Hmm. Now that I think of it I can probably guess what may be a right kind of argument.
The critical points are the troublesome point. If $x$ is some critical point of $f$, if I homotope $f$ itself to some $g$ so that $g$ is a submersion near $x$ (I vaguely remember one could do it, but I am not sure). Then since $f$ and $g$ are homotopic, preimage cardinality of regular values of both $f$ and $g$ in an nbhd $U$ of $x$ are the same.
In particular this seems to deal with critical values $x$ so that any nbhd $U$ of $x$ is disconnected by the critical set $S$, where my previous argument can never tell if two points however close to $x$ lying on different path components have the same preimage cardinality or not.
By critical point I meant critical value. Sorry about that.
when I say preimage cardinality, I really mean mod 2 cardinality by the way.
which is easy to prove. suppose $H : M \times I \to N$ is some homotopy between $f$ and $g$, and assume $x$ is a regular value of both $f$ and $g$. If $H$ moreover is a submersion near $x$, then $H^{-1}(x)$ is a cobordism (since preimage is a manifold) between $f^{-1}(x)$ and $g^{-1}(x)$. So it has to be a bunch of circles and a bunch of intervals. It's easy to see from this that $\#f^{-1}(x) = \#g^{-1}(x)$ mod 2.
Now, why should $H$ be a submersion in the first place?
Well, it need not be.
So take some nbhd $U$ of $x$ where both $f$ and $g$ are local submersions. Choose some regular value $p$ of $H$ in $U$.
$\#f^{-1}(x) = \#f^{-1}(p) = \#g^{-1}(p) = \#g^{-1}(x)$, the middle equality is mod 2.
If $f : M \to N$ is some submersion so that $f(y) = x$, then submersion theorem says that I can choose charts $U$ around $y$ and $V$ around $x$ and coordinates $(y^1, \cdots, y^n)$ and $(x^1, \cdots, x^m)$ so that $f$ is given by killing off the last $n - m$ coordinates.
I mean it appropriate local coordinates $f, g$ are given by killing off the last couple coordinates, i.e. it's a map $\Bbb R^n \to \Bbb R^m$ with $(x_1, \cdots, x_n) \mapsto (x_1, \cdots, x_m)$.
OK, I'm too frustrated to continue. Yes, the map is a submersion at every point of $f^{-1}(x)$. (You never once said that it's true at every point in the preimage; you should have.) Why is it true that there's an open nbhd $U$ of $x$ such that $f$ is a submersion on $f^{-1}(U)$?
Right, oops, I see now what I missed. Apologies. OK, if $f : M \to N$ is a map from a compact manifold $M$ then $f^{-1}(x)$, being a 0-manifold, is a finite discrete set. Since for every $y \in f^{-1}(x)$, there are nbhds $U_y$ of $y$ and $V_y$ of $x$ with $f|_{U_y}$ being submersions, I can take $A = \bigcap V_y$, which is open and $f : f^{-1}(A) \to A$ becomes a submersion, I think.
For noncompact domain this becomes a bit more messy, I suppose.
Because that's how we constructed $A = \bigcap V_y$ (which I suppose what you're calling $U$, I think)? Preimage $f^{-1}(a)$ for every $a \in A \subset V_y$ lies in the union of nbhds $U_y$ of $y \in f^{-1}(x)$. All of this works out because there are finitely many points in the preimage, hence $A$ is open, but I guess that is not the point you raise.
If you're still unconvinced, please let me know. I'll think about this carefully tomorrow morning knowing I am thinking something wrong right now due to brain being half-asleep.
Consider the circle as $\Bbb R/\Bbb Z$. Consider the obvious map from $(-\varepsilon,1+\varepsilon)$. Then at $[\varepsilon] \in S^1$, what you want to do does not work.
