Let $\varphi_{1},\varphi_{2}:\mathbb{S}^{1}\rightarrow\mathbb{R}$ be two
smooth general position (Morse) functions having the same set of critical
points $\left\{ p_{1},...,p_{n}\right\} \subset\mathbb{S}^{1}$ ($n$ is even)
and both $\varphi_{1}$ and $\varphi_{2}$ have a local maximum at $p_{1}$.
Suppose that $\varphi_{1}$ and $\varphi_{2}$ are *similar* in the following sense:

$\left( \varphi_{1}(p_{i})-\varphi_{1}(p_{j})\right) \left( \varphi _{2}(p_{i})-\varphi_{2}(p_{j})\right) >0$ for any $i\neq j$,

i.e. the critical level sets of $\varphi_{1}$ and $\varphi_{2}$ are in some sense similar.

Consider the corresponding two Dirichlet problems:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Delta u=0$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u|_{\mathbb{S}^{1}}=\varphi_{i}$ , $i=1,2$,

getting in such a way two harmonic solutions $u_{1},u_{2}:\mathbb{B}% ^{2}\rightarrow\mathbb{R}$.

Then is it true that the level lines portraits of $u_{1}$ and $u_{2}$ are the same up to topological equivalence, i.e. there is a homeomorphism $h:\mathbb{B}^{2}\rightarrow\mathbb{B}^{2}$ fixing all $p_{i}$ and sending the level lines of $u_{1}$ onto the level lines of $u_{2}$? Then, of course, $h$ is sending the critical set of $u_{1}$ onto the critical set of $u_{2}$.

In brief: does the similarity of the boundary conditions implies similarity between the solutions of the corresponding Dirichlet problems?

Note that we don't assume $\varphi_{1}$ and $\varphi_{2}$ to be close in any sense.

Let us consider polynomials as functions on $[0,1]$, and so define \begin{align*} \|f\|_2 &= \sqrt{\int_0^1f(x)^2\,dx} \\ \|f\|_\infty &= \max\{|f(x)|: 0 \leq x\leq 1\}. \end{align*} I am interested in the ratio of these norms. It is easy to see that $\|f\|_2\leq\|f\|_\infty$, with equality only for constant polynomials. In the opposite direction, put $$ f_d(x) = \sum_{i=0}^d \frac{(d+1+i)!}{(d-i)!i!(i+1)!}(-x)^i. $$ Experiments make it clear that $\|f_d\|_2=1$ and $\|f_d\|_\infty=(d+1)$ and that $f_d$ maximises the ratio $\|f\|_\infty/\|f\|_2$ among polynomials of degree $d$. These facts must surely be known. Can anyone point me to a reference? Do the polynomials $f_d(x)$ have a standard name?

What is the number of binary arrays of length $n$ with at least $k$ consecutive $1$'s? For example, for $n=4$ and $k=2$ we have $0011, 0110, 1100, 0111, 1110, 1111$ so the the number is $6$.

I'm having trouble to prove the following formula using Induction on $n \in \mathbb{N}$: $$\sum_{k=1}^n \binom{n}{k} \binom{n}{n+1-k} = \binom{2n}{n+1}.$$ I've tried all the usual identities, but they seem to lead nowhere. Is there any trick to this, or is it just not possible to prove this using induction?

I'm thankful for any tip or advice on how to approach this :)

For a pair of integers $(a,b)$, consider the conic in $\mathbb{P}^2$ given by

$$C_{a,b} : z^2 = ax^2 + by^2.$$

It is known that for most pairs $(a,b)$ the curve $C_{a,b}$ is not everywhere locally soluble, and thus, does not have a rational point. Given that $C_{a,b}$ has a rational point, it is then possible to find all rational points by simply considering all lines $L$ which go through a given rational point and then finding the other intersection point of $L$ with $C_{a,b}$.

