@TedShifrin Ah, that's not the argument I was making. But yes, every map must necessarily kill $\Bbb Z_2$-cohomology
(I was recalling that $H^1(X;\Bbb Z)$ is in bijection with $[X,S^1]$)
The question bounds helped me the most on was something like this... (warning: it was not this, and the actual integral I'm writing down probably doesn't evaluate to any of the given answers)
@TedShifrin It doesn't... all I was saying is that that's how I knew the answer for $n=1$
anyway, for that example... the numerator is bounded above by 1, and the denominator bounded below by 1; so that the integral must be at most $\pi/2$... but it's smaller than that, since it actually takes values smaller than 1. it's positive, because the integrand is nonnegative in that interval (and sometimes positive). so if it's any of those four, it must be b)
@TedShifrin That wasn't the actual problem, but there was a hideous problem like that about an integral of a rational function in trig functions. I could have spent 5 minutes doing the substitutions, or I could get it much faster with basic estimates like that
@TedShifrin Everything good here too, doubled down with complex analysis homework though. The prof has given 10 contour integrals as a homework, such a long and ardous task it is.
Hahahaha, I recently forgot to add the absolute sign on some of my inequalities in the complex analysis exam. (Typo rather than genuine mistake.) The prof was furious. :-P
Hi off topic: On page 175 of Munrke's topology, he proves that the distance from a point $x$ to a set $A$ given by $\inf{d(x,a):a\in A}$ is a continuous function on a metric space $(X,d)$ containing $A$. However, his proof of continuity seems to use the standard metric on the real numbers.
Wouldn't this only prove that the distance function is continuous with respect to the standard metric?
Let $K = (k_1,k_2,k_3,...k_N)$ be a vector in $\mathbb{R}^N$, consider the region $S_K$ consisting of all vectors $L = (l_1,l_2,l_3,...l_N)$ such that, $|l_i| \le |k_i| \forall i \in \{1,2,3,...N\}$. My question is, given $K$, is there a name for the region $S_K$, used in standard literature? If...
I should try and remember that second map; it's a nice argument.
@TedShifrin The same question as applied to $\Bbb{CP}^n$ is interesting. On the one hand, I would expect there aren't any such nontrivial maps (for $\Bbb{CP}$ or $\Bbb{RP}$). On the other hand... homotopy groups of spheres.
@TedShifrin You're right; first lift to $S^3 \rightarrow \Bbb{RP}^2$; this map would still be null-homotopic; and since $S^3$ is simply connected one can lift the homotopy to a homotopy of the Hopf map; so that this homotopes the Hopf map to one whose image is two points, thus one point, thus the null-homotopy lifts to a null-homotopy
@TedShifrin Oh, I thought you were assigning a new problem. No, the argument suggested worked, so we now have a nontrivial map from P^3 to P^2. I'm satisfied :)
@TedShifrin I would be very surprised if that worked, since the 1-dimensional case is probably too "easy".
@AndrewG The hopf map is the attaching map of the $4$-cell of $\Bbb{CP}^2$ onto the 2-skeleton, i.e. $\Bbb{CP}^1$, i.e. $S^2$. So it's a map $S^3 \rightarrow S^2$. It's not null-homotopic because if it were, $\Bbb{CP}^2$ would be homotopy equivalent to $S^2 \vee S^3$. But it's not, which can be checked by, say, calculating their cohomology groups.
@AndrewG Lemme know when you've got a handle on homotopic maps/homotopy equivalent spaces, and I can run through the details of the rest of what I said :)
@TedShifrin You should throw a copy of Bredon at him to sate him and keep him away from the really good stuff... ;)
@TedShifrin Hmm, I think my claim was incorrect. A null-homotopy of the Hopf map doesn't seem to give me a homotopy equivalence $\Bbb{CP}^2 \simeq S^4 \vee S^2$. Certainly I have a map $\Bbb{CP}^2 \rightarrow S^4 \vee S^2$, but I can't seem to draw an arrow in the other direction.
But it must be true, as Hatcher quotes (a more general fact) later.
OK, nvm, this is proposition 0.18.
@AndrewG So by Prop 0.18 in Hatcher (once you're there!) if the Hopf map is null-homotopic, then $\Bbb{CP}^2 \simeq S^4 \vee S^2$. But, by some super-fancy machinery (the cohomology ring), this isn't possible. So the Hopf map is a nontrivial map $S^3 \rightarrow S^2$.
