from f(x):R to (-1,1) and let the function defined by g(x)=cos x is it surjective? I suppose it must be surjective since it maps exactly to the range of a cosine function. How do you guys see this?
If you want to show something is surjective, you better know what it means to be surjective, in a very technical sense. But your idea is the right one. If the codomain is equal to the range, then the function is surjective.
A function f from A to Bis called onto, or surjective, if and only if every element b∈B there is an element a∈A with f(a) = b. A function f is called a surjection if it is onto
so you mean I say g is surjective using the definition?
I honestly don't know how to do it without the IVT, besides just saying "everything in the codomain is in the range". Perhaps there is something clever you can do with the exponential function.
The problem is that the usual definitions of 'cos' as a function are obnoxious to work with.
@MikeMiller No, they do not prove the Morse lemma.
Also they seem to be proving something weaker than density of Morse function. Namely, given a function $f : M \to \Bbb R$, $f_a = f + a_1 x_1 + \cdots a_n x_n$ is Morse for almost every $a = (a_1, \cdots, a_n) \in \Bbb R^n$.
Wythoff's game is as follows: there are two players $A$ and $B$ ( $A$ being the first player ) and there are $2$ piles of stones. When his turn a player can remove one or more stones from anyone pile or same number of stones from both the piles. A player unable to make a move loses. Also it is...
Does the run time complexity of Linear Programming or non linear programming depend upon the number of number of constraints? If yes does it increase or decrese?
@Ell Yeah, $x$ does not need to be $2\pmod7$. For example, $x=8$ (the number, not modulo anything) satisfies $\begin{cases}\phantom{{}^3}x\equiv 3\pmod5\\x^3\equiv1\pmod7\end{cases}$ and isn't $2\pmod7$.
@Balarka In particular, you can choose such an $(a_i)$ arbitrarily small. That's stronger than denseness: it gives some slight control kver the perturbations you need to make.
Can anyone tell me if my question will be deleted by the Community robot within a week now that the question has one downvote? The votes are at -1 and there are comments on the question. Will the comments protect it from the bot?
Is this polynomial root finding algorithm below known, and under what conditions for the choice of polynomial coefficients does it find at least one root?
Description of the algorithm:
Consider the polynomial:
$$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 = 0$$
Replace $x^n$...
for example, we have $\mathbb{Z}/5\mathbb{Z}$ and it involves classes with elements of the form $a+5b$ Is there exists finite sets which has elements of the form $a*b^5$? What are the names of these sets (if any)?
@BalarkaSen There are a few general function space results you should know, all in Hirsch's differential topology book, which I'll say a couple of. 1) $C^\infty(M,N)$ is dense in $C^0(M,N)$. This is also true relatively: if a map is smooth on a closed subset, you can homotope it to be smooth everywhere while fixing the map in the subset. This is also true relative to the boundary of non-closed manifolds.
(This implies that $C^\infty(M,N) \hookrightarrow C^0(M,N)$ is a homotopy equivalence.)
You can also smooth sections of a bundle through sections, or smooth maps from a fiber bundle so that if it's already fiberwise smooth, each $f_t$ is still fiberwise smooth. That sort of thing.
Oh, I think I told you the embedding result already, about denseness in function spaces. To repeat in case I didn't: if $\dim N > 2 \dim M$, then $\text{Emb}(M,N)$ is dense in $C^\infty(M,N)$.
how does it make sense that the order of $sr$ and $sr^2$ is $2$? These are elements of the dihedral group $D_{6}$ I get that $|r| = 3$ and $|r^2| = 3$ since $r^3 = 1$ and $r^6 = 1$ but $|sr| = 2$ doesn't make sense. $s^2 r^2 = r^2$ not $r^3 = 1$.
@BalarkaSen I've started reading algebraic geometry, and so far I proved that two affine plane curve intersect only finitely many points. Now I am thinking, if we consider the base field as $\mathbb C$ , then can we say that they intersect transversally or not?
It's just going to be literally "the intersection only occurs in the smooth locals and the tangent spaces sum to the tangent space of the ambient variety"
I still like topology over algebraic geometry, and only want to study the topological kind of algebraic geometry. But I guess I am not literally your friend.
@Anubhav.K In one sense, transversality in algebraic geometry is subtle. The thing is that algebraic varieties can be singular, and one can still ponder on intersection of singular algebraic varieties. In that case making them intersect in "general position" becomes a lot harder, unlike for smooth manifolds where transversality theorem says on can always make smooth submanifolds intersect transversally upto a homotopy. Just sayin'.
By which I remember I still have to learn transversality theorem.
@MikeMiller I don't know, that the homotopy can be chosen to deform the inclusion map to some other map which is arbitrarily close to the inclusion map?
In some appropriate norm on the function space (uniform norm, most likely).
Whoops, right, I should also care about the intermediates to be arbitrarily close to the original map too. Meh. Otherwise it can go all over the place.
