## general topology – Prove that \$ { prod_ {i in mathbb {N}} X_i} \$ and \$ { prod_ {i in mathbb {N}} Y_i} \$ are homeomorphic.

the problem posed comes from Munkrees. You marry my proof ?? I feel like I'm missing something. Thank you!

Suppose that for each $$i in mathbb {N}, X_i,$$ and $$Y_i$$ are homeomorphic. Prove it $$displaystyle { prod_ {i in mathbb {N}} X_i}$$ and $$displaystyle { prod_ {i in mathbb {N}} Y_i}$$ are homeomorphic.

$$textbf {Proof:}$$ assume $${X_i: i in mathbb {N} }$$ and $${Y_i: i in mathbb {N} }$$ are indexed families of topological spaces. Suppose that for each $$i in mathbb {N}$$, the spaces $$X_i$$ and $$Y_i$$ are homeomorphic. so $$displaystyle { prod_ {i in mathbb {N}} X_i}$$ and $$displaystyle { prod_ {i in mathbb {N}} Y_i}$$ are homeomorphic.

## Is it generally true that a space is not homeomorphic to the perforated version of this space?

For non-arbitrary spaces, we can discuss for such a case, like the number of components or other properties. But is this true for any space? It seems that if we have a homeomorphism $$f$$ of $$S$$ at $$S & # 39; = S – {p }$$, $$f (p) = q$$, but since a space is homeomorphic to itself, there are $$g (r) = q$$. Then there is no reverse if $$f$$ and $$g$$ coincide. But they don't have to and maybe $$f$$ is sort of homeomorphism since i can't infer more information from it.

If this is not true, a counterexample will be super useful! Thank you.

## If \$ X \$ is homeomorphic to a Lie group, is \$ X \$ a variety?

The question is quite simple, but I couldn't find an answer:

Let $$G$$ to be a Lie group (Hausdorff compact). Yes $$X$$ is homeomorphic to $$G$$ as a topological space, is $$X$$ then also a topological variety?

## continuity – Prove that \$ mathbb {N} \$ is homeomorphic to \$ {(1 / n): n ge 1} \$

Prove it $$mathbb {N}$$ (with the usual metric) is homeomorphic to $${(1 / n): number 1 }$$ (with usual metric)

I have defined a function $$f: n rightarrow 1 / n$$ which is clearly one-on and on. But I can not prove that $$f$$ And $$f ^ {- 1}$$ is continuous in the metric space.

Thank you!

## Geometry dg.differential – In a variety, \$ angle xpy> frac { pi} {2} \$, for \$ q \$ on \$ px \$ or \$ py \$, \$ B_q (r) \$ homeomorphic to \$ B_p (r ) \$?

Let M be a Riemannian variety with n dimensions, without border, with sectional curvature $$geqslant -1$$. For one point $$p in M ​​$$, suppose that there is $$l, delta> 0$$, $$x, y M$$ with $$d (p, x), d (p, y)> l$$ and a geodesic $$px$$ and $$py$$ with angle $$angle xpy> frac { pi} {2} + delta$$. Let $$q$$ to be a geodesic point $$px$$ or $$py$$, Question: is there $$r> 0$$, which depends only on $$n, l, delta$$ such as $$B_q (r)$$ is homeomorphic to $$B_p (r)$$?

Equally, we can ask the question as follows:

Let $$M_i$$ to be a sequence of Riemannian varieties with $$sec geqslant -1$$ and diameter $$leqslant$$. assume $$(M_i, p_i)$$ Gromov-Hausdorff converge (eventually collapse) towards $$(X, p)$$ (we know that this is an Alexandrov space). Suppose that there is $$l> 0, delta> 0$$, $$x, y in X$$ with $$angle xpy> frac {pi} {2} + delta$$. elevator $$x, y$$ at $$M_i$$, we have $$x_i, y_i in M_i$$. with$$angle x_i p_i y_i> frac { pi} {2} + delta$$. Let $$q_i$$ to be a geodesic point $$p_ix_i$$ or $$p_iy_i$$. Question: Is there $$r> 0$$, such as $$B_ {q_i} (r)$$ is homeomorphic to $$B_ {p_i} (r)$$?

