The fundamental groupoid is the groupoid whose objects are points in a path connected space X and morphisms are paths from one point to another, modulo path-homotopy.
@BalarkaSen That shouldn't surprise you: that's how the universal cover is usually constructed as a set (points + homotopy classes of loops based at that point), similarly to how the orientation double cover is constructed etc
But yeah, that's precisely how one constructs the universal cover, as path-homotopy classes of paths starting at a specific point etc. I guess I am more used to constructing it by gluing togather U x pi_1 with the monodromy informations.
@BalarkaSen At which point you're not really packaging any information together, you've just given me the original space again and told me to pay attention to paths!
so a function on $R\backslash \{2\}$ can still be a legit function, even tho it would have a gap if you try to plot in on $R$, because every $r\in R$ gets mapped to something?
Today I learned that if you don't take the axiom of choice you can find a model of set theory which has strictly amorphous sets: That is a set that is infinite but is not the disjoint union of any two other infinite sets. https://en.wikipedia.org/wiki/Amorphous_set
anyway, I did it right in my notes but for some reason $\nabla f(x,y) \cdot \vec{P_0P} = 0$ isn't giving me the eq. of a tangent plane that I desire. $P = (1,-2,1)$ so I just do $(6(1) + 2)(x-1) - (2(-2))(y+2) + 1(z-1) = 0$ ?
@AliCaglayan So if you don't let things have choice they can become scary. But there are also some things that are scary because of choice being available to them.
hmm but then it asks for the normal line on the surface at the point specified. Shouldn't the normal line just be $\nabla f$ parametrized? so $f_x\mid_{x=1} = 8 \to x(t) = 1 + 8t$ or something?
So I have seen that $\#_{i \in \Bbb N} S^2$ is just an open ended cylinder, but doesn't that mean $\#_{i \in \Bbb Z} S^2$ is a cylinder. I am having trouble understanding why the first and second constructions should be different
Or am I being stupid and they are both just one ended cylinders?
Indeed, even for just two manifolds you can take connected sum in two different ways (and get two nonhomeomorphic manifolds in that process sometimes).
@AliCaglayan I mean, as appropriate quotient spaces. There's not an actual formal meaning of $\#_{\Bbb N} S^2$. But I think it's obvious what you mean.
In this case it's "Let $D_i \subset S^2$ be a small disc around either the west pole or the east pole, $i = 0$ corresponding to west, $i=1$ corresponding to east. Consider $S^2 \times \Bbb N$, and in the $k$th factor, where $k>0$, delete the interiors of the $D_i$, and for $k=0$ delete only the east factor. Then glue them together in the obvious way."
@AliCaglayan I guess, sure. There is a map $\#_n S^2 \to \#_{n-1} S^2$ given by collapsing the last sphere you glued on. (But one has to be careful here because connected sum isn't functorial or anything.)
Then $\#_{\Bbb N}$ is the colimit of this. $\#_{\Bbb Z}$ involves you collapsing the outermost two spheres in a connected sum of $(2n+1)$ spheres at each stage.
I am not sure why $X$ is assumed to be compact in the Lemma below.
It seems to me that the proof would work fine if $X$ is any metric space (with metric $\varrho$).
Am I correct?
By the way $|g-f|$ is defined to be $\sup\{|g(x)-f(x)|:x\in X\}$, and $\mathscr I=[0,1]$, and $\aleph_0=\ome...
@AliCaglayan Probably not very exciting at a glance. Computing fundamental group of complement of a bunch of elliptic curves inside a complex abelian surface.
@MikeMiller Dunno. The person who gave it promised to elaborate once I actually gave it some thoughts; and I've been not thinking on it as long as I can remember. I guess I can just ask him.
It's probably not known or something.
I suppose it's more than just "well, computing fundamental group of random quasiprojective varieties is hard".
@saturatedexpo If you ever feel keen to try again, try to solve some of the exercises on Project Euler. They dig more at the computer science (i.e. mathy) side of programming and don't require you to be good at window programming.
The whole point of defining the Chern character is that, taken all together, it respects direct sum and tensor products of bundles. There's no other reason to write it down.
@Fargle my first experience was in Access (was some database stuff). i managed to program a game with it as a kid lol. then some pascal. and because of conveniency c++ at the end. But i wouldn't call myself experienced in any of those, just did what i wanted there ;)
if g isnt injective, f can' be. which is a contradiction to the givens. you mean like that?
@Alessandro let g be not injective. Then there exist $x,y\in M\backslash \{a\}$ with $x\not=y$ and f(x)=f(y). Because f is bijektive and therefore injective, this is a contradiction.
let g be not surjective. then there exist one $n\in N\backslash \{f(a)\}$ for that there is no $x\in M\backslash \{a\}$ with f(x)=n. This is clearly a condradiction but i fail to convince myself of it.
my thoughts are: f is bijective. therefore if $x\not=a$ then $f(x)\in N\backslash \{f(a)\}$
I am not sure why $X$ is assumed to be compact in the Lemma below.
It seems to me that the proof would work fine if $X$ is any metric space (with metric $\varrho$).
Am I correct?
By the way $|g-f|$ is defined to be $\sup\{|g(x)-f(x)|:x\in X\}$, and $\mathscr I=[0,1]$, and $\aleph_0=\ome...
