, I understand that the line would be disjoint if I took out a point, but I am completely lost, as I can show that a plane without a point cannot be the disjoint union of open sets.