I remember reading about locally flat stuff reading Samelson's proof back from the 60's sometime that justified (in a setting more general than smooth topology) that Poincaré duality corresponded to intersection.
When I started in grad school, I kept bugging everyone about why (in the smooth setting) Poincaré duality corresponded to transverse intersection. Stallings tried to do it at the end of our algebraic topology course, but flubbed it. Bott/Tu had a nice exposition later.
Ultimately, one cannot avoid the Thom class, I've concluded.
Me and Alessandro encountered local flatness while reading Freedman's notes on the proof of 4D topological Poincare conjecture. We barely started, the first few pages are about the Schoenflies problem
If we want to show two paths are homotopic, the definition requires that the the bottom and the top edge of the homotopy square should be equal to the paths. Now, if instead I construct a square in which the top and bottom edge are homotopic to the paths, it should still work, right?
I am thinking of stitching two square at the top and the bottom which take you to the original paths.
Hi, I have a simple question: If I have m + m^c for some c>0, then how to show that it is equal to m^r for some r with any bound. Do we eliminate m and we get r=c or do we work as following: m + m^c = m(1+m^m^{c-1}). or do we upper bounded and then it is easy to get m^r for some r, e.g. m + m^c <= m^{1+c}.
Hi guys, I'm looking for a non-famous theorms that have multiple-proofs. Those theorems should be solved with the knowledge of students in their first/second semster. Is it possible to get some suggestions? I thought of asking them to prove that $3^n=\sum_{K=0}^{n}{n\choose k}2^k$ but I could only think of two solutions (binomial and combinatorics).
Suppose $X$ has an open cover $(U_i)_{i\in I}$, which admits no lebesgue number. In particular, for each $n$, there exists a sequence of non-empty sets, $A_n$, for which, $A_n$ is not contained in any of the $U_i$.Considering AOC, we choose $x_n\in A_n\backslash U_i$. Then, $x_n$ has a convergent subsequence, $(x_{n_k})$ such that $x_{n_k}\rightarrow x$ for some $x\in U_m$. Since $x_{n_k}$ converges, there must exists $K$, for which, $k\geq K$ implies $x_{n_k}\in U_m$. Contradiction.
The way to go about it is to put every vertex on a different element of the standard orthogonal basis and then start going "if I have an edge between two vertices, then I take convex combinations of the corresponding basis vectors. If I have a face between 3 edges then I take convex combinations of the 3 vertices etc.", right?
@BalarkaSen Because I want $|K|$ to come with an easy to work with metric I think
I'm deciphering a proof of Gromov, so this $\ell^2$ business is clearly the best way to do things
The result by Gromov is that $X$ has asymptotic dimension $\leq n$ iff for all $\varepsilon>0$ there is a uniformly cobounded $\varepsilon$-Lipschitz map into a simplicial complex, the idea is that you start with a nice cover of the space given by asymptotic dimension $\leq n$ and then you map $X$ to the nerve of the cover by sending points in a single set of the cover to a vertex, things in two of them to the corresponding edge etc.
It's obviously $\varepsilon$-Lipschitz in the standard simplicial metric (since you're collapsing small open sets in the cover to points), but I see it's an issue to write that down lol
We know what $x\notin A$ express. I am wondering if this proposition: $\neg(x\in A)$ can be proven or is a definition which is an equivalence from the first one, or perhaps is a property called "Negation equivalence". What do you think? Thanks!!
@Thorgott thanks! A student asked "How do I prove that the complement of the negation (of a set) is the same as the negation of the proposition?", so I ran into your help
Oh I'm going to ask a silly question. So for example if we have to show that $A\subseteq A\cup B$, we all know that we start from for all $x$, $x\in A$. Then $x\in A\lor x\in B$, hence $x\in A\cup B$. This "$x\in\text{some set}$", all of them are NOT consider a proposition?
