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Jan
19
comment What's your favorite proof accessible to a general audience?
A good way to spice this proof up* is to change necklaces to pizzas and beads to pizza toppings. (*pun not intended, but then reconsidered and intended after all.)
Jan
9
accepted A rank-nullity theorem between $\mathbb Z^n$ and $\mathbb Z^k$
Jan
8
comment A rank-nullity theorem between $\mathbb Z^n$ and $\mathbb Z^k$
I didn't know, but now I read about it. So as I understand: This sequence trivially left-splits; We use the splitting lemma to conclude $\mathbb Z^n = \ker \phi \oplus \operatorname{im} \phi$; and then we trivially get the union of bases as we wanted. So, the heart of the proof is in the proof of the splitting lemma. Right?
Jan
8
comment A rank-nullity theorem between $\mathbb Z^n$ and $\mathbb Z^k$
Is it correct that this short exact sequence is simply a formulation of the first isomorphism theorem? How does the rank equation follow from this short exact sequence?
Jan
8
comment A rank-nullity theorem between $\mathbb Z^n$ and $\mathbb Z^k$
I don't know the meanings of "flat", "split" and "$\otimes$" :(
Jan
8
asked A rank-nullity theorem between $\mathbb Z^n$ and $\mathbb Z^k$
Dec
9
awarded  Caucus
Nov
29
revised Find $g \circ f$ with $f$ and $g$ given.
LaTeX
Nov
29
answered If a series converges, then the following sequence converges to $0$
Nov
13
comment Is a diffeomorphism's image automatically open?
@zibadawatimmy, $\varphi : I \times \mathcal S^{n-1} \to \mathbb R^n$ is a smooth map with $X$ as its image.
Nov
13
comment Can Lagrange multipliers be used to give a good bound on the number of critical points?
@Titus With the help of my advisor, I ended up noticing that for my purposes, I can take the linear constraint to be a line parallel to the standard axes, i.e. of the form $x = c$, $y = c$, etc. This makes everything easier: The restricted gradient is simply the original gradient with the constant coordinate removed.
Nov
11
comment Open mathematical questions for which we really, really have no idea what the answer is
This is a special case of the following conjecture: Let $P(x)$ be a polynomial with integer coefficients. Then it is prime infinitely often unless it has a stupid reason not to be. ("Stupid reasons": $P$ is reducible; the coefficients aren't coprime; etc.) This general conjecture might be a better example of what OP is looking for.
Nov
11
comment Strengthening the intermediate value theorem to an “intermediate component theorem”
I meant $f(x_0) < 0$ where $x_0$ is the center of the ball and $f(x) > 0$ where $x$ is any point on the boundary of the ball. At any other point there are no assumptions.
Nov
10
awarded  Nice Question
Nov
8
asked Is a diffeomorphism's image automatically open?
Nov
5
comment Does the compact manifold $f=0$ resist small perturbations?
This example is more or less what I meant when I said I know how to construct a counterexample if you only assume continuity. The issue with adding a "tiny bump" is that the perturbation would have to change very quickly in a small area which means its gradient cannot be small.
Nov
4
asked Does the compact manifold $f=0$ resist small perturbations?
Nov
2
comment Strengthening the intermediate value theorem to an “intermediate component theorem”
I'd be particularly interested in the case where $f \ne 0$ has exactly two components. Then of course it suffices to show that $f = 0$ is connected. This is a sort of converse to Jordan-Brouwer, which I don't know if it is hard.
Nov
1
asked Strengthening the intermediate value theorem to an “intermediate component theorem”
Oct
24
comment Cutting a torus enough times disconnects it
@Dan, I'd be interested in that answer. I'm learning homology theory right now just to solve this problem.