## general topology – Show that \$ widehat {(0,1)} \$ is homeomorphic to \$[0,1]/ sim \$ where \$ x sim y iff x = y text {or} {x, y } = {0,1 } \$.

$$widehat {(0,1)} = (0,1) cup {p }$$, or $$p not in (0,1)$$. A set $$U$$ in $$widehat {(0,1)}$$ is open if ($$U not nor p$$ and $$U$$ is open in $$(0.1)$$) or ($$U ni p$$ and $$(0,1) setminus U$$ is a compact subset of $$(0.1)$$).

CA watch $$widehat {(0,1)}$$ is homeomorphic to $$[0,1]/ sim$$ or $$x sim y ssi x = y text {or} {x, y } = {0,1 }$$.

A set of representatives of $$[0,1]/ sim = {[x] mid x in[0,1] }$$ is $$bigcup_ {x in (0,1)} underbrace {[x]} _ {= {x }} cup underbrace {[0]} _ {=[1]= {0,1 }}$$.

Let $$p:[0,1] twoheadrightarrow [0,1]/ sim: x mapsto [x]$$ to be the quotient map, so $$U subset[0,1]/ sim$$ is open if $$p ^ {- 1} (U) subset [0,1]$$ is open.

I have to prove that
begin {align *} f colon[0,1]/ sim & longrightarrow widehat {(0,1)} \ [x]& longmapsto x text {if} x in (0,1) \ [0]=[1]& longmapsto p end {align *}

is a homeomorphism.

It is clear that it is a bijection. I have problems to prove that it is continuous.

Let $$U subset widehat {(0,1)}$$ open. assume $$p not in U$$then $$U$$ is open in $$(0.1)$$, So $$U = (0,1) cap B$$ ($$B$$ open in $$mathbf {R}$$). then $$f ^ {- 1} (U)$$ is open in $$[0,1]/ sim$$, since $$p ^ {- 1} (f ^ {- 1} (U)) = U$$which is open in $$[0,1]$$, since $$U = underbrace {((0,1) cap B)} _ { text {open in} mathbf {R}} cap [0,1]$$.

Now assume $$to U$$. then $$(0,1) setminus U$$ is a compact subset of $$(0.1)$$ (I guess that means by Heine Borel that he's closed $$(0.1)$$?) So $$(0,1) setminus U = (0,1) cap C$$ ($$C$$ shut in $$mathbf {R}$$). Now $$(0,1) setminus ((0,1) setminus U) = (0,1) cap underbrace {( mathbf {R} setminus C)} _ { text {open in} mathbf {R}}$$. It means that $$p ^ {- 1} (f ^ {- 1} (U))$$ is the union of an open subset of $$(0.1)$$ and $${0,1 }$$, which is not open in $$[0,1]$$.

I think this problem is difficult for me because I do not feel familiar with the topology of the quotient yet. Could someone provide some help or a simpler way to handle this?

## Topology gt.geometric – Are triangulations with common refinements homeomorphic?

Are there simple triangulations $$K_1$$ and $$K_2$$ of a topological variety $$M$$ such as $$K_1$$ and $$K_2$$ have a common subdivision but they are not homeomorphic PL? Ideally, I would like an example in dimension 4.

From what I understand from PL homeomorphism, there is a common subdivision $$L$$ and embarkations $$phi_i$$ of $$K_i$$ to a Euclidean space that takes simplexes of both $$K_i$$ and of $$L$$ to linear simplexes. Note that there is a triangulation $$L$$ 3 simplex $$K$$ which contains clover in its 1-skeleton with only 3 edges. As the number of clover sticks is 6, there is no incorporation $$phi: K to mathbb {R} ^ n$$ which is linear on both simplexes of $$K$$ and $$L$$.

