In the lecture we had the following statement. Since our Prof. said it is trivial we didn't proved it, but since I want to get used with the notations and methods I would like to prove it. So the statement is the following.
Let $p:Y\rightarrow X$ is a covering such that $p(y)=x$. Then if $\sigma$ is a loop at $x$ and $\sigma^+$ is the unique lifting of $\sigma$ to a path startign at $y$ then $\sigma^+$ ends at $y$ iff $[\sigma]\in p_*\left(\Pi_1(Y,y)\right)$ where we define $$p_*:\Pi(Y,y)\rightarrow \Pi(X,x);\quad[\sigma]\mapsto [p\circ \sigma]$$
Now my Idea was the following:
$\Rightarrow$ Let us assume that $\sigma^+$ ends at $y$ so it is a loop in $Y$ based at $y$ then by the lifting property and by uniqness $$\sigma=p\circ \sigma^+\\ \implies [\sigma]=[p\circ \sigma^+]\\\iff [\sigma]\in p_*(\Pi(Y,y))$$ $\Leftarrow$ Let us assume here that $[\sigma]\in p_*(\Pi(Y,y))$, this means that $[\sigma]=[p\circ \psi]$ for some loop $\psi$ based at $y$. Now we can take $\psi=\sigma^+$ and thus we get $[\sigma]=[p\circ \sigma^+]$. But in my opinion this means that $\sigma$ is homotopic to $p\circ \sigma^+$ with fixed endpoints. But then how can I say that $\sigma^+$ ends at y?
And is the rest $\Rightarrow$ this implication so correct?
Thanks for your help.
Concerning the $\Leftarrow$ implication, note that since $\sigma$ and $p\circ\psi$ are homotopic, and since $\sigma^+$ and $\psi$ have the same initial points, $\sigma^+$ and $\psi$ are homotopic too. But $\psi$ is a loop. Therefore, $\sigma^+$ is a loop too.