@Mike in example 0.9, we can collapse each disk to a point without changing the homotopy type of the overall space (or rather, I guess, of the CW complex?) because disks are, in fact, contractible, that is, homotopy equivalent to points? But I couldn't do this if we were using circles instead, etc. (which is why we still have that loop hanging out in Z and W)
@AndrewG The CW-complex structure doesn't really matter here - the space you get from collapsing a disc to a point will always be homotopy equivalent to the original space. (This is the theorem he proves later, but is stated above Ex 0.7)
@AndrewG Well, most things you naturally think of are CW complexes. Every (compact? don't remember) smooth manifold is a CW complex, and every compact topological manifold is homotopy equivalent to one.
There's also some really really really nice things about CW complexes that you'll learn if you get to Ch.4 down the road :)
Can the Euler characteristic of two homotopically non-equivalent spaces be the same? Hatcher says the converse is false; if they're homotopically equivalent, they have the same Euler characteristic. So if I want to distinguish two spaces (/CW complexes), and they have different Euler characteristics, that's enough...but if they have the same characteristic, can I conclude they're homotopically equiv.?
I'm guessing the answer is no or homotopy theory would end pretty quickly lol
$S^n$ is not homotopy equivalent to $S^m$ for $m \neq n$, but their Euler characteristics are either $2$ or $0$ depending on whether $m$ is even or odd.
(That they're not homotopically equivalent follows from calculating their homology - that happens in chapter 2.)
You'll never find a perfect invariant (i.e., you look at it and it can tell apart two non-homotopically equivalent spaces) that's not completely useless.
Like, "homotopy type of $X$" is a complete invariant. But, like, come on.
@Gustavo, like, a single way to measure something's (insert property here). If there were such a thing for homotopy type, there wouldn't be much of a theory there, I guess.
To clarify, I don't mean that a definition would be really complicated. I mean that I don't know a definition, and even if there was one, I'm just being colloquial.
@MikeMiller null-homotopic means homotopic to the identity map, right? So for Prop 0.18, f is my attaching map and g is the identity, $A$ is, what, $\mathbb{C}P^1 \approx S^2$ and $X_0$ is what?
@AndrewG $X_0 \cong \Bbb{CP}^1$; $A$ is $\partial D^4 = S^3$; $f$ is the Hopf map (your attaching map), and $g$ is the map sending $S^3$ to the 0-cell.
@AndrewG It took me two weeks to get through that chapter and do the exercises, and that was the time alotted in my topology course. I wouldn't worry about getting through it in one night lol
If two Cauchy sequences have the same limit, then they are equivalent. Am I right?
@robjohn
My book defines equivalence this way: Two Cauchy sequences $\{x_k\}$ and $\{y_k\}$ are said to be equivalent if for all $\epsilon>0$, there is $k(\epsilon)$ such that for all $j\ge k(\epsilon)$ we have $d(x_j,y_j)<\epsilon$.
@Sush yes. Suppose that $\lim\limits_{k\to\infty}x_k=a$ and $\lim\limits_{k\to\infty}y_k=a$. Then given $\epsilon\gt0$, there is a $k(\epsilon)$ so that for all $k\ge k(\epsilon)$, $|x_k-a|\le\epsilon/2$ and $|y_k-a|\le\epsilon/2$. Then $|x_k-y_k|\le|x_k-a|+|y_k-a|\le\epsilon$.
@Vrouvrou You mean that X is the whole space and it can be divided into two disjoint closed sets? If this is the case, then $X_{1,2}$ are both closed and open. So $X_i$ is equal to its interior.
So you can write $X=\overline{X_1}\cup\overline{X_2} =\operatorname{Int} X_1 \cup \operatorname{Int} X_2 = X_1\cup X_2. In each case it is unionn of two disjoint sets.
Given that $v = DC = \lambda EB$, prove that $\lambda v = CB + ED$.
Whatever I try seems to end up with $CB + ED = (\frac {1}{\lambda} - 1)v$, ie:
$$CB + ED = CD + DE + EB + ED = EB - DC = EB - v$$ so $$ CB + ED = \frac {1}{\lambda}v - v = (\frac {1}{\lambda} - 1)v $$
@Vrouvrou $X_1$ is open since the complement $X\setminus X_1=X_2$ is closed.
I see I have forgotten to close math mode, when I wrote: $X=\overline{X_1}\cup\overline{X_2} =\operatorname{Int} X_1 \cup \operatorname{Int} X_2 = X_1\cup X_2$.
Given that $v = DC = \lambda EB$, prove that $\lambda v = CB + ED$.
Whatever I try seems to end up with $CB + ED = (\frac {1}{\lambda} - 1)v$, ie:
$$CB + ED = CD + DE + EB + ED = EB - DC = EB - v$$ so $$ CB + ED = \frac {1}{\lambda}v - v = (\frac {1}{\lambda} - 1)v $$