At a glance, I would note that $(-\epsilon, 1 + \epsilon)$ is not compact, which was my hypothesis. But I really need preimage of $[\epsilon]$ to be finite, which this seems to satisfy. I believe this means I should go to sleep. You seem to be right.
actually whoops we can't use H as the group identifier for $D_4$ !
call it $K = \{R_0, R_{180}\}$ instead, sorry
So, $G/K = \{K, R_{90}K, HK, DK\}$
(and even though it's kinda obvious I did manually check that $K$ is normal to $D_4 = G$)
I should probably just say $D_4 / K$ really
anyway, I get that the elements that are in the cosets "collapse"/compress to the identifier of the coset e.g. $HK = \{H, V\} = VK$ (where the left/right cosets are identical for all elements of $K$)
Or, that $H \sim V$ right?
(sorry I am learning this so I am not able to concretely / efficiently articulate it)
You see that a map like $x\mapsto xK$ doesn't really have anything to do with groups, if by $xK$ you just mean the equivalence class that $x$ is in, under $\sim$.
So we can always define this map (usually called a 'canonical projection')
Maybe some users who answer many questions on odes and pdes could say what they thing about usefulness of a new tag synonym for initial-value-problems which was suggested in this post.
@BenjaminR For the benefit of other users I will add link to that post. While this tag does not exists, I usually put function-composition in such questions. (Although I am aware that it is not entirely correct.)
Weird. Adjoint always makes me think of $(Ax, y) = (x, A^*x)$, and that's generally the context in which I see the word "adjoint" being used (adjoint maps, adjoint functors).
@BalarkaSen Yeah, that is also the way to term is usually used now
I should get some writing done. I had a nice idea over the weekend that allows me to extend some formulas for dimensions of Hom-spaces to hold for all modules, rather than just those with good filtrations.
By adjoint I mean the matrix that is formed by taking the transpose of the cofactor matrix. By the cofactor matrix I mean the matrix formed by writing the cofactors of each element.@TobiasKildetoft
There's nothing special about 3. If you have a square matrix in general then the det represents volume of the parallelpiped spanned by the column vectors.
Can you describe what cofactor mean using this interpretation?
Hmm so the cofactor of an element in a square matrix is basically the element times the determinant of the matrix formed by deleting the row and the column containing the element . So it would just mean a scalar multiple of the volume of the structure spanned by the vectors in the matrix formed after deleting the rows and coloumns
@MaryStar This is a general property of polynomial rings. Any map defined on the variables and which is the identity on the base ring uniquely determines a homomorphism.
Ah ok... I got it!! Thank you!! :-) After that I want to show that if $I$ is an integral domain then each automorphism of $I[x]$ that is trivial in $I$ is of the above form.
We have that since the mapping $x\rightarrow ax+b$ is the unique automorphism that is identity in the commutative ring $R$ and since the integral domain is a commutative ring the automorphism must be of the form $x\rightarrow ax+b$ because of uniqueness.
@MaryStar No, that is not the uniqueness you have shown
You have shown that the form describes a unique homomorphism and that it is an automorphism. You have not shown that no other automorphisms are possible
We want to show that the automorphism of $I[x]$ is of the form $x\rightarrow f(x)=ax+b$, right? So, we have to show that $T:I[x]\rightarrow I[x]$ with $p(x)\mapsto (p\circ f)(x)$ is an automorphism, or not? But we have to show that **each** automorphism is of that form... @TobiasKildetoft
@TobiasKildetoft Well, the context is, I have a polynomial of degree $d$, $f:\mathbb{C}\rightarrow \mathbb{C}$ and I'm calculating the branching index with the help of the Hurwitz formula. I get that it's $(d-1)$, so my text goes on to say that a given equation $f(x)=y$ will have multiple "roots" for $(d-1)$ values of $y$.
@MaryStar please spend a bit more time thinking about this. You already know what the answer is supposed to be, so you just need to figure out why it cannot be something else.
Suppose that the automorphism is of the form $x\rightarrow p(x)$, where $p(x)$ is a polynomial of degree $i>1$. Let $\phi:q(x)\mapsto q(p(x))$ be the mapping. Then $\phi$ must be onto: For each $y$ we must find a $z$ such that $\phi (z(x))=y(x)\Rightarrow z(p(x))=y(x)$. Set $y(x)=x$. Then $z(p(x))=x$. Since the degree of $p$ is $>1$, that cannot be true. Is this correct? @TobiasKildetoft
If $k\ge0$, there is one solution. If $-1/e<k<0$, there are two solutions. If $k=-1/e$, there is one solution, $x=-1$. If $k<-1/e$, there are no solutions.
