@Krijn Here's something interesting about cohomology which does not hold in homology (well, does not hold literally). $\varphi, \psi$ be $k$ and $l$-cochains, i.e., elements of $C^k(X; R)$ and $C^l(X; R)$. Then define a $(k+l)$-cochain $\varphi \smile \psi$ as follows: $(\varphi \smile \psi)([v_0, \cdots, v_{k+l}]) = \varphi([v_0, \cdots, v_k]) \psi([v_k, \cdots, v_{k+l}])$.
@Krijn You can in fact prove this descends down to a map $H^k \times H^l \to H^{k+l}$ at the cohomology level. Thus, you can define the graded ring $H^*(X) := \bigoplus H^n(X; R)$ with ring structure this map (called the "cup product")
So cohomology is a stronger invariant than homology, because it admits a ring structure.
Interesting, I never noticed that the mouse pointer goes like that when one hovers over the formulas in W|A. Hence naturally assumed you set it as default.
Also, the argument that Hatcher uses about the difference between $X \to X \times X$ and $X \times X \to X$ makes sense, which hints at a difference in structure
The one I have satisfied myself with is the above. "Formal things cannot be multiplied, but ring-valued functions over formal things can (by multipying values in the ring)"
@Krijn So, e.g., RP^2. H_2(RP^2; Z) = 0, H_2(pt; Z) = 0. This doesn't tell you whether RP^2 is homotopy eq to pt or not. But you can tell the difference looking at H^2 itself but with Z/2 coeff. H_2(RP^2; Z/2) = Z/2.
Okay, $\forall \epsilon > 0$ $\exists \delta > 0$ such that for all $x \in \mathbb{R}$ satisfying $0 < |x - 0| < \delta$, $|1/x - L| < \epsilon$ if this limit existed. The negation should be something like $\exists \epsilon > 0$ such that for all $\delta > 0$, there is an $x$ satisfying $0 < |x| < \delta$ which implies that $|1/x - L| < \epsilon$ for all $L \in \mathbb{R}$?
Well, actually, I guess. You can indeed get the other homology with coefficients from homology with $\Bbb Z$ coefficients. But not in a straightforward way.
I'm not seeing the point here. Obviously it's not defined at $0$, so you're basically saying that since it's not defined there, it has no limit by default? @MikeMiller
@Clarinetist: OK, no, you're asking two different things. You can still say "What is the limit $\lim_{x \to 0^+} 1/x$? That makes sense. (The limit does not exist, or it's $\infty$, if you like.) But that's not important, because continuity says "For every $x$ in the domain, epsilon delta blah blah"
$0$ is not in the domain. So limits as things go to zero do not make any difference on continuity.
@MikeMiller So you're basically telling me that rather than proving $\lim_{x \to 0}1/x$ doesn't exist using the negation of $\epsilon$-$\delta$ definition, it makes more sense to prove inequality of the one-sided limits?
Oh, one last thing @BalarkaSen or @MikeMiller, I see how we could use cohomology to distinguish different spaces, what are some other uses of cohomology?
If I can show that a function $f(x)$ is a solution to a differential equation, then why should I have to show that it is the solution to a differential equation with initial value problem $y(o) = -1$?
Eh, it's technical. You use local orientations. If $M$ is a $n$ dim topological manifold, then given a point $x \in M$, local orientation at $x$ can be thought of as a small sphere rotating clockwise or counterclockwise about it. Rigorously, it's the generator of $H_n(M, M - x)$ (isomorphic to $\Bbb Z$ by excision, so precisely 2 choices for orientation). And then you choose local orientation at each point continously.
I always thought (could be wrong) that you're just replacing the tangent bundle by the orientation sheaf or something. But @MikeMiller can feel free to pick on me if this is not right.
Uh, alright. Well. Since the equation (solution) was $y = sinxcosx - cosx$, then am I to simply show that $y(0)$ is indeed $-1$, since it asks to show that it is the solution with initial value $y(0) = -1$?
