## fa.functional analysis – Is there a notion of „flatness” in point-set topology?

In algebraic geometry, flat morphisms are usually associated with the intuition that they have „continuously varying fibers”. Is there a notion in topology formalizing the same intuition? Consider for example a map $$p colon P to X$$ of good (locally compact Hausdorff, say) topological spaces and assume for the moment that it is proper (i.e. has compact fibers). If the fibers vary continuously, then the following property should be true for example:

(*) If $$f colon P to mathbf{R}$$ is a continuous function, then the induced function $$widetilde{f} colon X to mathbf{R}, qquad widetilde{f}(x) := sup_{q in p^{-1}({x})}f(q)$$
is also continuous.

This is true for example if the fibers are constant, i.e. if $$p$$ is a trivial bundle $$p colon X times F to X$$. In that case, the map $$f colon X times F to mathbf{R}$$ induces a continuous map $$X to C(F, mathbf{R})$$, the space of continuous functions equipped with the compact-open topology (which is the topology of uniform convergence) and the map $$widetilde{f}$$ factors as $$X to C(F, mathbf{R}) xrightarrow{sup} mathbf{R}$$.

But there are also examples where the fibers are not (locally) constant: For example the map $$mathbf{C} to mathbf{C}$$, $$z mapsto z^2$$ has continuously varying fibers in the intuitive sense (as $$z$$ approaches the origin, the two points in the fiber get closer to each other). This is also a flat map in algebraic geometry.

For an example which is not flat in the intuitive sense and where also the above property is clearly wrong, consider the first projection
$$p colon {(x,y) in mathbf{R} times (0,1) mid x cdot y = 0 } to mathbf{R}.$$
(This example is of course also inspered by algebraic geometry.)

I should perhaps add that I am aware of the related questions The topological analog of flatness? and Flatness in Algebraic Geometry vs. Fibration in Topology. But I am interested in the purely point-set-topological setting and in fact I care specifically about the property (*) (I have a specific example of a continuous map $$p$$ in mind for which I hope (*) to be true and I’m trying to gain some intuition.)

## gt.geometric topology – Open manifolds which have stable \$pi_1\$ at infinity but are not inward tame

Let $$M$$ be a $$1$$-ended open manifold. An important result of Siebenmann states that (in dimension $$geq6$$) if $$M$$ is $$(i)$$ inward tame, i.e. for every closed neighborhood of infinity there exists a homotopy which pulls it into a compact subset, $$(ii)$$ the end is pro-$$pi_1$$ stable and $$(iii)$$ a certain obstruction vanishes, then the manifold is collarable, i.e. it admits a collar neighborhood of infinity (which is a neighborhood of the form $$partial N times (0, +infty)$$). There are various papers, due mainly to Guilbault, Gu and their coauthors which try to understand what still holds if we drop the hypothesis $$(ii)$$ and try to replace it with softer conditions. On the other hand, I have not found any example of (say $$1$$-ended) manifolds whose end is indeed pro-$$pi_1$$ stable but that are not inward tame: in fact, the only manifolds I know that are not inward tame are very far from being pro-$$pi_1$$ stable (for example, the Whitehead manifold). Can you provide me any reference or examples?

## gt.geometric topology – Quasi-isometric rigidity of surface groups and commensurability

Yes (I assume that by “virtually isomorphic” you mean commensurable modulo finite kernels, which is a nonstandard misleading use of “virtually”). This is because surface groups have Serre’s property D$$_2$$ meaning that each 2-cohomology class (in a finite abelian group with trivial action) it trivial on some finite index subgroup.

(Also a side remark: the known result on QI rigidity is stronger, since it says that every group QI to this surface group is, modulo a finite kernel, isomorphic to a cocompact lattice in the isometry group of the hyperbolic plane. I.e., there’s a structural statement which does not require passing to a finite index subgroup.)

## general topology – Union of open sets is open

The proof is generally pretty straightforward, but my only difficulty is with a small subtlety. The statement is:

Let $$(X,d)$$ be a matrix space and $${U_i}_{i in I}$$ a collection of open sets. Prove that $$bigcuplimits_{i in I} U_i$$ is open.

My attempt is:

If $$I = emptyset$$, then $$bigcuplimits_{i in I} U_i = emptyset$$, which is open, so we proceed under the assumption that $$I neq emptyset$$. Furthermore, if $$U_i = emptyset$$ for all $$i in I$$, then $$bigcuplimits_{i in I} U_i = emptyset$$, which is open, so assume $$U_i neq emptyset$$ for at least one $$i in I$$. So, let $$x in bigcuplimits_{i in I} U_i$$. Then $$x in U_{i_0}$$ for some $$i_0 in I$$. As $$U_{i_0}$$ is open, there exists $$epsilon > 0$$ such that $$B_{epsilon} (x) subset U_{i_0} subset bigcuplimits_{i in I} U_i$$, so $$x$$ an interior point of $$bigcuplimits_{i in I} U_i$$. Therefore, $$bigcuplimits_{i in I} U_i$$ is open.

Every proof of this fact I have every seen completely omits the first two cases I wrote down, where $$I = emptyset$$ or $$U_i = emptyset$$ for all $$i in I$$. Are these even necessary to consider? Are they just too trivial to comment on?