Any suggested references would also be welcome.

## General Topology – The punctured balloon is not homeomorphic to the Euclidean space

Just another ordinary day with another (big) ordinary math conversation. A friend and I asked this:

Problem. Prove it $$B (0, r) setminus {0 } subseteq mathbb {R} ^ n$$ is not homeomorphic to open bullets for $$r> 0$$.

I have not taken a topology course yet and my friend has just started a topology course. So we tried to find as basic a solution as possible.

It seems easy enough, even though we struggled …

• It is enough to prove that it is not homeomorphic $$mathbb {R} ^ n$$.
• Most elemental topological invariants do not work.

The best we have found is to compute the fundamental group and take generalizations of the fundamental group for $$n> 2$$. This should probably work.

So here's my question: Is there another way to prove this result?

## Algebraic topology – Are the level sets of a smooth map of a topological space homeomorphic to a collection of cells?

This question is related to the definition of Euler-Integral transformations, page 147 of Michael Robinson's book "Topological signal processing".

## Definition: characteristic of Euler and constructable function

The euler characteristic $$chi$$ is a valuation of a cell complex $$X_f$$ at $$mathbb {Z}$$ Defined by:
$$chi (X_f) = sum_ {c in X_f} ^ {} {- 1 ^ {dim (c)}}$$

$${ bf Note:}$$The definition of $$chi$$ allows to define it for an arbitrary collection of cells. For example, a sub-collection $$X_f$$ of cells in $$X_f$$ is not always a cellular complex, but we will admit that $$chi (X_f & # 39;)$$ is well defined. Even though the author does not say so clearly in the book, this hypothesis seems to be widely used.

A whole-valued constructible function is a function of a topological space $$f: X mapsto mathbb {Z}$$ such that there is a non-unique cellular complex $$X_f$$ For who:

a) There is a homeomorphism $$h: X mapsto X_f$$

b) the function $$f circ h$$ is constant on every cell of $$X_f$$

This can be seen as a generalization of the piecewise constant function on topological spaces having a complex cellular skeleton.

## Statement of the problem

Given a topological space $$X$$ and a smooth map $$P: X mapsto mathbb {R}$$. We suppose that $$X$$ is such that there is a homeomorphism $$h$$ enter $$X$$ and a cellular complex $$X_f$$ . We define the sets of levels of $$P$$:
$$X_c = {x in X, P_x (x, bullet) = s }$$

The question is: can we calculate $$chi (h (X_c))$$ ?

The author seems to do it. But I am troubled by the fact that the Euler feature is only defined for cell collection and that the set defined above might very well not be a collection of cells.

For example, if $$Y$$ is the ball centered in o in $$mathbb {R} ^ 2$$ with the complex cellular structure made of a $$0$$-cell, a $$1$$-cell and a $$2$$-cell. Let us $$P: y mapsto | y |$$.

The level set of $$P$$ for $$s$$ are the intersection between a circle of radius s and $$Y$$. Three cases can occur:

1. For $$s = 0$$, the defined level is the single point (0,0).
2. For $$0 Y_s is a closed circle inside $$Y$$.
3. For $$s = s_ {max}$$ the level defined is the border of the $$1$$-cell attached to the $$0$$-cell.

It seems that apart from the third case, the defined levels are neither cells nor a set of cells. This explains my inability to understand how the author can apply the characteristic euler on these sets.

I wonder if anything has escaped me in $$h$$ to be a homeomorphism, or by $$P$$ to be smooth, as I do not seem to use any of these assumptions. I would appreciate some clues as to what the author actually tries to do (for those who read the book) or on the accuracy of the Euler features on these sets.

## Two complex manifolds homeomorphic non diffeomorphic

Is there a closed topological variety supporting two non-diffeomorphous smooth structures, both of which have a complex, compatible structure? Same question, but for the symplectic structure.