@MikeMiller are there multiplicative analogues to a mod b, where there is some function S such that $a S b$ has elements $a*b^n$ n is integer? What are the names of these structures in general?
If you're interested, you should explore its properties yourself to see why it doesn't do very much.
@BalarkaSen: For instance, you can use this to set up a homology theory for which the intersection product is visible at the chain level, or the ability to take fiber products at the chain level.
Let $P$ be a smooth manifold with corners. Say a geometric chain is a smooth map $\sigma: P \to M$; and two geometric chains are isomorphic of there's a diffeomorphism $P \to P'$ taking $\sigma$ to $\sigma'$. Say a geometric chain is small if $\sigma(P)$ is in the image of a smooth map from a manifold of smaller dimension. Say a geometric chain is negligible if both $[P]$ and $[\partial P]$ are small.
OK, now define $C_*(M)$ to be the $\Bbb Z/2$-vector space generated by isomorphism classes of geometric chains, modulo $\sigma \sim 0$ if $\sigma$ is negligible, and modulo the relation $[\sigma] + [\sigma'] = [\sigma \sqcup \sigma']$.
The boundary operator $\partial$ makes this into a chain complex. The homology of this chain complex is naturally isomorphic to singular homology. This is not too hard to prove. We needed to quotient out by negligibility or else $H_*(pt)$ would be wrong (and we needed to pass to isomorphism classes lest we not have a set of generators).
this notion of negligibility is old; it's been around since people tried to define singular homology via cubes, and in that setting one needed to add a notion of nondegeneracy for the same reason
we could in addition define the chain complex $C_*^U(M)$ to be the geometric chains transverse to a submanifold $U$. The transversality-denseness result from above more or less straightforwardly implies that $C_*^U(M) \to C_*(M)$ is a chain homotopy equivalence, whence this also defines singular homology.
We thus have a literal intersection product $C_k^U(M) \to C_{k-\dim U}(M)$ given by intersecting with $U$. Passing to homology we get intersection product.
if $M = U \cup V$, pick a function $f$ such that $f|_{U - V} = 0$, $f|_{V - U} = 1$. pick a regular value of $f$; say $1/2$. define $W = f^{-1}(1/2)$. then the mayer-vietoris boundary map is given on the chain level by $C_k^W(M) \to C_{k-1}(U \cap V)$.
similarly making a transversality assumption one can make chain-level fiber products. these are nice tools.
anyway it's nontrivially more work to work oriented so I won't do that but the construction is nice. i learned this from (and think this is due to) max lipyanskiy, geometric homology.
@BalarkaSen this works for Frechet manifold codomain. in particular, because every countable CW complex is homotopy equivalent to a Hilbert manifold, you can define homology for any space you think about that way.
I was checking an answer in a post from 2013, I saw a mistake in the answer (there was a sec instead of a csc). I edited the answer and waited for the approval of a moderator. I return one hour later and a user with lots of points and I guess some mod privileges edited the answer. That is just hilarious.
@EinsL Any user with over 2000 reputation can edit a post. When you suggest an edit, it goes into a queue that everyone with that reputation can see. What presumably happened is someone opened the queue, saw your edit, and saw further improvements they could make.
not that it's entirely surprising: If I don't have to spend a lot of effort on it, then 1) it's probably on a problem that's lower level and of wider interest, and 2) I can get it out faster and therefore avoid having someone else post first
Both $S^3\times \Bbb CP^\infty$ and $\left(S^1\times \Bbb CP^\infty\right)\big/\left(S^1\times \{x_0\}\right)$ have cohomology ring isomorphic to $\Bbb Z[a]\otimes \Lambda[b]$ with $|a|=2$ and $|b|=3$, as can be seen from Künneth and cellular cohomology. Thus, the cohomology ring structure can't ...
is it alright or generally accepted for someone to write a paper-sized answer to a question that is broad, incredibly complex, or simply inspiring to the answerer?
I've noticed this in a few answers here and there and I'm not sure whether or not it's the accepted norm or if those people are doing something not normally approved of.
Of course, this is a question for opinion, not actual policy.
it's not the norm, but it's fine. sometimes the problem requires a long answer, and other times there may be some insight gained by a long circuitous argument
Gausse's Law is $ \phi = \frac{Q}{\epsilon_0} = \oint_{S} \vec{E} \cdot \mathrm{d} \vec{A} $. The surface area of a sphere is $ A = 4 \pi r^2 $. So for a point charge $ \phi = \frac{Q}{\epsilon_0} = \oint_{S} \vec{E} \cdot \mathrm{d} \left(4\pi r^2 \hat{r} \right) $ and $ \phi = \frac{Q}{\epsilon_0} = 4 \pi r^2 \oint_{S} \vec{E} \cdot \mathrm{d} \hat{r} $. But how do I get from Coulomb's law from here?