How to show that $\phi$ is homomorphism? Prove  $\pi(S^1,1) \cong     \mathbb{Z}$
My attempt :  Define  $\phi :\pi(S^1,1) \to     \mathbb{Z}$ by $\phi([f])=deg(f)$   where deg$(f)$ mean  degree of $f$
$Deg(f)= \widetilde {f}(1)$  where $\widetilde{f}$  is the  unique  lift of  $f$  with  $\widetilde{f}(0)=0$
Now  we have to show  that
$\phi $ is  injective, surjective and homomorphism
To show $\phi $ is   surjective :
Given $n \in \mathbb{Z} $. let $\widetilde{f}:[0,1] \to \mathbb{R}$ defined by  $\widetilde{f}(t)=nt$
then $f : [0,1] \to S^1 $  is closed  path  based at $1$. Since  $\widetilde{f}$ is lift  of $f$   with   $\widetilde{f}(0)=0$
we have  $\phi([f])= deg(f)= \widetilde{f}(1)=n$
This implies  that $\phi$  is surjective
To show $\phi $ is   injective :
Assume $\phi ([f])=0$ i,e  $deg(f)=0$
This means that  the  lift $\widetilde{f}$ of $f$ satisfy $\widetilde{f}(0)=\widetilde{f}(1)=0$
Since $\mathbb{R}$ is contractible  so we can define  homotopy  by $F: [0,1] \times [0,1] \to \mathbb{R}$ with  $F(s,t)=(1-s)\widetilde{f}$
$\implies  F(0,t)=\widetilde{f}(t) , F(1,t)=0$ and  $F(t,0)=F(t,1)=0$
Now we  will  used this theorem: let $p : E \to B$ be  a map . If $f$ is a  continious mapping of some  space $X$ into $ B $,a lifting of $f$  is  map $ \widetilde{f} : X \to E$  such that $p\circ \widetilde{f} =f$
So $p \circ F :[0,1] \times [0,1]  \to S^1 $  with $p \circ F(0,t)=f(t) , p \circ F(1,t)=1,p \circ F(t,0)=p \circ F(t,1)=1$
This implies $[f]=1  \in \pi (S^1 ,1)$  which prove that  $\phi$ is injective
My confusion : How  to show that  $\phi$ is homomorphism ?
 A: First note that your proof of injectivity relies on the fact that $\phi$ is a homomorphism.
To prove that $\phi$ is a homomorphism, recall that $[f] \cdot [g]$ is defined by
$$[f] \cdot [g] = [f \cdot g] ,$$
where
$$(f \cdot g)(t) = \begin{cases} f(2t) & t  \le 1/2 \\ g(2t-1) & t \ge 1/2 \end{cases}$$
Let $\tilde f$ and $\tilde g$ be the unique lifts of $f$ and $g$ such that $f(0) = 0$ and $g(0) = 0$. Define
$$(\widetilde{f \cdot g})(t) = \begin{cases} \tilde f(2t) & t  \le 1/2 \\ \tilde f(1) + \tilde g(2t-1) & t \ge 1/2 \end{cases}$$
This is a well-defined continuous map (note that $\tilde f(2 \cdot 1/2) = \tilde f(1) = \tilde f(1) + \tilde g(2 \cdot 1/2 -1)$).
With $p : \mathbb R \to S^1, p(t) = e^{2\pi it}$, we have
$$(p \circ (\widetilde{f \cdot g}))(t) = \begin{cases} p(\tilde f(2t)) = f(2t) & t  \le 1/2 \\ p(\tilde f(1) + \tilde g(2t-1)) = p(\tilde f(1)) \cdot p(\tilde g(2t-1)) =  g(2t-1) & t \ge 1/2 \end{cases}$$
Thus $\widetilde{f \cdot g}$ is the unique lift of $f \cdot g$ with $(\widetilde{f \cdot g})(0) = 0$. Clearly
$$\deg(f \cdot g) = (\widetilde{f \cdot g})(1) = \tilde f (1) + \tilde g (1) = \deg(f)  + \deg(g) .$$
