Tag: rightarrow

algebraic topology – Induced group homormophism $f_{*} : pi_1( S^1,p) rightarrow pi_1( S^1,f(p))$ due to $z mapsto z^2$.

Let $f: S^1 rightarrow S^1$ s.t. $z mapsto z^2$. This map induces a group homormophism $f_{*} : pi_1( S^1,p) rightarrow pi_1( S^1,f(p))$ . How can we explicitly write this map? I know that the mapping is that a loop gets mapped to 2 loops. But I am not sure how to write it?

Morevover, is there any bijective map for which the iduced map between the fundamental group is not an isomorphism?

real analysis – What is the inverse of $G: (x,y) rightarrow (x, 1-ye^x+xe^y)$

How can I calculate the inverse of
$$
begin{align*}
G: mathbb{R}^2 &rightarrow mathbb{R}^2\
(x,y) &mapsto (x, 1-ye^x+xe^y)
end{align*}
$$
I actually just need the inverse in a neighborhood of the point $(0,1)$, and I know that the inverse exists in that neighborhood because
$$
(D_pG) = left(begin{matrix}
1 & 0\
-ye^x+e^y & xe^y-e^x
end{matrix}right)_p
implies
(D_{(0,1)}G) = left(begin{matrix}
1 & 0\
e-1 & -1
end{matrix}right)
$$
which is invertible, and by the inverse function theorem I know that there is $G^{-1}:U subset mathbb{R}^2 rightarrow mathbb{R}^2$ where $G(0,1) in U$ open set.

Here’s what I was able to achieve:

One thing to note is that when $v rightarrow 0$ we have that

$$
G^{-1}(v) sim D_{(0,0)}G^{-1}(v) + G^{-1}(0,0)
$$

But $G^{-1}(0,0) = G^{-1}(G(0,1)) = (0,1)$ and $D_{(0,0)}G^{-1} = D_{G(0,1)}G^{-1} = (D_{(0,1)}G)^{-1}$, therefore we can conclude that when $v rightarrow 0$ it follows that

$$
G^{-1}((v_1,v_2)) sim (D_{(0,1)}G)^{-1}((v_1,v_2)) + (0,1) = (v_1,-v_1-v_2+v_1e+1)
$$

Can someone please check my work and give me some directions?

Thanks!

real analysis – $epsilon,delta$-proof for the limit of $log sinh{(x^2)}-x^2$ for $x rightarrow infty$

I want to use a $epsilon,delta$-proof for the existence and value for the limit of $$log sinh{(x^2)}-x^2$$
for $x rightarrow infty$.

Now, I know the definition for such proof to be $forall epsilon > 0 exists c forall x > c: left | f(x)-L right |<epsilon$.

I am struggeling to approach this problem as I am not familiar with the method for such proofs. What if the limit as $x rightarrow infty$ is $infty$ then this definition would crumble, right?

general topology – If f: $mathbb{R}_{l} rightarrow S_{Omega}$ is continuous, then f is not injective.

If f: $mathbb{R}_{l} rightarrow S_{Omega}$ is continuous, then f is not injective.

I’ve been trying to solve this problem for a few days, but I haven’t been able to see how can I do it. First, $mathbb{R}_{l}$ refers to $mathbb{R}$ with the lower limit topology (generated by intervals of the form $(a,b)$ ) and $S_{Omega}$ refers to the uncountable well-ordered set with the order topology (generated by (a,b) and rays). Until now, I’ve tried to use the fact that both of those topologic spaces are Hausdorff and suppose that f is injective to get a contradiction, but I don’t see where can I get it.

Sufficient condition for uniform convergence of $f_{epsilon}(x, cdot)$ to $f(x, cdot)$ implies that $f_{epsilon} rightarrow f$ uniformly.

Consider functions $f_{epsilon}: X times U rightarrow mathbb{R}$ for all $epsilon >0 $ and $f: X times U rightarrow mathbb{R}$. $X subset mathbb{R}^n$ and $U subset mathbb{R}^m$ are compact spaces.
Suppose that given $x in X$, $f_{epsilon}(x, cdot)$ and $f(x, cdot)$ are continuous (on $U$).

