@Kasmir sadly I know nothing but run the question by anyway and let's see?
@Perturbative yeah, the idea is this, if you take $S^1$ and you identify antipodal points, it's the same thing as taking the semicircle and identifying the endpoints
Ie, an operation which, when applied to a function and then applied again to the resultant function, should give the derivative of the original function.
@alxchen because they are inherent to the extension (i.e. independent of the basis you choose)
and they tell you, to a certain extent, the structure of the extension
and a more high-level answer is that global and local class field theory classifies field extensions using things including the norm map and it is the most celebrated number theoretic result in the 20th century
an example is that [C:R]=2 and so is 2 the index of the image of the norm map N:C*->R* in R*
@Kasmir you can map your matrices to GL_2(K)^2 by taking only the first upper left 2x2 block for the first term and the lower right 2x2 block for the second term
@MatheinBoulomenos isn't a principal localization a non-surjective finite type epimorphism? Say $\mathbb Z\to \mathbb Z_2\cong \frac{\mathbb Z[x]}{(2x-1)}$?
Suppose $Q\in M_{3\times3}(\Bbb R)$ is a matrix of rank 2. Let $T:M_{3\times3}(\Bbb R)\to M_{3\times3}(\Bbb R)$ be the linear transformation defined by $T(P)=QP$. The the rank of $T$ is .....
How many pairs of a,b,c exists such that m is maximum
m = ((((5 moda) modb) modc) mod5)
Given: The value of a,b,c lies between 1 and 7 ( both inclusive ).
@Astyx But we are also taking mod5 in the end. If we have a,b,c >5, I think answer would be zero. For a,b,c >5 , (((5 moda) modb) modc) will be 5 and ( (((5 moda) modb) modc) mod5) will be zero as 5 mod5 =0
@Astyx I have following in mind. Please correct me If I am wrong. To get max_remainder (less than the number itself) we should divide the number n by n/2+1. For .e.g for 50 we get max remainder when we divide it by 26. I think if we have a,b,c =n/2+1 (i.e 3 ) we can get max value.
@TobiasKildetoft Well I guess in terms of how much structure we have we have a progression like $\text{Groups} \to \text{Rings} \to \text{Modules} \to \text{Algebras}$ at least that's how I'm thinking of it in my head, so are there commonly used algebraic objects that have more structure than algebra's is what I'm asking
This paper of Atiyah's was one of the first challenging papers I tackled to lecture on as a first-year graduate student. I still don't understand all of it, but it has many beautiful ideas in it. I've mentioned it here several times.
@TedShifrin That paper seems to have stuff from a whole lot of intersecting fields in it, like algebraic geometry/topology, differential geometry, complex analysis etc
Isn't the part of the theorem that says "if there is a relative homeomorphism" vacuously true though since $e$ is defined to be a homeomorphic copy of $D^n \setminus S^{n-1}$, so there always exists a homeomorphism $Y_f \cong e \cup Y$?
Ooooh okay, @MikeMiller, so a continuous map $g : (X, A) \to (Y, B)$ is a relative homeomorphism if $g|_{(X \setminus A)} : X \setminus A \to Y \setminus B$ is a homeomorphism. And $g : (X, A) \to (Y, B)$ being continuous means that $g : X \to Y$ is continuous and $g(A) \subseteq B$
Ok, so the issue is there's no reason a homeomorphism from the open disc extends to a continuous map from the whole disc.
I can come up with examples where that's not true without too much difficulty.
I guess it's a little rude to keep them to myself. Take the closed topologist's sine curve in $\Bbb R^2$. Its complement has two components: the interior and exterior. The interior $D$ is simply connected, and hence there is a homeomorphism $\Bbb R^2 \to D$. But if $D^2 \to \Bbb R^2$ is an extension of this homeomorphism to the closed disc, then $f(S^1) = \overline D - D$, which is the closed topologist's sine curve.
But there is no surjective map from $S^1$ to the closed topologist's sine curve because the latter is not locally connected.
here the closed topologist's sine curve means what wikipedia calls that (the closure of the graph of $\sin(1/x)$ above $(0,1)$) and then close up the loop by taking the right endpoint and making a circle back to the 'bad side' of this.
Okay I get what you're saying @MikeMiller. Also I think I made a pretty big error earlier since there's no reason to expect $e \cong D^n \setminus S^{n-1}$ to imply a relative homeomorphism $\Phi : (D^n, S^{n-1}) \to (e \cup Y, Y)$ since a homeomorphism $f$ that sends $D^n \setminus S^{n-1}$ to $e$ will not satisfy $f(S^{n-1}) \subseteq Y$ (because well $f$ isn't even defined there), and so even though $f : D^n \setminus S^{n-1} \to e \cup Y \setminus Y = e$ is a homeomorphism,
it isn't a relative homeomorphism $f : (D^n, S^{n-1}) \to (e \cup Y, Y)$
It is easy to prove that if $f: (D^n, S^{n-1}) \to e \cup Y$ sends the interior of $D^n$ homeomorphically to $e$ then it sends $S^{n-1}$ into $Y$. The issue is getting a continuous extension.
Ohh well that follows basically through set theory, i.e. we have $f : D^n \to e \cup Y$ and if it restricts to a bijection from $D^n \setminus S^{n-1}$ onto $e$ then we must $f[S^{n-1}] \subseteq Y$, and as you said even if we send the interior of $D^n$ homeomorphically to $e$ we don't know if we'll get a continuous extension to all of $D^n$
Also I guess intuitively that must force $e$ and $Y$ to be really 'close' to each other in $Z$ to preserve continuity of a function $g : D^n \to e \cup Y$
you could just choose the map which is a bijection on the interior to do whatever you want on the boundary
Here is how I would argue, I imagine it can be made cleaner. Suppose $z$ is in the interior of $e$ with $f(x_0) = z$, and let $x_n \to x$ be a sequence of elements of the interior of $D^n$ converging to a boundary point. Then because $f$ is a homeomorphism, it is in particular an open map, which implies that a neighborhood of $x_0$ maps to a neighborhood of $z$.
Now if $f(x) = z$ then all $x_n$ for $n$ large lie in this open neighborhood, which means it must have been something in the image of that small neighborhood of $x_0$. Because the map is a bijection on the interior of the disc, this is a contradiction.
That's a nice proof! @MikeMiller But doesn't that show that we can never get a continuous extension to all of $D^n$ if we have a homeomorphism $f : D^n \setminus S^{n-1} \to e$?
It shows that any continuous extension sends $S^{n-1}$ to $\overline e \setminus e$.
Certainly it's not true that there is never an extension. Take $e$ to be the open unit disc in $\Bbb R^2$ and $Y$ the unit circle, and your homeomorphism the identity.
How many pairs of a,b,c exists such that m is maximum
m = ((((5 moda) modb) modc) mod5)
Given: The value of a,b,c lies between 1 and 7 ( both inclusive ).
