The 1-dimensional "homology" group of the circle becomes impressive, (though I think the 1-dimensional "homology" group of $\Bbb R^2\setminus\{0\}$ stays at $\Bbb Z$)
Conjecture: Call the cycles-modulo-boundary groups of dimension $n$ created when we restrict attention to injective simplices $G_n(X)$. Then, for large enough $m$, we have $H_n(X)=G_n(X\times\Bbb R^m)$.
for the u substitution of sin(x)*cos(x)/(1+sin^2(x)) dx, if taking 1+sin^2(x) as u, then du = 2sin(x)*cos(x) dx...how would I substitute that in when I need to substitute du for half of that
My own rant: I have at least four different ways I could split up this computation, and all of them feel pretty terrible. So I need to figure out if one of them is somehow minimally terrible.
"Every countable, metrizable space without isolated points is homeomorphic to the rational numbers $\Bbb Q$", theorem 2 in this paper, I think I'm missing something, is $\mathbb Q\cap[0,1]$ homeomorphic to $\Bbb Q$?
Comment ça va? @user1952009 many thanks to you and this site, are very generous. If in the future you want to explain more mathematics about $\zeta(s)$ you are welcome if tell me here in this Chat, or in a comment or answer. My best wishes to this 2017.
I think it's probably equivalent to singular homology.
First prove that X x I has the same injective homology as X, by hand. Once you have that, take your homotopy through injective maps to be as a map X x I -> Y x I, doing the usual chain homotopy argument to see that the induced maps of the endpoints of X are the same. The invert the map(s) Y -> Y x I on homology.
At least the canonical map from simplicial homology to injective homology is injective, and the canonical map from injective homology to singular homology is surjective.
Hey guys, if I we want to write a simple formula like $\varphi(x,y)$: term=term (simply stating some term including variables x and y is equal to itself), is there any way to avoid writing the same term twice? I mean like the way we use a symbol for a function
while the injective homology of $S^1$ is the same as the set of injective cycles (since there are no 2-simplices so nothing's a boundary) and so would be a really huge group.
No. If X is a finite-dimensional CW complex, there is no injective map of a higher-dimensional simplex into it. Your argument shows that there's always a plethora of top-dimensional injective cycles.
I still want to teach the topology of noncompact surfaces, and I think Kevin Carlson and I could teach a class called "infinity categories for normal people"
it's the right place to think of a lot of ideas in homotopy theory - the category of spaces is an infinity category, 2-morphisms are homotopies, 3-morphisms are uomotopies between homotopically
A lot of the time you don't want to say "blah up to homotopy", you want to say "blah up to coherent homotopy". For instance, you might have a space with a product map G x G -> G.
You'd like that map to be homotopic to the map you get when you swap the factors
So choose such a homotopy. That will give you in the end a bunch of maps G x G x G -> G
Those should all be homotopic, pick the jomotopies. Etc
@Semiclassical I like the aspect that (homotopy type of) any topological space is an infinity-category. objects are points, morphisms are paths between points, 2-morphisms are homotopies of paths, 3-morphisms are homotopies of path-homotopies etc
if you truncate that till level n (so stop at n-morphisms) you get the fundamental n-groupoid. for n = 1 that's like an elaborate thing which keeps track of a bunch of fundamental groups pi_1(X, pt) where pt varies over X.
well, yeah. but technically speaking isomorphisms pi_1(X, x1) \cong pi_1(X, x2) depend on a choice of a path from x1 to x2. those are precisely the morphisms of the fundamental groups in the fundamental groupoid and...
yeah forget this. i am leaving the realm of my own understanding
Ok there are some key difference in Wheels and ordinary rationals: 1. The only element that is our additive identity is now 0/1. In usual rationals, 0/y for any y is our identity due to 0 being a multiplicative absorber. 2. Reciprocal operation becomes distributive, which is not allowed in general in usual rationals 3. The new element, 0/0 becomes an additive absorber
Distributive law then result in the following interesting properties: 1. Zero terms are neutral wrt addition and division by reciprocals wrt the nonzero element y and z, thus they kinda function like identities for these elements 2. All reciprocals 0/y where $y\neq 1$ is now a zero term. In usual rationals, they are all lumped into the additive identity,
(I still yet to figure out (x+0y)z=xz+0y however...)
This bubbling out the zero terms behaviour is something I found very nice about wheels compared to most zero term algebra I have seen so far
Note even in our definitions, 0/c is a zero term, thus it is in general a nonzero element, so Wheel is actually an example of a zero term algebra
i should have said: (2,4) ~ (3,6) because (3*2,3*4) = (6,12) = (2*3,2*6)
The first step in constructing wheels is the change the equivalence relation from $sr=sr'$ where $r,r'$ are ordered pairs denoting the numerator and denominator of a rational number, into a more general one: $sr=s'r'$.
meanwhile the zero reciprocals n/0 and 0/n are distinct from each other as no s, s' can be found to relate an arbitrary n/0=m/0
whereas in ordinary rationals, all reciprocals of the form 0/n are equivalent since 0x=0 for all x. Therefore Wheels extend commutative rings with reciprocal elements such that they are all involutive
@AkivaWeinberger the sources I saw use $x_1,...,x_n$ for the variables, and in that order one has to read the solution set. Why do we notate the solution set then as a set of vectors?