What are some numerical methods not involving differentiation that approximates the solution of an equation that cannot be solved analytically? e.g. $x^2 \sin (x) + e^x = 0$
Let $G = \mathbb{Z}_8 \oplus \mathbb{Z}_4$, and $H \lhd G = \langle(2,2)\rangle = \{\,(0,0)\ ,\ (2,2)\ ,\ (4,0)\ ,\ (6,2)\,\}$ So, $\mid G \mid$ $= 32$ and $\mid H \mid$ $= 4$, and $\therefore$ the order of the quotient group $G$ / $H \ =\ 8$
Then the orders of the elements of the quotient group $G$ / $H$:
A good response would take more care in explaning the notion of projective space, but here's a brief one. Your equation, in projective space, is $xw^2 = z^3$; the points you have so far are those with $x=1$. The points at infinity are those with $x=0$.
So what you gather is that the points at infinity are those (rather, the one) of the form $[0:w:0]$.
I take a branching point to be a point of a Riemann Surface such that in a local normal form (ie, with the "right" coordinates) is the image of a map $z\mapsto z^n$
(maybe I'm mixing Riemann Surfaces here unnecessarily)
@Sebgr I see. Unfortunately I am not familiar with how to do this. Also note that $x^2 - y^3$ is not smooth at the origin, so you cannot have local coordinates there - so I suppose you're looking at points away from origin.
I really stayed awake waiting for some validation that I had understood the quotient group of $\langle(2,2)\rangle$ in $Z_8 \oplus Z_4$ but I guess people thought maybe I was trolling, or it was too inane to answer?
I think I've told you before the only $K(G,n)$s I actually "know how to write down" (eg, that are some explicit space that doesn't involve an infinite, non-constructive process) have $n=1$ or are $K(\Bbb Z,2)$ or $K(\Bbb Z/2,3)$.
let $\text{TOP}(n)$ be the topological group of germs of homeomorphisms $(\Bbb R^n,0) \to (\Bbb R^n,0)$. let $\text{PL}(n)$ be defined similarly, but for PL homeomorphisms. there is an obvious inclusion map here, and an obvious way of sending each of these to $(n+1)$ [that commutes with inclusion], so I can define the limit space $\text{TOP}/\text{PL} = \lim \text{TOP}(n)/\text{PL}(n)$
it's a theorem of Kirby and Siebenmann that $\text{TOP}/\text{PL}$ is a $K(\Bbb Z/2,3)$. this, plus their surgery theory, implies that for every topological manifold $M$, there is a well-defined class, which i guess maybe i'll call $\alpha(M) \in H^4(M;\Bbb Z/2)$, that vanishes if and only if $M$ has a PL structure; and a class in $H^3(M; \Bbb Z/2)$ that determines whether or not this PL structure is unique
that last statement deserves clarification but i won't do that
@robjohn My guess is that sometimes we need to study groups of similar (or not necessarily similar) sums and consider them together to get some cool forms.
@iwriteonbananas I more or less have no idea what the differentials on that page look like. You would either want a good understanding of McCleary (his discussion of transgressions comes to mind, whatever those are) or a good understanding of the CW structure of the fibration; and if you had that...
When solving for eigenvalues within a Matrix, what order do they form on the diagonal? Are the ones closest to (0,0) always the greatest, and the smallest closest to (n,n)?
hm, you're right. OK, so the complication isn't in the spectral sequence, it's in the fact that you then have to reassemble the groups from the filtration, I think
Like when you calculate the $E^\infty$ page it's not a literal list of groups or whatnot
with regards to the comment to this answer, I'm oh-so-tempted to just say "No, you can't." (at least to the extent that i really doubt there to be any more elementary answer than the one already provided)
