Q/ Show that any homomorphism $\phi:\pi_1(S^1)\rightarrow \pi_1(S^1)$ can be realised as the induced homomorphism $f^{*}$ of a map $f:S^1\rightarrow S^1$.
A/ $f$ induces $\phi$ if $f^*([\alpha])=[f\alpha]=\phi([\alpha])$ for all $[\alpha]$. Note that $\pi_1(S^1)=\{[\omega_n]\;|\;n\in \mathbb{Z}\}$ where $\omega_n : I\rightarrow S^1$ is defined by $\omega_n(t)=e^{2\pi i nt}$.
Since $\pi_1(S^1)\cong \mathbb{Z}$ there is an associated homomorphism
$\varphi :\mathbb{Z} \rightarrow \mathbb{Z}$ st $\phi([\omega_n])=[\omega_{\varphi(n)}]$
My idea was to construct some sort of f using this function on the integers to map $\omega_n$ onto $\omega_{\varphi(n)}$ but then I realised that such an $f$ would be dependent on n too so we would have $f:S^1 \times \mathbb{Z} \rightarrow S^1$ which wouldn't make sense and then having thought about it for a while I'm not particularly sure where to go next. Am I on the right sort of lines or have I missed something completely?
Hint: Any homomorphism from $\mathbb{Z} \cong \pi_1(S^1)$ is uniquely defined by where it sends $1$. Thus, you can solve your problem by showing you can send the equivalence class $[t\mapsto e^{2\pi i t}]$ (the element $1$ in $\mathbb{Z}$) to any other equivalence class in $\pi_1(S^1)$ (element of $\mathbb{Z}$).