## at.algebraic topology – What is the homotopy category of the sphere spectrum?

Since the sphere spectrum is connective, we may view it as an $$mathbb{E}_{infty}$$-group in spaces, namely $$QS^0$$. This space, as any simplicial set, has a homotopy category $$mathsf{Ho}(QS^0)$$, and since $$mathsf{Ho}$$ is symmetric monoidal, $$mathsf{Ho}(QS^0)$$ acquires the structure of an $$mathbb{E}_{infty}$$-group in categories, making it into a “grouplike” symmetric monoidal category (more commonly called a symmetric groupoidal category or an abelian $$2$$-group).

Is there a known explicit description of the symmetric groupoidal category $$mathsf{Ho}(mathbb{S})overset{mathrm{def}}{=}mathsf{Ho}(QS^0)$$?

## at.algebraic topology – What happens to a closed manifold to ensure it is homeomorphic to a torus \$T^{n}\$?

For dimensions $$n geq 5$$, the answer is yes. First, note that $$M$$ is homotopy equivalent to a torus since it must be a $$K(mathbb{Z}^n,1)$$. Second, Hsiang-Wall show in “On Homotopy Tori II” that in such dimensions, homotopy tori are in fact actual tori (i.e. homeomorphic to tori).

## at.algebraic topology – Every homeomorphism isotopic to one with finitely many fixed points

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## general topology – Show that spaces \$X\$ and \$tilde X\$ are not homeomorphic

I am a beginner in algebraic topology. I am solving the following exercise from my book:

Problem: Consider space $$X:=mathbb{R^2} textbf{/} {(0,0)}$$. Define space $$tilde{X}$$ as follows:

$$tilde{X}={(gamma): gamma$$ is a path in $$X=mathbb{R^2} textbf{/} {(0,0)}$$ starting grom a point $$x_0in X$$ $$}$$. I want to show $$X$$ and $$tilde{X}$$ are not homeomorphic.

My work:

My plan of attack is to find some invariant topological property that holds in one space but does not in other.

Any hint or a reference will be a great help!!

## general topology – Show that if \$M ≥ 0\$ and \$|f(z)| ≤ M\$ for all \$z ∈ ∂V\$ , then \$|f(z)| ≤ M\$ for all \$z ∈ V \$

Suppose that V is a bounded open subset of the plane and $$f ∈ C(overline V) ∩ H(V)$$ i.e. $$f$$ is continuous on $$overline V$$ and $$frestriction_V$$ is holomorphic on $$V.$$

Show that if $$M ≥ 0$$ and $$|f(z)| ≤ M$$ for all $$z ∈ ∂V$$ , then $$|f(z)| ≤ M$$ for all $$z ∈ V .$$
I am not understanding how to approach this. I tried using Liouville’s Theorem and Cauchy’s Estimate.

By Cauchy estimates, for any $$R > 0, |f ′(z_0)|< frac {M}{R} →0$$ as $$R →∞$$.
I am not going anywhere with this. I do not think this is the right way to show this. Can anyone please help?

## gn.general topology – Pareto-optimal front in \$m\$-dimensional space can not have more than \$mathbf{H}_{m-2}\$ homology groups

I need to prove that a Pareto-optimal front in $$m$$-dimensional space (i.e. $$m > 1$$) can not have more than $$mathbf{H}_{m-2}$$ homology groups.

This is my proof strategy:

In a geometrically realized $$k$$-simplex $$sigma_k$$, the $$k+1$$ points $$mathbf{u}_1,mathbf{u}_2, ldots ,mathbf{u}_{k+1} in mathbb{R}^k$$ are affinely independent. Where $$mathbf{u}_i = (f_1, f_2, ldots, f_m) in mathbb{R}^m$$ and $$m > 2$$. Then, the simplex determined by them is the set of points
begin{align} sigma_k=left{theta_1 mathbf{u}_1 + theta_2 mathbf{u}_2 + ldots + theta_{k+1} mathbf{u}_{k+1} mid sum_{i=1}^{k+1} theta_{i}=1 text { and } forall, i,,theta_{i} geq 0right} end{align}
Let’s exclude an arbitrary point $$mathbf{u}_j$$ from $$sigma_k$$ and construct a new simplex $$sigma_{k-1}$$:
begin{align} sigma_{k-1}=left{theta_1 mathbf{u}_1 + theta_2 mathbf{u}_2 + ldots + theta_{j-1} mathbf{u}_{j-1} + theta_{j+1} mathbf{u}_{j+1} + ldots + theta_{k} mathbf{u}_{k} mid sum_{i=1}^{k} theta_{i}=1 text { and } forall, i,,theta_{i} geq 0right} end{align}

Now what I want to show that all the points in $$sigma_{k-1}$$ Pareto-dominates $$mathbf{u}_j$$. If I can do that then I can say in a Pareto-optimal front the highest dimensional simplex is $$sigma_{k-1}$$.

Hopefully from there the rest of the proof about $$mathbf{H}_{m-2}$$ should be easy.

But I can’t sort this out. Also not sure if I am even going on a right direction.