If $f_{epsilon}(x, cdot)$ converges uniformly to $f(x, cdot)$, then does it imply that $f_{epsilon} rightarrow f$ uniformly?
I guess that from the uniform convergence of $f_{epsilon}(x, cdot)$ to $f(x, cdot)$,
we may choose $overline{epsilon}(x) > 0 $ for any $epsilon’ >0$ such that $|f_{epsilon(x)}(x, u) – f(x, u)| < epsilon’, forall epsilon(x) leq overline{epsilon}(x), forall (x, u) in X times U$. Then, $epsilon := min_{x in X}overline{epsilon}(x)$ implies the desired result.

Actually, I’m not sure that such $epsilon$ exists. Is it correct? If not, are there any sufficient conditions to derive similar conclusions?

complex numbers – Is $T : mathbb{C} rightarrow mathbb{C}: a + bi mapsto overline{a + bi} = a – bi$ a linear image?

So a question in my handbook for linear algebra is the following; Is the image T
$$T :mathbb{C} rightarrow mathbb{C}: a + bi mapsto overline{a + bi} = a – bi$$

an linear image if:

  • $mathbb{C}$ is interpreted as a real vector space
  • $mathbb{C}$ is interpreted as a complex vector space

I know how I would solve the question itself (I would check the axiom of linear image: if $L: V rightarrow W$ is a linear image then $L(lambda_1v_1 + lambda_2v_2) = lambda_1L(v_1) + lambda_2L(v_2)$) but I’m stuck at what they mean with the interpretation.

real analysis – if $lim _{xrightarrow infty}{f'(x)}=0$ then does $lim_{x rightarrow infty}{f(x)}$ exist in the broad sense

Let $f$ be a differentiable a function in $mathbb{R}$, and let $lim _{xrightarrow infty}{f'(x)}=0$

Does $lim_{x rightarrow infty}{f(x)}$ exist in the broad sense?

I’m really lost here. This exercise is from a section on MVT, and intuitively it seems to be correct, but I can’t seem to find a lead. If someone could just give me a hint that would be great.

So far my best shot has been using Heine’s definition of the limit, but no dice.

differential geometry – $f: B^2 rightarrow B^2$ be a continuous map such that $f(x) = x$ for every $x in S^1$, show $f$ is surjective

Let $B^2 = {x in mathbb{R}^2; ||x|| leq 1}$ be the unit disk inthe plane. Let $f: B^2 rightarrow B^2$ be a continuous map such that $f(x) = x$ for every $x in S^1 = {x in mathbb{R}^2; ||x|| = 1}$. Show that $f$ is surjective.

This my following attempt at a solution:

If $f$ isn’t surjective, then there exists a point $y in B^2$ such that $f(x) – y$ is never $0$ for all $x in B^2$. Thus $frac{f(x) – y}{|f(x) – y|}$ is well defined and exists on $S^1$.

however, I’m sure this is a dead end. Some hints would be greatly appreciated!

$lim_{n rightarrow infty} frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1}$. Am I correct?

I have to find this limit:

begin{align} lim_{n rightarrow infty}
frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1} end{align}

My attempt:

Since we know that $-1leq sin (x) leq 1$ for all $x in mathbb{R}$ we have:

begin{align}
lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdot(-1)}{n+1}underbrace{leq}_{text{I have doubt if this is correct}}&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{(1)^{n}cdotsqrt{n}cdot(1)}{n+1}\ \ lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{n}}{n+1}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{n}}{n+1}
end{align}

My doubt is because of the $(-1) terms$. If everything is correct, then we have

begin{align}
lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{frac{1}{n}}}{1+frac{1}{n}}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{frac{1}{n}}}{1+frac{1}{n}}\ \ Rightarrow 0 leq &lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq 0 \ \ therefore& lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1}=0
end{align}

Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!