Given any point $p \in S^{1}$ let $\gamma_{p}$ be the loop that starts at $p$ and goes counterclockwise around $S^{1}$ (Here we think of $S^{1}$ as the unit circle in $\mathbb{C}$ and the patch can be given by $\gamma_{p}(t)=p \cdot e^{2 \pi i t} ).$ We can take $[\gamma_{p}]$ to be a generator of $\pi_{1}\left(S^{1}, p\right) \cong \mathbb{Z}$.
Show that given any path $\lambda:[0,1] \rightarrow S^{1}$ the isomorphism $\phi_{\lambda}: \pi_{1}\left(S^{1}, \lambda(1)\right) \rightarrow \pi_{1}\left(S^{1}, \lambda(0)\right)$ send $\gamma_{\lambda(1)}$ to $\gamma_{\lambda(0)}$ . Hint: path lifting might be useful
I have been told to use the concept of path lifting, and also the following sketch:
Take $\eta$ a path from $p$ to $q$. We have to show that $\eta \star [\gamma_q] \star\bar{\eta}$ is (homotopic) to $[\gamma_p]$. Then take $\eta_s \star [\gamma_q] \star\bar{\eta_s}$ where $\eta_s = \eta(st)$ and taking $\eta_0(t) = \eta(0) = p$ $\forall t$ and we are done.
Basically I do not understand why we are done with this. (Maybe we are not and I am missing something)
Edit: I would be glad to read another solution
There is no other solution. The isomorphism $\phi_{\lambda}: \pi_{1}\left(S^{1}, \lambda(1)\right) \rightarrow \pi_{1}\left(S^{1}, \lambda(0)\right)$ is given by $\phi_{\lambda}([\omega]) = [\lambda * \omega * \lambda^{-1}]$. We have to show that $\phi_{\lambda}([\gamma_{\lambda(1)}]) = [\gamma_{\lambda(0)}]$. This means that the paths $\lambda * \gamma_{\lambda(1)} * \lambda^{-1}$ and $\gamma_{\lambda(0)}$ are homotopic. You can do this directly by writing down a homotopy, but it is easier to use unique path lifting (covering projection $p : \mathbb R \to S^1, p(t) = e^{it}$).
Lift $\lambda$ to a path $u$. Then $p(u(0)) = \lambda(0), p(u(1)) = \lambda(1)$. Lift $\gamma_{\lambda(1)}$ to the path $v(t) = u(1) + 2\pi t$. Note that also $w(t) = u(t) + 2\pi$ is a lift of $\lambda$. We have $u(1) = v(0)$ and $v(1) = u(1) + 2\pi = w(1) = w^{-1}(0)$. Thus $\alpha = u * v * w^{-1}$ is a lift of $\lambda * \gamma_{\lambda(1)} * \lambda^{-1}$. We have $\alpha(0) = u(0)$ and $\alpha(1) = w^{-1}(1) = w(0) = u(0) + 2\pi$. The map $\beta(t) = u(0) + 2\pi t$ is a lift of $\gamma_{\lambda(0)}$ such that $\beta(0) = u(0)$ and $\beta(1) = u(0) + 2\pi$. The homotopy $H(t,s) = (1-s)\alpha(t) + s \beta(t)$ is a homotopy of paths from $\alpha$ to $\beta$. But then $p \circ H$ is the desired homotopy from $p \circ \alpha = \lambda * \gamma_{\lambda(1)} * \lambda^{-1}$ to $p \circ \beta = \gamma_{\lambda(0)}$.