Are all contractible spaces semilocally simple connected? I think so. All contractible spaces are simply connected. Given $x \in X$, $X$ is an open neighborhood of $x$ such that the inclusion-induced map $\pi_1(X,x) \to \pi_1(X,x)$ is trivial, because $\pi_1(X,x)$ is the trivial group. Is this right?
I have a very simple equation that I'm having trouble with. Consider $x,y,p \in [0,1]$. If $px+(1-p)y = 0$ must hold for every $p$, what are the possible solutions to $x$ and $y$?
user280247
3:21 PM
any book or website of maths puzzles for amateurs you recommend?
user280247
hellos guys, any book or website of maths puzzles for amateurs you recommend?
I'm thinking about an exercise, among other things it asked me to decide whether $\mathrm{Spec}\Bbb Z[i]$ is a normal and regular scheme. Normal is easy since the stalks are all PIDs (being localizations of a PID) hence integrally closed, but I'm unsure about the regularity, I guess there is some CA trick to avoid actually checking it?
It probably shouldn't be too hard to check regularity by hand, but maybe think about what's nice about $\mathbb{Z}[i]$ (think about some of the adjectives that one might use to describe $\mathbb{Z}[i]$)
Sanity check: If $X=\mathrm{Spec}A$ is a regular affine scheme and $A$ is a Noetherian ring of finite Krull dimension $n$ then the Zariski (co)tangent space has dimension $n$ at all closed points of $X$ because if $\mathfrak p$ is a closed point, hence a maximal ideal, $\dim A=\dim A_\mathfrak p$ and by regularity this is also the dimension of the (co)tangent space
"Let f be a ring homomorphism from a ring R to a ring S. If R has a unity 1, S$\ne${0}, and f is onto, then f(1) is the unity of S." Why do we need onto ?
Is it because $f(x)=0$ is a possible ring homomorphism? Or any other reason?
@AlessandroCodenotti I'm not sure if $\dim A = \dim A_{\mathfrak{p}}$ is always true, when $\mathfrak{p}$ is a maximal ideal. I think stacks.math.columbia.edu/tag/02JE is a counterexample
@AlessandroCodenotti I think the link I attached seems to give an example? where the localisation of $B$ (in the link) at two different maximal ideals has different dimensions
It seems to work for nice enough rings though, which is reassuring (also I'm not surprised at all to find Nagata's name associated to yet another ugly counterexample :P)
But yeah I was still thinking of $\Bbb Z[i]$ where it works
This exercise wanted to know the dimension of the (co)tangent space in $(i+1)$ and while it can be computed explicitely I was trying to avoid doing so :P
Next it asks to determine $\pi_\ast\mathcal O_X$ and whether it is locally free, where $X=\mathrm{Spec}\Bbb Z[i]$ and $\pi_\ast$ is induced by $\Bbb Z\to\Bbb Z[i]$, but I think I got this, I'll write it down later
No, sorry, my doubt wasn't on smooth, but on "Dedekind domain implies smooth", what's the precise statement?
(we only defined smooth for a scheme $X$ over a field $k$ though where we defined it as locally of finite type, constant dimension $n$ and $\Omega_{X/k}$ locally free of constant rank $n$)
We also saw that if $k$ is perfect regular and smooth are equivalent, and Dedekind domains are regular, is that what you meant?
@user193319 Alright: So is the following statement true only when the two spaces are of the same dimension: A linear map from V to W (both finite dimensional) is invertible iff rankT=dimV?
I believe what you say is correct. A $T : V \to W$ is a linear operator between vector spaces of the same dimension is invertible iff it is injective iff it is surjective. This essentially follows from the rank nullity theorem.
On page 64 of Hatcher's book on Algebraic Topology, he writes the following:
...if the map $\pi_1(U) \to \pi_1 (X)$ is trivial for one choice of basepoint in $U$, it is trivial for all choices of basepoint since $U$ is path-connected.
I interpret this as the following claim:
Let $\iot...
@AkivaWeinberger no lol its okay. I really haven't studied fractals yet, so I don't know anything about them. It was just a question I was curious about
Does the study of fractals and diff geo overlap at some point?
can you have a fractal where you can define differentiability?
@topologicalmagician people use geometric measure theory to study diff geo and to study fractals but the crowds doing one or the other don’t tend to be the same
(because differential geometers low key don’t care about fractals)
generalizing Akiva's example, in the dihedral group $D_{n,2n}$ you have $\rho \sigma$ which has order $2$, and $\sigma$ still of order $2$, but $(\rho\sigma)(\sigma) = \rho$ having order $n$
I have a conceptual question about convex conjugates for example, in the equation in suvrit.de/teach/ee227a/lect4.pdf slide 24 of 103, what is a good interpretation of the variable "z" ?
On page 64 of Hatcher's book on Algebraic Topology, he writes the following:
...if the map $\pi_1(U) \to \pi_1 (X)$ is trivial for one choice of basepoint in $U$, it is trivial for all choices of basepoint since $U$ is path-connected.
I interpret this as the following claim:
Let $\iot...
