i have to say, i've always been uncomfortable about resorting to geometry to prove an analytic result. for that reason alone, i think defining sine by its (complex) taylor series is logically the best way to go.
@anon David Mitra has only beaten Arturos weekly output once -- a week where Arturos rank dropped to #14; he must have been on vacation. But he's quite frequently #2.
Well, plus enough upvotes to cap several times over, of course. So 766 -- there was some downvotes that apparently were not offset (which I didn't think was possibly under the new system).
If $f:X\to Y$ is a continuous map between topological spaces, and $A$ is a subset of $X$, does it follows that $f(A)$ is in bijective correspondence with $A$?
Can someone clear up a bit of confusion I'm having with some basic topology notation?
Well, it's really more than just notation I guess, but I think that's what's really hanging me up.
If $f:X\to Y$ is a continuous map between topological spaces, and $G=\{(x,f(x)):x\in X\}$, if $A\subseteq X$, does it follow that $A\times (f(A))\subset G$?
This guy... he asked one question upon registration, and today he asked like 14 questions about topology. Some makes no sense, and some - I am certain - are duplicates.
@DavidK Not at all, if $f(a)\neq f(b)$ for $a,b\in A$ then both $(a,f(a))$ and $(a,f(b))$ are in $A\times f(A)$ but $(a,f(b))\notin G$.
@AsafKaragila That's what I thought. I'm trying to show that if $A\times U$ and $A\times f(A)$ are disjoint, then $A\times U$ and $G$ are disjoint. Here, $U\subset Y$.
I know that $(A\times f(A))\cap G\neq\varnothing$, but that's as far as I can get.
i refined it to be quite specific and gave an (IMHO good) answer myself which at least demonstrates that it is answerable in the vague sense. question was answered quite specifically in the comments post-closure, which i might have accepted had it been an actual answer.
i've been pondering it for the last month and i really think it is a neat coincidence. too bad nobody had much chance to chime in.
in this case i think my own answer was definite enough for my own edification, but i would have liked the question to remain open to hear other perspectives.
also @Didier's comments were useful but the question was closed.
I don't know. Really. I didn't vote to close your question. And I don't have anything against reopening it. But to understand why it was closed we need to speak to someone who voted to close.
@DavidWheeler Before I forget: @DavidWheeler I see you've requested admission to the Comm. Alg. room. Are you planning to do Comm. Alg. with us? Or was that just an accident?
Someone remind me, please, if $G$ is a group and $\pi:G\to A$ is a homomorphism into an abelian group then $\ker\pi$ is contained in the commutator subgroup, or does it contain it?
I am supposed to find the Maclaurin expantion of
$ \log\left( \frac{1+x}{1-x} \right) $
So I noticed the obvious that $\log (1-x) - \log(1+x)$
Then Maclaurin polynomial of $\log (1+x)$ so
$ \displaystyle P(x) = \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\f...
@JonasTeuwen cool. Sorry, the student came - I have office hours now. So, you've done with your integral? The function appeared not to be so ugly :) in that case Mathematica would help, it does symbolic computations quite nicely and it know lots of weird special functions
I think there is a mistake in the lecture notes. When they define free $R$-module they say "Let $S$ be any set. Then the free $R$-module generated by $S$ is $F(S) := \oplus_{s \in S} R \cdot s$. To me it looks as if $S$ should be a subset of $R$.
@MattN Why should $S$ be a subset of $R$? $S$ is the basis. When you have a basis of a vector space over a field, the basis is not a subset of that field.
I just have to finish this question on the Hurewicz homomorphism, and show that the fundamental group of a covering space is what I want it to be to finish another question.
Fine, I'll just show that commutator thing by hand.
@DylanMoreland Wishful thinking to avoid more bookkeeping.
@tb Well the commutator is trivial and the kernel of this map has only countably many elements. By taking a quotient of the universe by the countable sets $\sigma$-(class)-ideal we yield that my claim is true. Huzzah. :|
@AsafKaragila still waiting for this big sigh of relief that will go around the world when people in this chat room read from you: "I just handed in the algebraic topology homework"
Can you perhaps help me distill something from Spanier? I need to finish the proof that every subgroup of the fundamental group is a fundamental group of a covering space.
I have defined the obvious covering space by homotopy equivalence classes mod the concatenation (+inverse) being in the subgroup. I have the obvious topology by pulling up the topology from the space. I just need to show that last part - the fundamental group of what I have defined is the needed subgroup.
@HenningMakholm Hi, Henning, indeed. We had disjoint times of appearance in this chat room. It's going to improve in the near future, though... How are you doing?
@AsafKaragila well, can't you go a step further and say you go to the universal covering. The subgroup of the fundamental group acts on the universal covering space (by deck transformations -- or by concatenation of paths) and what you have defined is a simply connected space modulo the action of the subgroup of the fundamental group...
@HenningMakholm well, then you're doing way better than me. I don't manage to get any questions on the site answered recently...
i always thought it went like this: universal cover<-->trivial fundamental group, cover(quotient of universal cover)<--->subgroup of $\pi_1(X)$, X<--->full fundamental group
@tb Well, I know that given a loop in the subgroup I can find a path in the covering space whose projection gives me the loop. So now I need to show that if I have a path in the cover it projects to a loop in the subgroup.
I have a serious lapse of understanding the way tagged partitions work. Can someone tell me how to use these animals and find a bound for $S(f,\dot P)$ in case of functions.
@AsafKaragila well, yes, but think about it this way: the universal covering space can be constructed as homotopy classes (with fixed end points) of paths $\gamma: [0,1] \to X$ from x_0 to x, the covering projection being given by $[\gamma] \mapsto [\gamma](1)$. The (subgroup of the) fundamental group is given by homotopy classes of loops $[\ell]$ based at $x_0$ and the action is given by concatenating $[\gamma] \mapsto [\gamma] \ast [\ell]$.
I am supposed to find the Maclaurin expantion of
$ \log\left( \frac{1+x}{1-x} \right) $
So I noticed the obvious that $\log (1-x) - \log(1+x)$
Then Maclaurin polynomial of $\log (1+x)$ so
$ \displaystyle P(x) = \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\f...