Since $C_{a,b}$ has genus 0, it is possible to parametrize all rational points by quadratic forms, whenever a rational point exists. That is, there exist binary quadratic forms $f,g,h$ with integer coeffcients such that the map

$$\displaystyle (u,v) \mapsto (f(u,v), g(u,v), h(u,v))$$

parametrizes the points on $C_{a,b}$; that is, we have the equality

$$\displaystyle h(u,v)^2 = a f(u,v)^2 + b g(u,v)^2.$$

Let $(f,g,h)$ be an *admissible* triple of binary quadratic forms if the above holds. Let $\delta_1 = \min\{|\Delta(f)| : (f,g,h) \text{ admissible}\}$ and $\delta_2 = \min\{|\Delta(g)| : (f,g,h) \text{ admissible}\}$. Is there a known way to compute $\delta_1, \delta_2$ from $(a,b)$?

The Clenshaw-Curtis quadrature rule approximates an integral $I=\int\limits_{-1}^{1} f(x) \, dx$ by $$I\approx I_n = \sum\limits_{j=1}^N f(x_j)w_j \, ,$$ where the $x_j$'s are the roots of the $N$-th order Chebyshev polynomial, and and $w_j$'s their respective weight. To prove the accuracy of this integration formula, one usually goes by either Fourier representation of $f(x)=f(\cos (\theta))$, or by the "Fourier" expansion of $f$ in the Chebyshev polynomials. See e.g., in the Wiki page.

**My Question:** Is there a way to prove the accuracy of this formula, which does not rely on spectral/Fourier theory? Specifically, to show that it is exact ($I=I_n$) for polynomials of degree $\leq n$, and to bound its error for $f\in C^n$.

Earlier, I had asked a similar question but that was not the correct problem where I got stuck. After a few quick answer, I realized that and I apologize for that.

Let $B_m$ be the space of all skew-symmetric matrices of size $m$ over the finite field $\mathbb{F}_q$ of $q$ elements. Let $E$ be a subspace of $B_m$ of dimension $r$ containing atleast one rank $2$ matrix. Write $E$ as $E= E_1 \bigoplus E_2$ with $ \dim E_i= r_i$ for $i=1,2$ and $E_1$ is a maximal subpace of $E$ containing only rank $2$ matrices. Now for a given rank $4$ matrix $Q\in E_2$ how many matrices $P\in E_1$ exist such that $P +Q$ is again of rank $2$. My Guess is $q^2$ and also that $q^2$ is a strict upper bound.

I am reposting the second question from here (after clarifying it) on the recommendation of user "GH from MO".

Let $b_1,b_2,\dots$ be an enumeration of $\mathbb Q$.

**Question 2:** Suppose I define $$G(x,y) = a_0(y) + a_1(y)(x-b_1) + a_2(y)(x-b_1)(x-b_2) + \dots$$ where the $a_k(y)$ are polynomials in $y$ and $g(x) = G(x,x)$.

Suppose the $a_k(y)$ are not identically zero for $k$ large enough. Otherwise, we clearly get polynomials. Is this the only way to get a polynomial? That is, if $g(x)$ is equal to a polynomial function, then is $a_k = 0$ for $k \gg 0$?

Consider a non-singular, completely positive, unital map $\Psi: \mathbf M_k(\mathbb C) \to \mathbf M_h(\mathbb C)$. This map will have one or more retractions $\Phi: \mathbf M_h(\mathbb C) \to \mathbf M_k(\mathbb C)$. Does there exist a choice of retraction $\Phi$ such that, for some $n > 0$ and some $E \in \mathbf M_h(\mathbb C) \otimes \mathbf M_n(\mathbb C)$, we have $$ \mathrm{sdiam}\Bigl( \bigl(\Phi \otimes \mathrm{id}_n\bigr)(E) \Bigr) > 2 \lVert E \rVert$$ where $\mathrm{sdiam}(X) = \lambda_{\max}(X) - \lambda_{\min}(X)$ is the spectral diameter? For instance: in the case $h = k$ (in which $\Phi = \Psi^{-1}$ would be unique), are there $\Psi$, $n>0$, and $E \in \mathbf M_h(\mathbb C) \otimes \mathbf M_n(\mathbb C)$ for which $$\begin{align} \lambda_{\max}\Bigl( \bigl(\Phi \otimes \mathrm{id}_n\bigr)(E) \Bigr) &> \lVert E \rVert, \tag{1}\\ \quad\text{and}\quad \lambda_{\min}\Bigl( \bigl(\Phi \otimes \mathrm{id}_n\bigr)(E) \Bigr) &< - \!\!\;\lVert E \rVert ? \tag{2}\end{align} $$

(This question is a follow-up to a previous question, in which it was established that there are maps $\Psi$ and operators $E$ for which every retraction $\Phi$ satisfies Eqn. (1) above.)

