Proving that endomorphisms can be induced by pointed maps?

Question

How do you prove that every homomorphism $\phi:\pi_{1}(S^1,1)\rightarrow \pi_{1}(S^1,1)$ is induced by some pointed map $f :(S^1,1)\rightarrow (S^1,1)?$

Answer 1

In this case, you just need to show that there is a continuous map of the circle of degree n for reach integer n. This can be done by regarding $S^1 \subset \mathbb C$ and taking $z \mapsto z^n$.

However, the theory here is a bit more general, and it has to do with eilenberg MacLane spaces. There is a treatment in Hatcher chapter 1 appendix B.

Answer 2

Every homomorphism of a cyclic group is determined by where a generator is sent. Suppose you have a homomorphism sending the generator $[1]$ to the element $[f]$, well this homomorphism is clearly achieved by $f_*$ since composition of the identity with $f$ is $f$.

Answer 3

The group $\pi_1(S^1,1)$ consists of homotopy classes of pointed maps $(S^1,1)\to (S^1,1)$, and the identity map $\operatorname{id}$ is a representative of the generator.

Any homomorphism $\varphi\colon \pi_1(S^1,1)\to \pi_1(S^1,1)$ is determined by where it sends $\operatorname{id}$, so write $$\varphi(1)=\varphi([\operatorname{id}])=[f].$$

Now we claim that the representative $f\colon (S^1,1)\to (S^1,1)$ induces $\varphi$. By definition, $$f_*(1)=f_*([\operatorname{id}])=[f\circ\operatorname{id}]=[f],$$ and since $f_*$ agrees with $\varphi$ on a generator, $f_*=\varphi$.