That's correct, but we should be able to argue this precisely. Suppose $[\sigma] \in p_* \pi_1(\widetilde{X}) \leq \pi_1(X)$, then certainly $\sigma$ is image of some loop in $\widetilde{X}$ by $p$, let us call that loop $\Sigma$.
Suppose there is another lift to a path $\gamma$ in $\widetilde{X}$ which is potentially a non-loop.
Both $\gamma$ and $\Sigma$ start at $\widetilde{x_0}$ by construction (I fudged the basepoint earlier). But we know a loop downstair lifts to a unique path upstairs starting at a given point.
So then endpoint of $\gamma$ is the endpoint of $\Sigma$ is $\widetilde{x_0}$.
Here we are using the "unique path lifting property" of covering spaces, notably.
The uniqueness of lifting essentially tells you there is only one lift once you fix a starting point, alternatively, $p_*$ is an injection (this is a little more, because you're saying homotopic loops lift to homotopic loops, but this homotopy is also unique, etc)
@feynhat There is a unique lift of $f \circ \gamma$ and $f \circ \gamma'$ in $\widetilde{X}$ starting at $\widetilde{x_0}$. Your argument tells that $(f \circ \gamma) * \overline{f \circ \gamma'}$ lifts to a loop in $\widetilde{X}$. That forces the unique lifts of $f \circ \gamma$ and $f \circ \gamma'$ to have equal endpoints, right?
The endpoint of $\widetilde{f \circ \gamma}$ and $\widetilde{f \circ \gamma'}$ being the value of $\widetilde{(f \circ \gamma) * \overline{f \circ \gamma'}}$ at time $t = 1/2$.
@BalarkaSen I see. Is the following true: Suppose $p : (\widetilde X, \widetilde x_0) \to (X,x_0)$ is a covering map. $\alpha$ and $\beta$ are paths starting at $x_0$ and have same endpoints, so that $\gamma = \alpha * \bar \beta$ is a loop. Suppose the lift $\widetilde \gamma$ of $\gamma$ is also a loop, then $\widetilde \alpha(1) = \widetilde \beta(1)$.
ugh... Mathjax has problem with bars and tildes. The second $\beta$ was with a bar, and the third one was with a tilde. I hope you were able to make that out @BalarkaSen. I can't see any difference in my browser.
The British government has frozen mortagage payments but haven't frozen rent for renters, so those people who own a house get a few months mortagage payment free while those renting the houses still have to pay their landlord, and AT THE SAME TIME haven't done anything to increase sick pay in the UK, but still forcing everyone to take time off work on 94 GBP a week
What area of interest did you tell them? Did you give them a preference for you wanted to work under? I don't see anyone there who works in algebraic topology. @BalarkaSen
is there a sequence $(f_n)$ defined on $[a,b]$ such that every $f_i$ is Riemann-integrable and $\lim f_n=f$ exists pointwise at every point in $[a,b]$ but $f$ is not Riemann-integrable on $[a,b]$?
The problem is that I don't have a deadline for the PDEs exam with the coronavirus situation, so I don't feel the urge to study because time is running out
(If you replace $w$ with $k$, the forward transform amounts to going from a position-space representation to a momentum-space representation. Pretty sure it preserves the L2 norm, because that amounts to the normalization condition.)
I was asked to prove the following: Suppose $A$ is a "diagonally dominant" square matrix, i.e., $\|a_{ii}\| > \sum_{j = 1, j \neq i}^n \|a_{ij}\|$ for all $1 \leq i \leq n$. Then $A$ is invertible. This is trivial from the norm stuff because $A = D + B$ where $D$ is the diagonal part, then $A = D(I + D^{-1} B)$ and $\|D^{-1} B\|_{L^1} < 1$. Now if $\|M\|_{L^1} = \varepsilon < 1$, then $I + M$ is invertible since $I - M + M^2 - \cdots$ is bounded in norm by $1/(1 - \varepsilon)$
We can use a Cauchyness argument to prove it converges.
$Ax=b$ is equivalent to $x=D^{-1}(b-Bx)$, which is a fixed point equation for the affine linear map $x\mapsto D^{-1}(b-Bx)$, which is a contraction, because $\lVert D^{-1}B\rVert<1$, hence conclude by Banach fixed point theorem
Take a complex square matrix, look at one row, sum up the absolute values of all the elements in that row except the diagonal one, take that as radius of a circle whose center is the diagonal element in that row
the union of these circles contains all eigenvalues of the matrix
same works with columns if you look at the transpose instead