I'm curious if it is possible to formulate cup products and prove that they exist in a general way which would subsume a lot of examples: e.g. group cohomology, sheaf cohomology for sheaves on topological spaces, quasicoherent sheaf cohomology, things like etale cohomology where we look at sheaves on more general sites, etc.

In the sources I've looked at (e.g. Cassels-Fröhlich, Lang's *Topics in Cohomology of Groups*), the existence of cup products is proven in terms of cochains, Čech cohomology, etc., even when more abstract definitions and uniqueness theorems are given.

What I'm looking for is something like this: Let $\mathscr{A}$ is an abelian category with a symmetric monoidal structure $\otimes$ such that $(\mathscr{A}, \otimes)$ satisfies certain conditions (e.g. enough injectives, existence of $\mathrm{Hom}$-objects, whatever other features are common in practice) and $\{H^i\}$ is a universal $\delta$ functor from $\mathscr{A}$ to another abelian symmetric monoidal category $\mathscr{B}$ (assuming whatever we want for $\mathscr{B}$, even $\mathscr{B} = \mathbf{Ab}$, the category of abelian groups). If $\phi \colon H^0(M) \otimes H^0(N) \rightarrow H^0(M \otimes N)$ is an additive bi-functor, then there is a unique sequence of additive bi-functors $\Phi^{p,q} \colon H^p(M) \otimes H^q(N) \rightarrow H^{p+q}(M \otimes N)$ such that:

- $$\Phi^{0,0} = \phi$$
- $\Phi$ is a "map of $\delta$-functors separately in $M$ and $N$": if \begin{equation}\tag{1} \label{s1} 0 \rightarrow{A'} \rightarrow A \rightarrow A'' \rightarrow 0 \end{equation} is an exact sequence in $\mathscr{A}$ with \begin{equation} \tag{2}\label{s2} 0 \rightarrow A' \otimes B \rightarrow A \otimes B \rightarrow A'' \otimes B \rightarrow 0 \end{equation} still exact, then $\Phi^{p +1 ,q} \circ (\delta_1 \otimes H^0(\mathrm{id}_B)) = \delta_2 \circ \Phi^{p+1, q}$. Here, $\delta_1 \colon H^p(A'') \rightarrow H^{p+1}(A')$ and $\delta_2 \colon H^p(A'' \otimes B) \rightarrow H^{p+1}(A' \otimes B)$ are the maps provided by the $\delta$-functor structure on $H$ via the sequences (\ref{s1}), (\ref{s2}). Similarly, if we swap the roles of $A$ and $B$, we require that $\Phi^{p +1 ,q} \circ (\delta_1 \otimes H^0(\mathrm{id}_B)) = (-1)^{p} (\delta_2 \circ \Phi^{p+1, q})$.

The answers to this question shed some light on this matter: Suppose we are in a setting where $H^0(M) = \mathrm{Hom}(O, M)$ for some object $O$ of $\mathscr{A}$ (e.g. group cohomology where we can take $O = \mathbf{Z}$, sheaf cohomology where we can take $O = \mathscr{O}_X$, etc.) Then $H^p(M) = \mathrm{Ext}^p(O, M)$, so we should get a pairing $H^p(O) \otimes H^p(O) \rightarrow H^{p+q}(O)$ induced by the 'composition' mapping $\mathrm{Hom}(O, O) \otimes \mathrm{Hom}(O, O) \rightarrow \mathrm{Hom}(O,O)$. I'm not sure exactly how to prove this part in general either, but I've at least seen it discussed in terms of classes of extensions of modules (I'm not sure how generally the result that $\mathrm{Ext}$ describes extension classes holds). This also doesn't allow general group objects, and I'm not sure how to do the extension.

The above question also discusses a more homotopical/$\infty$-categorical way to think about cup products, but I'm not familiar enough in that language to really get what's going on: I'd much prefer an argument working in ordinary abelian categories.

We say that a projective variety $X$ is of general type if the Kodaira dimension is equal to the dimension of $X$., i.e. $\text{kod}(X)=\dim X$.

