@PedroTamaroff I have been trying to figure this one out for ages and I reckon I can't do it since I don't understand the geometric picture behind the snake map : Consider the homology long exact sequence of $(\Sigma_g, A)$ where $A$ is the punctured torus the rest of $\Sigma_{g-1}$ is squashed with. The sequence gives $0 \to H_2(A) \to H_2(\Sigma_g) \to H_2(\Sigma_{g-1}) \stackrel{\delta}{\to} H_1(A) \to H_1(\Sigma_g) \to H_1(\Sigma_{g-1}) \to 0$.
The whole thing will come down nicely if I could prove $\delta = 0$, but I have no idea how to do this.
@robjohn I think I might be interpreting the plots in your answer incorrectly. Does it show that if you're on $W_{0}$ and circle around $z = - \frac{1}{e}$, you can actually end up on either $W_{-1}$ or $W_{1}$?
OK, @Pedro, so essentially you have to pick something in the kernel, pullback via B --> C, pushdown to B', checking that it pushes to 0 via B' --> C' and then pullback via A' --> B'.
I am not entirely sure that the condition you're talking about is enough to ensure the snake map is zero, @Pedro, but I am going to think about it. I have to go now.
@infinitesimal The two forces should be $$\overrightarrow{F}_1=\frac{F}{2}(-\overrightarrow{i}-\overrightarrow{j}) \\ \overrightarrow{F}_2=\frac{F}{2}(\overrightarrow{i}-\overrightarrow{j})$$
@MaryStar The force along the slope is $\vec F \sin \theta$ and perpendicular to that is $\vec F \cos \theta$ where $\theta$ is the angle made with the horizontal.
@ParthKohli Ok... But for example at $\overrightarrow{F}_1=\frac{F}{2}(-\overrightarrow{i}-\overrightarrow{j})$ how do we get the vector $-\overrightarrow{i}-\overrightarrow{j}$ ??
We know that the magnitudes of the two vectors are $F/\sqrt{2}$. Our job is half-done. Now we just need to find the direction vectors for both of them.
Looking at your answer, these guys have defined the negative y-axis to be along the force.
@RandomVariable $\mathrm{W}_0$ is the red sheet in all the images. However, all three sheets meet at $-\frac1e$, though the branch cut between $\mathrm{W}_{-1}$ and $\mathrm{W}_1$ contains $-\frac1e$ so that point is in the interior of neither of the blue or green sheets.
@RandomVariable sorry, I was IAW working on an answer.
@robjohn The plots show only 2 of the sheets meeting at $z= - \frac{1}{e}$. But you just said that all 3 sheets meet there. Did you mean that all 3 branches meet there?
I am looking at the following exercise:
Let $I=(0,1)$. Find the solution $\phi$ that has a continuous derivative in $\mathbb{R}$ and satisfies :
$$y''=0 \text{ in } I \\y''+k^2y=0 \text{ apart from } I, \text{ where } k>0$$
and furthermore $\phi$ has the form:
$\phi(x)=\left\{\begin{matrix}
...
@BalarkaSen Can you give me an explicit description of the map from $H_1(A)$ to $H_1(\Sigma_g)$? You take the class of a closed chain in $A$ and what do you do to it?
Suppose you have a closed chain in $A$ and that it is homologous to $0$ in $\Sigma_g$. You have to show it is homologous to $0$ in $A$. Can you do this?
@BalarkaSen: Relative homology classes are represented by relative cycles $\alpha \in C^n(X)$ such that $\partial \alpha \in C^{n-1}(A)$. Then $\partial [\alpha] = [\partial \alpha]$.
If you inductively know an explicit description of $H^n(\Sigma_g, A)$, it should not be terribly difficult to calculate the boundary map using the above.
@robjohn The function $f(z)=\frac{z}{z-a}$ is given and I have to prove that the image of the unit circle with center the origin is also a circle with center the point $(\frac{1}{1-a^2},0)$. Any hints?
