@iwriteonbananas for the cone thingy you have posted, I guess all you need is the homology long exact sequence for triples.
you need to exploit the fact that $H_n f$ is an isomorphism by saying "kernel of this map and image of this map is zero" instead of looking at $H_n(pt)$s.
What does $[\Bbb Z/(n)][x]$ refer @Balarka? Polynomials taking values in $\Bbb Z_p$ even when $p$ is not prime, or should I assume we are talking about solely when this is a field(p prime)
Why does Munkres say that a topology on a set $X$ must be closed under any union of the subsets of $X$ in it, but then say that it must be closed under finite intersections only?
the topology on $\Bbb R$ is no longer a topology. These definitions are inspired by metric spaces. It will be hard to understand the motivation for anything if you don't know much about metric spaces.
I actually did do a bit of Simmons earlier, because finding a pdf of Munkres without weird lines through the pages was hard. I got as far as Schroeder-Bernstein (and I typed it up in LaTeX for fun).
@MikeMiller care to tell me why the map induced by the inclusion $S_1 \sqcup S^1 \to S^1 \times I$ is not trivial? it should be the map $R \oplus R \to R, (a,b) \mapsto a+b$
I looked at a few topology books that day (because, apart from Munkres, there are few "definitive" recommendations). I'd been wanting to understand Schroeder-Bernstein for a while, so I did the proof and forgot about the book.
i guess the map $i_*$ is the same as the composition $$H_1(S^1 \times I) \to H_1(S^1 \times \{1/2\}) \to H_1( S^1 \times \{0\} \cup S^1 \times \{1\})$$ where the 1st map is an isomorphism and the 2nd map is $(id, id)$
Each map $S^1 \times \{pt\} \to S \times I$ is a homotopy equivalence, hence induces an isomorphism on homology. The map from the disjoint union is the direct sum of those maps.
For some reason, "one-to-one" and "onto" are seen as more "intuitive" terms for beginners, and so they've gained popularity in intro-to-proofs courses and various texts, and they've caught on a bit from there.
For everyone who's interested in the Sage CAS: the Sage Proposal at Area 51 needs some more questions with 10 up-votes before moving to the next phase. (If that doesn't happen before june 1st, the proposal will be closed...)
First chapter of Simmons over. Metric spaces appear now. That's why I didn't know.
user147690
8:55 AM
Many of my old highschool friends are getting married and/or having/have-had a kid. This seems very strange to me. Pretty much people either went to uni or the above happened. Is this the same in America??
I just checked and see that Chris's sis has not been here for 3 days and r9m has not been here for 5 days. Did something happen? I was gone from last Friday.
@SohamChowdhury metric closure of $\varprojlim \Bbb Z/p^i\Bbb Z$ (topology being the subspace topology from the product topology on $\prod \Bbb Z/p^i\Bbb Z$) is homeomorphic to $\Bbb Z$ with the $p$-adic metric.
@BalarkaSen Somewhat alright. Still having trouble getting my head straight on all of these things, PIDS, UFDs, EDs, rings, principal/maximal/prime ideals
@Soham $\mathcal{C}([0, 1])$ be the set of all continuous functions from $[0, 1]$ to $\Bbb R$. Prove that this a metric space equipped with $d(f, g) = ||f - g||$, where $||f|| = \sup_{x \in [0, 1]}\{|f(x)|\} < \infty$ (prove that it even make sense).
1. "If all are positive . . . " 2. "If one is positive, wlog, let it be x . . . " 3. If two are positive, negate them and see 2. 4. If all are negative, negate them and see 1.
You have to prove that $\sup_{x \in [0, 1]}\{|f-g|\} \leq \sup_{x \in [0, 1]}\{|f - h|\} + \sup_{x \in [0, 1]}\{|f - h|\}$. To do that you have to prove a simple lemma about sup.
@BalarkaSen For a space $P = \{p\}$, there must exist an open ball centered on p with radius > 0 for the space to be open, so there must be points in $P$ such that $d(x,p) > 0$, but there is no such x.
