I'm proposed the following problem:
Show that every group homomorphism $\Upsilon : \pi_1(S^1,1) \to \pi_1(S^1,1)$ can be written as $\Upsilon = f_\ast$ where $f : S^1 \to S^1$ is a continuous map.
I'm considering $S^1$ as a subset of the complex numbers $\mathbb{C}.$ Every path is defined in the unit interval $I = [0,1].$ We will denote the class of a path $\alpha$ (in $S^1$ based at $1$) as $[\alpha] \in \pi_1(S^1,1).$
What I've tried is the following:
Consider the group isomorphism $\iota : \pi_1(S^1,1) \to \mathbb{Z}$ defined as $\iota([\alpha]) = \operatorname{deg}(\alpha),$ for every path $\alpha$ in $S^1$ based at $1.$ Since the path $\alpha_n(t) := \operatorname{exp}(2\pi int)$ has degree $n,$ we can say that $\iota^{-1}(n) = [\alpha_n].$
We get a group homomorphism $\eta := \iota \circ \Upsilon \circ \iota^{-1} : \mathbb{Z} \to \mathbb{Z}.$ It's well known that every group homomorphism $p$ from $\mathbb{Z}$ to itself verifies $p(n) = n + \dots + n = kn$ for some integer $k := p(1).$ Hence, exists some $k \in \mathbb{Z}$ such that $$\eta(n) = n + \dots + n,$$ i.e. add $n$ to itself $k$ times.
We have \begin{align*} \Upsilon([\beta]) &= (\iota^{-1} \circ \eta \circ \iota)([\beta]) \\ &= \iota^{-1}(\eta(\iota[\beta])) \\ &= \iota^{-1}(\eta(\operatorname{deg}(\beta))) \\ &= \iota^{-1}(\operatorname{deg}(\beta) + \dots + \operatorname{deg}(\beta)) \\ &= [\alpha_{\operatorname{deg}(\beta)}]\dots[\alpha_{\operatorname{deg}(\beta)}] := [\alpha_{\operatorname{deg}(\beta)}]^k = [\beta]^k. \end{align*}
Since the map $\Sigma_k : S^1 \to S^1$ defined by $\Sigma_k(z) = z^k$ induces the homomorphism $(\Sigma_k)_\ast : \pi_1(S^1,1) \to \pi_1(S^1,1)$ given by $(\Sigma_k)_\ast[\alpha] = [\alpha]^k,$ we can conlude that $\Upsilon = (\Sigma_k)_\ast,$ where $k$ is the integer given by $\operatorname{deg}(\Upsilon[\alpha_1]).$
Actually, I'm using whithout mention well-known results about the fundamental group of $S^1$ as:
The map $\iota$ is an isomorphism of groups,
that $i^{-1}(n) = [\alpha_n]$ is consistent follows from the fact that $\alpha \sim \beta$ if and only if $\operatorname{deg}(\alpha) = \operatorname{deg}(\beta)$ for paths in $S^1$ based at $1.$
I want to verify if my answer is correct, and if it is, I want to know if there are unclear points on the proof or if there are more direct or clever ways to get the solution.
Thanks for everyone!
Your proof is correct, but you can do it easier. The fundamental group of of a based space $(X,x_0)$ is usually introduced as the set of equivalence classes of closed paths beginning and ending at $x_0$ ("closed paths based at $x_0$"). The equivalence relation is homotopy rel. $\{ 0,1 \}$ ("path homotopy").
There is a 1-1-correspondence between closed paths $u : [0,1] \to X$ based at $x_0$ and basepoint-preserving maps $\lambda : (S^1,1) \to (X,x_0)$. In fact, let $p : [0,1] \to S^1$ be the quotient map $p(t) = e^{2\pi i t}$. Then for each basepoint-preserving $\lambda : (S^1,1) \to (X,x_0)$ the path $\lambda \circ p$ is a closed path at $x_0$. Conversely, for each closed path $u$ based at $x_0$ there exists a unique basepoint-preserving $\lambda : (S^1,1) \to (X,x_0)$ such that $\lambda \circ p = u$. Similarly you can show that path homotopies of closed paths based at $x_0$ are in 1-1-correspondence with basepoint-preserving homotopies of basepoint-preserving maps $(S^1,1) \to (X,x_0)$. Therefore we can naturally identify $\pi_1(X,x_0)$ with the set of basepoint-preserving homotopy classes of basepoint-preserving maps $(S^1,1) \to (X,x_0)$. If $f : (X,x_0) \to (Y,y_0)$ is basepoint-preserving map, then clearly the induced $f_*$ on fundamental groups is given by $f_*([\lambda]) = [f \circ \lambda]$.
Now let $\Upsilon : \pi_1(S^1,1) \to \pi_1(S^1,1)$ be a homomorphisms. We know that $\pi_1(S^1,1)$ is infinite cyclic generated by the equivalence class $1 = [id]$ of the identity map on $S^1$. Take any $f : (S^1,1) \to (S^1,1)$ representing $\Upsilon(1)$. Then $$f_*(1) = f_*([id]) = [f \circ id] = [f] = \Upsilon(1) .$$ This suffices to prove the claim because the homomorphisms $f_*, \Upsilon$ agree on the generator.