When $K_X$ is positive then by the result of S.T.Yau we have a Kaehler-Einstein metric on $X$, i.e., $Ric(\omega)=-\omega$,

For varieties of general type, we have the finite generation of the canonical ring by recent breakthrough prize winners and hence they solved that we still have Kaehler-Einstein metric on varieties of general type i.e., the extension of S.T.Yau result.

**OK**, Which type of questions can be solved by using finite generation of the canonical ring in the geometric analysis? (a summary of important results could be enough!)

Does there exist a (onedimensional) integral functional of calculus of variations (with $f$ finite everywhere) $$ F(y)=\int_a^b f(t,y(t),y'(t))\,dt $$ such that $$ \inf_{y\in Lip([a,b])}F(y)<\inf_{y\in C^1([a,b])}F(y) $$ that is, it shows the Lavrentiev phenomenon between $C^1$ and Lipschitz.

I wish to understand the following integral, $$E(u,\Sigma)=\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2$$ Where $u\in C^2(\Omega)$ is a solution of some elliptic equation, so we have $D^2u$ is semi-positive, $\Omega\subset R^n$ is a open set, $\Sigma$ is a arbitrary smooth surface equipped with the induce metric from $R^n$, and $\Sigma\subset \Omega$, $\{e_1,e_2\}$ is a pair of orthogonal biases of $\Sigma$.

I wish to understand if there is a similar result like Newton-Leibniz formula in the one dimensional case, which is the following: $$\int_{\gamma}\partial_{e_1e_1}ude_1=\int_{\partial \gamma}u=\partial_{e_1}u(\gamma(1))-\partial_{e_1}u(\gamma(0))$$ Where $\gamma: [0,1]\to \Omega$ is a $C^2$ curve and $e_1$ is the gradient direction along the curve.

So my problem is following:

**problem**
Can we get some integral expression for $E(u,\Sigma)=\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2$ use just $u_1=\partial_{e_1}u,u_{2}=\partial_{e_2}u $, and the integral domain of the expression is $\partial \Sigma$? more precisely I wish we could find a functional $\hat E(u,u_1,u_2)$ such that,
$$E(u,\Sigma)=\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)?$$
$\hat E$ is a functional only relate to $u,\nabla u$. Because of $\hat E$ only related with $u,\nabla u$, I would like to say it is the lift of $E(u,\Sigma)$. And may be this is only true for some special domain $\Sigma$, for example $\Sigma$ is a Ball, this is exactly what I excepted.

**Motivation**

Now I need to explain my motivation why I except this is true and why I need this or it variation is true.
The motivation is come from the 1 dimensional version is true, and which is crucial to establish **the mean value principle** for Laplace equation,

$$\frac{1}{\mu(B)}\int_{\partial B(r,x_0)}u(x)dx=u(x_0)$$

I wish to generated the mean value principle to some special nonlinear elliptic equation by this way, although this mean value principle may be not exist, I still wish to explain why the mean value property failed by this way.

**Attempt**

I have four ways to attempt this problem, but there always emerge difficulties I could not settle.

- Use the identity $$exp(tr(A))=det(exp(A))$$ We try to solve the equation $D^2 u=exp^{tr (A)} ...(*)$, we pretend it could be solved then we have: $A=log(D^2(u))=\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{k}{D^2(u)}^k$, so we have, $$E(u,\Sigma)=\int_{\Sigma}e^{tr(\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{k}{D^2(u)}^k)}de_1\wedge de_2$$ It seems much easy to fine $\hat E(u)$ use stokes theorem with $E(u)$ under this form. But there exists two problem, one is that the solvable of $(*)$ and there exists infinity many different solution of $(*)$ if my insight is right and I do not know how to proof the identity we could proved in this way is independent with the choice.

2.

We discretization the problem and consider it in $\mathbb Z^2$ which is a two dimensional affine subspace of $\mathbb Z^n$.

