Okay so let $q : X_n \sqcup_{\alpha}e_{\alpha}^k \to X_n \sqcup_{\alpha}e_{\alpha}^k / \sim$ be defined by $q(x) = [x]$ ($q$ is the canonical map sending each $x$ to its equivalence class in the quotient space)
Then the open sets of the quotient space are those subsets $U$ such that $q^{-1}[U]$ is open in each $e_{\alpha}^k$ and open in $X_n$
:) I think the rest is thrashing out the rigorous terminologies. Notice that "$q^{-1}(U)$ is open in $X_n$" means $q^{-1}(U) \cap X_n$ is open in $X_n$. Now, we understand the map $q$: it sends $q(x) = x$ if $x \in X_n$ and $q(x) = \Phi_\alpha(x)$ if $x \in e_\alpha^k$ where $\Phi_\alpha$ is the characteristic map corresponding to that cell.
So, uh, $q^{-1}(U) \cap X_n$ should be the same as $U \cap X_n$ by definition of $q$.
It's just inclusion $X_n \subset X$ on the $X_n$ factor of the disjoint union of the spaces.
So for a fixed $n$, "$U \subseteq X$ is open iff $U \cap X_n$ is open in $X_n$" is nothing other than a reparsing of the definition of quotient topology: "$U$ is open iff $q^{-1}(U)$ is open". Do this for all $n$, and you're done.
Thanks, Balarka! I'm only having some slight trouble understanding why $q(x) =X$ if $x \in X^n$, and why $q(x) = \Phi_{\alpha}(x)$ is $x \in e_{\alpha}^k$, I think it's because of your definition of $\sim$, could you define it slightly more rigorously if possible?
@Perturbative Ah, yes, so let's try to understand one skeleton up. $X_{n+1}$ is obtained from $X_n$ by attaching $(n+1)$-cells $e_\alpha$ on $X_n$; $X_{n+1} = X_n \bigcup_\alpha e_\alpha/\sim$ where $\sim$ is defined by $x \sim y$ iff $x \in e_\alpha$ and $y \in X_n$, $\varphi_\alpha(x) = y$, $\varphi_\alpha$ being the attaching map of some cell $e_\alpha$.
@LeakyNun, I look at the function $$g_r(s) = \frac{f(re^{2\pi is})/f(r)}{|f(re^{2\pi is})/f(r)|}$$ with f being the said polynominal. And $$ s \in S^1 $$
I am trying to figure out what $g_r(s)$ looks like and what happens, if I let r vary.
If $q : X_n \bigsqcup_\alpha e_\alpha/\sim \to X_n$ is the quotient map corresponding to quotienting by this equivalence relation, then notice that $q(\text{int} \, e_\alpha) \subset X_{n+1}$ are the open $(n+1)$-cells in $X_{n+1}$, so $q$ restricted to $e_\alpha$ is exactly the characteristic map $\Phi_\alpha : D^{n+1} \to X_{n+1}$ of $e_\alpha$. On the other hand since $x \sim y$ for $x, y \in X_n$ iff $x = y$, $q|_{X_n} : X_n \to X_{n+1}$ is merely the inclusion map $X_n \subset X_{n+1}$.
Guys I'm not sure if this is the right place for this but, is there something like a function maker function which, where we can put multiple input(x) and output(x) for the result of a function that takes the given input to the given output?
I'm planning to try to create a function for bisectors in triangles, I'm know there are better ways such as using other formulas but I just wanted learn if there is such a way that I just explained.
But let me explain a little. So when I was in grade 9 I took this AP art class and this guy I knew was in the class. We had another friend we would always hang out with and one day we were out on the quad and my non art friend was looking at some art online and said "woah look at that" and my art friend said "by the time we're in grade 12, we'll be able to do that!" so I said "not necessarily, it takes a lot of hard work and natural born talent to get that good at art."
@Dodsy You know how a functions takes an input and gives an output. I have the inputs and outputs and I'm looking for a way other than the trial-error way.
@Jasper Oh since I've started learning programming I think functions as thing that you put arguments in to return something. I didn't think of it as input output pairs.
Angular
Let me show you with a picture what I'm trying to figure out
hello, i am trying to prove that the roots of minimal polynomial of a linear operator T is the same as characteristic roots of T on a finite dimensional vector space V. i was able to prove the converse, for the other part, i let p(x)=a0 +a1 x +a2 x^2+...+am x^m be the minimal polynomial of T and c be a root of p(x). Then p(c)=0 also p(T)=0. let v $\ne 0$ be a vector in V, then i calculated p(T)v-p(c)v = a1 (T(v)-cv) + a2 (T^2(v)-c^2v)+...+am (T^m(v)-c^mv)=0.
in the hope of proving T(v)=cv. from here, is it possible to conclude that T(v)=cv?
i know that i can say that characteristic polynomial annihilates T, and p is the least degree, monic, such polynomial and p is the generator of the annihilating set, so every root of p is a root of char. poly. of T
but what i did above, can i conclude that T(v)=cv, by "something"
So this has nothing to do with minimal polynomials. You're asking why a root of the characteristic polynomial is an eigenvalue — i.e., that there's a corresponding eigenvector?
hello, i am trying to prove that the roots of minimal polynomial of a linear operator T is the same as characteristic roots of T on a finite dimensional vector space V. i was able to prove the converse, for the other part, i let p(x)=a0 +a1 x +a2 x^2+...+am x^m be the minimal polynomial of T and c be a root of p(x). Then p(c)=0 also p(T)=0. let v $\ne 0$ be a vector in V, then i calculated p(T)v-p(c)v = a1 (T(v)-cv) + a2 (T^2(v)-c^2v)+...+am (T^m(v)-c^mv)=0.
If $X$ is a CW-Comples, is a subcomplex of $Y$ of $X$ a subspace $Y \subseteq X$ that is a union (or disjoint union?) of cells of $X$, such that if $Y$ contains a cell, it also contains its closure
Suppose you have $\kappa = \Bbb N$ and one element has entries in all the odd slots and the other has entries in all the even slots. How do you get the addition of those two vectors to work out?
I don't see how it works to have one basis element for each subset of $\kappa$. How would you do the vectors $(1,0,1,0,1,0,\dots)$ and $(1,0,2,0,3,0,\dots)$ in that direct product?