Guys question: suppose we have to prove $C\subseteq A\cup B\to(C\setminus B)\subseteq A$. Is the same as?: $$C\subseteq A\cup B\to\forall x((C\setminus B)\subseteq A)\tag{1}$$ or?: $$\forall x(C\subseteq A\cup B\to(C\setminus B)\subseteq A)\tag{2}$$
this is the only euler platonic polyhedron with all dihedral angles being right angles and therefore. Now gotta extend this to non-regular case
okay I think the answer is yes akiva
wait no it's gotta be no
because if you embed the cube into $\Bbb R^2$ via the net and look at the preservation of the dihedral angles. This is the only polyhedron that preserves the angles after embedding into $\Bbb R^2$
say you have a compact 2-manifold that evolves, and satisfies a differential equation.
for example bessel functions on a disk satisfying the bessel differential equation.
once you solve your diff eq. then it's great. but what if at the boundary of the disk, the diff eq is undefined because the oscillations just blow up so to speak
@topologicalorientablesurface second countable compact Hausdorff spaces are metrizable and in metrizable spaces compactness and sequential compactness are equivalent
Let $X$ be a topological space. $A\subseteq X$ is connected if and only if for any disjoint, open sets $U,V\subseteq X$, we have $A\subseteq U\cup V$ implies $A\subseteq U$ or $A\subseteq V$.
alright, for the backwards direction, I suppose for a contradiction, $A$ is disconnected, so $A=U_A\cup V_A$ where $U_A\cap V_A=\varnothing$ and $U_A=U\cap A$ and $V_A=V\cap A$ for some open sets $U,V$ in $X$
I mean the results in Riemannian geometry are awesome. But trying to read a proof of one and my eyes are rolling off the sockets sideways and dangling off my face
Let $X$ be a topological space. $A\subseteq X$ is connected if and only if for any disjoint, open sets $U,V\subseteq X$, we have $A\subseteq U\cup V$ implies $A\subseteq U$ or $A\subseteq V$.
hints for backwards direction?
I suppose $A$ is disconnected. So, $A$ can be expressed as the disjoint union, of two non-empty open sets, $U_A,V_A$ i.e. $A=U_A\cup V_A$ In particular, $U_A=U\cap A$ and $V_A=V\cap A$ where $U, V$ are open in $X$. So $A\subseteq U\cup V$.
I'm not good at coming up with good questions for this site so I'll throw my newest question out here: is there such a thing as an inverse cofinality function? What I mean is that if $cf(\aleph_\omega)$ goes to $\aleph_0$, is there a function from $\aleph_0$ to the next cardinal that is cofinal with $\aleph_0$, and so on? So we would have $icf(\aleph_0) = \aleph_\omega$ and then the next number in the sequence would be a number that is cofinal with $\aleph_\omega$.
For example, there is a global nowhere-vanishing $2$-form on $S^2$, but it cannot be written as the wedge product of two global $1$-forms. Ah, but it can be written as the sum of such, using partitions of unity. So this makes me think what you said should work in the smooth category but not in the analytic/holomorphic category.
Notice that nLab does this only in the smooth category. What they say is the definition of the deRham complex is not the way I've ever seen it. And in the holomorphic category, you should be able to write down a counterexample easily.
Quick question, in the assertion "A module $M$ over a commutative ring with $1$ is flat iff every relation on $M$ is trivial", what is the definition of a "trivial relation"?
I came up with this when reading Lemma 10.38.11, Stacks Project.
Ted, I have a potentially non-question. I forget how the topological degree proof of the fundamental theorem of algebra goes, but something along the lines of letting $p(z) = \sum_{i=0}^n. a_i z^i$; lifting $p$ to. $\hat{p}$ by letting $\hat(p)(\infty) = \infty$, we get a continuous mapping. $\hat{p}: S^2 \rightarrow S^2$. Finally, $\hat{p}(z)$ is homotopic to $\hat{z}^n$, the latter being $n$ to $1$, thus the prior having the same degree, thus hitting $0$ in the codomain $n$ times.
But IIRC, there was a $S^2 \setminus \{0\}$ somewhere?