The advantage of discretization is that we can explicate calculate $u_{11},u_{12},u_{21},u_{22}$ now, in fact,

$$u_{11}(x,y) = h^2(u(x+2h,y)+u(x,y)-u(x+h,y)-u(x+h,y)) $$ $$u_{12}(x,y) = h^2(u(x+h,y+h)+u(x,y)-u(x+h,y)-u(x,y+h)) $$ $$u_{21}(x,y) = h^2(u(x+h,y+h)+u(x,y)-u(x+h,y)-u(x,y+h)) $$ $$u_{22}(x,y) = h^2(u(x,y+2h)+u(x,y)-u(x,y+h)-u(x,y+h)) $$

and we could use this to calculate $det(D^2u)$, but after calculate I do not find general principle and what shape should $\Sigma$ be to make the identity, $$\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)$$ make sense in this way.

3. Investigate the Frobenius integrable condition, which is just mean:

$$L_iL_ju(x)-L_jL_iu(x)=\sum_k c_{ij}^k(x)L_ku(x)$$ should be true, I tried to split $E(u,\Sigma)$ into several parts, and every part of it satisfied the Frobenius integrable condition, i.e.

$$E(u,\Sigma)=\sum_{i=1}^k E_i(u,\Sigma)$$

and $E_i$ satisfied the Frobenius integrable condition. And we investigate each $E_i$ first and combine the result we got together to establish a result for $E(u,\Sigma)$.

But the difficulties comes from that I do not know how to decompose $E(u)$ at all!

4. The last strategy could only get a part of result(instead of identity, we could only get a inequality). thanks to the elliptic condition we know $D^2u$ is semi-positive and the Principal minors of $D^2u$ is also semi-positive, so $D^2u|_{\Sigma}$ is semi-positive. we could consider the function $f(x)=Det(D^2(u)|_{\Sigma})^{1/2}$, which is a concave function so use Jensen inequality we could get following result:

$$\frac1{{\rm vol}\Sigma}\int_{\Sigma}(\det D^2u|_{\Sigma})^{1/2}\le\det\left(\frac1{{\rm vol}\Sigma}\int_{\Sigma} D^2u|_{\Sigma}(x)\right)^{1/2}.$$

May be according this could gain a mean-value inequality but I am not very sure.

May be all of these approaches are useless. In any case, I wish some result could be establish, whatever positive answer or negative answer. I will appreciate to any valuable advice or new idea, thank you very much!

The injective hull for a module always exists, however over certain rings modules may not have *projective covers*. I have a question.

If $A$ is an Artinian module on a Noetherian local ring $R$ then $A$ has projective cover? If not, please give a counter example.

Let $A$, $B$, and $C$ be positive, invertible $4 \times 4$ complex matrices. So we have three nondegenerate "sesquilinear quadratic" forms $\langle Av,w\rangle$, $\langle Bv,w\rangle$, and $\langle Cv,w\rangle$. Say that $v,w \in \mathbb{C}^4$ are *good* for $A$ if they are orthogonal and have the same norm relative to the quadratic form given by $A$, i.e., $\langle Av,w\rangle = 0$ and $\langle Av,v\rangle = \langle Aw,w\rangle$.

Can we always find two nonzero vectors which are simultaneously good for $A$, $B$, and $C$?

I can do this if $A$, $B$, and $C$ commute. Here is the argument, in case it helps. First find an orthonormal basis $\{e_1, e_2, e_3, e_4\}$ of $\mathbb{C}^4$ consisting of simultaneous eigenvectors for $A$, $B$, and $C$. Say $Ae_i = a_ie_i$, etc. Then find a nonzero $\vec{d} = (d_1, d_2, d_3, d_4) \in \mathbb{R}^4$ satisfying $\langle \vec{d}, \vec{a}\rangle = \langle \vec{d},\vec{b}\rangle = \langle \vec{d}, \vec{c}\rangle = 0$. Since the $a_i$ are positive, $\vec{d}$ has at least one strictly positive component and at least one strictly negative component. Then $v = \sum \sqrt{d_i}e_i$, taking the sum over $i$ with $d_i > 0$, and $w = \sum \sqrt{-d_i}e_i$, taking the sum over $i$ with $d_i < 0$, have the desired properties.

This is a followup to the question here: How to show that the following function isn't a polynomial over Q?.

As before, let $b_1,b_2,\dots$ be an enumeration of $\mathbb Q$. The question might be sensitive to the enumeration (but probably not).