The function $f(z)=\frac{z}{z-a}$ is given and I have to prove that the image of the unit circle with center the origin is also a circle with center the point $(\frac{1}{1-a^2},0)$. @Axoren
@Mike @Pedro I think it was about exploiting the naturality. Probably using the commutative square at the snake maps induced by $(\Sigma_g, A) \subset (B, S^1)$ where B is genus $g-1$ surface minus a disk. Essentially it boils down to the zeroness of $H_2(B) \to H_1(S^1)$
@RandomVariable I had introduced the blue and green versions of W only because they made things nicer along the negative real axis, but W is not smooth there anyway, so we might as well not worry about looking nice there and put the branch cut there.
@RandomVariable The definitions for the sheets of W are fine as defined everywhere.
@RandomVariable However, The two sheets meet at the branch cut and the point $-\frac1e$ is on that branch cut.
@user159870 The unit circle is centered at $(0, 0)$, correct? So, we are aware that the image of this unit circle is a circle which is not-necessarily unit. However, it's necessarily a circle, so it has an origin. When you apply that map to the unit circle, it's origin should map to its new origin.
@user159870 It's a much easier way than proving that there's a point $c$ for which every point in the image lies equidistant from it and then proving that that point is $\left(\frac{1}{1-a^2}, 0\right)$
@PedroTamaroff @robjohn @Axoren I am mixed up. How can we write the function $f(z) = \frac {z} {z - a}$ as a composition of translations and inversions???
@user159870 That is first an inversion ($z\mapsto1/z$), followed by a scale ($z\mapsto-az$), followed by a translation ($z\mapsto1+z$), followed by an inversion ($z\mapsto1/z$)
@user159870 $f(z)$ doesn't represent a circle. It represents a transformation that preserves circleness. When we apply that transformation to every point on the unit circle, (when we take the image of the circle over that function), we get a new circle.
The reason we know it's a circle is because the transformation involves a sequence of steps and each of those steps by themselves will turn a circle into another circle.
@robjohn My question was, "Does it [the plot] show that if you're on $W_{0}$ and circle around $z=−\frac{1}{e}$, you can actually end up on either $W_{−1}$ or $W_{1}$?" You didn't respond with either a "yes" or a "no." And that's leading to quite a bit of confusion on my part.
@user159870 The definition of the image is as follows, which might be what you've been missing this whole time. Given a set $S$ which is a subset of $X$ and a function $f : X \to Y$, then $f(S)$ is the image of $S$ over the function $f$. $f(S) = \{\ f(x)\ |\ x \in S\ \}$
The image is just the set of all the possible outputs if you choose every possible input from your starting set.
So, the set of all possible inputs for the unit circle is just every point that lies on the boundary of the circle.
What do we take as argument of the function to show that the image of the unit circle with center the origin is also a circle with center the point (1/(1-a^2),0)????
@user159870 since the line $at$ intersects the unit circle with center at $(0,0)$, at right angles and analytic functions are conformal, the $\mathbb{R}$ must intersect the image of the circle at right angles.
That is why I computed the points above ($\frac1{1-|a|}$ and $\frac1{1+|a|}$). they are antipodal points of the resultant circle.
@user159870 we know that the circle is mapped to a circle, if we can find two antipodal points, their average is the center of the circle
@user159870 we have two antipodal points: $\frac1{1-|a|}$ and $\frac1{1+|a|}$ and their average is $\frac1{1-|a|^2}$ so that is the center of the unit circle after $f$
@user159870 from the answer you said you wanted to get, I assume that $a\in\mathbb{R}$. The answer I gave works for any $a\in\mathbb{C}$
I am looking at the following exercise:
Let $I=(0,1)$. Find the solution $\phi$ that has a continuous derivative in $\mathbb{R}$ and satisfies :
$$y''=0 \text{ in } I \\y''+k^2y=0 \text{ apart from } I, \text{ where } k>0$$
and furthermore $\phi$ has the form:
$\phi(x)=\left\{\begin{matrix}
...