@Clarinetist Now what you could say is "There is no continuous extension of $1/x$ to a function $f: \Bbb R \to \Bbb R$". This is true, because if there were, then $\lim_{x \to 0} f(x) = f(0)$, but the thing on the left doesn't exist
@Krijn This is in some sense not far from asking "What are the applications of partial derivatives?" The answer is almost "Literally everything is an application of cohomology". Finding obstructions to the existence of sections of a fiber bundle. Classifying vector bundles. Intersection numbers.
@Vrouvrou Let $f(x)=1\forall x$. According to your definition, since $f$ is clearly continuous at $x=0$, take a neighborhood $]-a,a[$ of $0$ and we'd have $\forall x\in]-a,a[,f^{-1}(x)\subset],a,a[$
@Krijn $[X, S^1]$ be the set of all (based) maps $X \to S^1$ upto (based) homotopy. This is naturally a group : $f, g : X \to S^1$ be two maps. Then define $f \times g : X \to X \times X \to S^1 \times S^1 \to S^1$, where the first map is inclusion into diagonal and the third is the usual multiplication coming from top. group structure on $S^1$ (check this is well-defined upto homotopy).
@Vrouvrou Then what does "bounded" mean? There are definitions of "bounded" for metric spaces, and for topological vector spaces, but as far as I know, not for arbitrary topological spaces.
@BalarkaSen: It's probably worth it for you to know that when $G$ is not abelian, there is still a notion of $H^1(X;G)$; it's a set, not a group. (To define this all it takes is care when writing down what a cocycle is and what it means to be a coboundary.) Then one still has bijections $[X,K(G,1)] = H^1(X;G) = \text{Hom}(\pi_1(X),G)$.
@Vrouvrou By continuity we have $\overline{A}\subset f^{-1}(\overline{f(A)})$. Now all you need is that for a bounded set $B\subset F$, the closure $\overline{B}$ is also bounded.
@Vrouvrou It may be that that is already known. But it's not hard. In a metric space, recall what the definition of a bounded set is. It is straightforward to show that the closure of a bounded set is bounded.
@BalarkaSen: I have to go, and this isn't worth thinking about forever. I told you this before. For the "immersion" case (that is, when $df_p$ is injective), the correct assumption is that the image is complemented.
Completely metrizable topological vector space. You are free to assume that the metric is translation-invariant (that is, $d(v,w)=d(v+h,w+h)$
Then you still have a notion of derivative, same way you do for Banach spaces, just using a choice of metric instead - derivative doesn't depend on the choice of metric
Consider the function $f: (0, 1) \to \mathbb{Q}$ defined by
$$f(x) = \begin{cases}
0, & x\text{ irrational} \\
1/q, & x = p/q\text{ in lowest terms.}
\end{cases}$$
The problem is to show that $\lim\limits_{x \to a}f(x) = 0$, using $\delta$-$\epsilon$, for all $a \in (0, 1)$. That is, for all $\ep...
Not the book I was thinking of, but it works fine. I wouldn't learn from it. It is sometimes amusing to check back on it to see the relevant statement for Banach manifolds.
He does not cover Frechet manifolds.
You have a guess for me on the above so I can hit the road?
The answer was "There isn't one, at least without massive changes to the assumptions". See here. Given massive changes to the assumptions, there is one, which is extremely useful, known as the Nash-Moser theorem. See here.
Look back at the proof of the inverse function theorem at some point and see how crucially it uses the norm at various points that wouldn't work if you just had a metric.
the graph of a step function leaps every now and then. Is there a known way to remove those jumps while preserving the rest of the functions rise and rate
This is why people don't usually develop the differential topology of Frechet manifolds: it's much harder to do. The only place I know of where this is done seriously is some notes by Andrew Stacey, and Peter Michor's big book.
Basically between $n$ and $n+1$, your initial function goes up by $n/2$ @TheGreatDuck so I did what you suggested, I added the pieces together $f_1(x)=f_1(E(x))+(x-E(x))x$
Actually the change in each jumps is n. I'm aware of the solution for that particular example. I'be been trying to shoehorn these together. My question is rather if there is a general operator or property that has been well defined somewhere. I.E. I can read something up on it.
The example was just a quick example of a step function with breaks