**Question 1:** Suppose I define $$f_3(x) = (x-b_1)^3 + (x-b_1)^3(x-b_2)^3 + \dots,$$

how do I show that this function is not a polynomial? I do not believe the answers to the old question extend to this case since they all seem to use the positivity of squares.

Can we show that $f(x)$ is not a polynomial?

There is a even more general version (and the one I was interested in from the beginning) that I was interested in, however I forgot to omit the trivial cases/there was some ambiguity in the second question, so I asked it again here:

Let $X$ be a smooth variety over a perfect field $k$ with $X(k) \neq \emptyset$. Then is the natural map \begin{equation} \mathrm{Pic}(X) \to (\mathrm{Pic}(X_{\bar{k}}))^{\mathrm{Gal}(\bar{k}/k)} \qquad (1) \end{equation} surjective?

Remarks:

- If $X$ is a proper, then the map (1) is in fact an isomorphism. This is usually proved using the Hochshild--Serre spectral sequence.
- The map (1) need not be injective in general. Take $X$ to be the complement of a closed point of degree two in $\mathbb{P}^1_k$. Then $\mathrm{Pic}(X) = \mathbb{Z}/2\mathbb{Z}$ but $\mathrm{Pic}(X_{\bar{k}}) = 0$.

Let $p$ be an odd prime. Let $K$ be a finite extension of $\mathbb{Q}_p$ and $\mathcal{O}$ its ring of integers. Let $k$ be the residue field of $\mathcal{O}$.

Let $A$ be an abelian variety over $K$ which has semistable reduction. Let $A_k$ be the special fiber of the Néron model of $A$. Then, there is an exact sequence $$ 0 \to A^0_{k} \to A_{k} \overset{\pi}{\to} \Phi \to 0 $$ where $A^0_k$ is the connected component of the identity and $\Phi$ is the component group.

Here is my question: For simplicity, the component group is cyclic of order $pt$, for some $t$ not divisible by $p$. Let $x \in A_k$ so that $\pi(x)$ generates the $p$-torsion of $\Phi$, which we denote by $\Phi[p]$. Suppose that $x$ is of exact order $p^2$, i.e., $p^2 x=0$ but $ax \neq 0$ for any $1 \leq a < p^2$. Then, is it possible that the map $$A_k [p] \to \Phi[p]$$ induced by $\pi$ surjective? (If it is necessary, we may restrict the case with $e(K/\mathbb{Q}_p)<p-1$.)

For more information, this assumption can be satisfied if $A=J_0(N)$ the Jacobian variety of the modular curve $X_0(N)$ when $N=Mp$ with $(M, p)=1$. If $p \equiv 1 \pmod {12}$, then the component group of $A_p$ is cyclic. The cuspidal divisor $x=0-\infty$ is a rational torsion point and hence maps to $A_p$. Moreover it generates the component group by the map $\pi$. If $M=1$ then the order of $x$ and the order of the component group is same by Mazur but if $M=qr$ with two primes $q, r$ such that $q\equiv -r \equiv 1 \pmod p$, then the above situation can occur.

Let $W$ be a cobordism between $n$-manifolds $M$ and $N$, and $f\colon W \mapsto X$ be a map to some manifold $X$.

Does anybody know of any nice examples of general relationships between the images of the maps $g_*\colon H_*(M)\mapsto H_*(X)$ and $h_*\colon H_*(N)\mapsto H_*(X)$ induced by the restrictions $g=f\mathrel{|M}$ and $h=f\mathrel{|N}$, based on some given data about about these manifolds?

For Example, if $M=N$ and $W=[0,1]\times M$, then $f$ is a homotopy between $g$ and $h$, so $g_*=h_*$. Also, if our manifolds are oriented and compact, $X$ is a connected $n$-manifold, and $f$ is smooth, then the degree of $f\mathrel{\mathrel{|}\partial W}=f\mathrel{\mathrel{|}M\coprod N}$ is zero, so the degrees of $g$ and $h$ are equal up to sign, as are the images of $g_*$ and $h_*$ in degree $n$ homology.

let $\pi:\omega\to\omega$ be permutation and $\mathcal{F}$ is Ramsey ultrafilter on $\omega$. There are uncountable many growing subsequences of $\pi$. Can one proof that one of them has domain in $\mathcal{F}$ ?