I came across this proof (in a set of notes by J. Wu) while searching for various proofs of $\pi_{1}(S^{1})\cong\mathbb{Z}$. The proof uses the so-called relative simplicial approximation theorem.
Relative Simplicial Approximation Theorem: Let $K$ and $L$ be a simplicial complexes and let $A$ be a subcomplex of $K$. Let $f:|K|\longrightarrow|L|$ be any continuous map such that $f|_{|A|}$ is a simplicial map. Suppose that there are finitely many simplices of $K\setminus A$. Then there exists $N$ and a simplicial map $g:\mathrm{sd}^{N}(K,A)\longrightarrow L$ such that $g|_{|A|}=f|_{|A|}$ and $g$ is homotopic to $f$ relative to $|A|$. Here, $\mathrm{sd}^{N}(K,A)$ is the $N$-th barycentric subdivision of $K$ relative to $A$.
The proof of $\pi_{1}(S^{1})\cong\mathbb{Z}$ that is given is as follows:
Let $K$ be the boundary of a $2$-simplex with vertices $w_{0},w_{1},w_{2}$ in the order of counterclockwise along the circle, with $w_{0}$ as the basepoint. Let $f:|K|\longrightarrow|K|$ be any pointed continuous map. By the relative simplicial approximation theorem, we only need to consider the simplicial maps from a subdivision $K'$ of the circle $K$ to the circle $K$. Let $K'$ have vertices $v_{0},v_{1},\dots,v_{n}$ in the order of counter-clockwise along the circle, where $v_{0}$ is regarded as the basepoint and $f\left(w_{0}\right)=v_{0}$, so that $f|_{\left\{w_{0}\right\}}$ is a simplicial map. Let $g:K'\longrightarrow K$ be a pointed simplicial map. Then $g$ maps the circle $\left(v_{0},v_{1},\dots,v_{n},v_{0}\right)$ (in order) into the sequence \begin{align*} \left(g\left(v_{0}\right),g\left(v_{1}\right),\dots,g\left(v_{n}\right),g\left(v_{0}\right)\right), \end{align*} where $g\left(v_{i}\right)$ is one of $w_{j}$ for $j=1,2$ with $g\left(v_{0}\right)=w_{0}$. If $g\left(v_{i}\right)=g\left(v_{i+1}\right)$, then $g$ is constant on the $1$-simplex $\left[v_{i},v_{i+1}\right]$ and so, up to homotopy, we can remove $g\left(v_{i+1}\right)$. For the subsequence $\left(g\left(v_{i}\right),g\left(v_{i+1}\right),g\left(v_{i+2}\right)\right)$, if $g\left(v_{i}\right)=g\left(v_{i+2}\right)$, then $\left(g\left(v_{i}\right),g\left(v_{i+1}\right),g\left(v_{i+2}\right)\right)$ is a path from $g\left(v_{i}\right)$ to $g\left(v_{i+1}\right)$ and backwards to $g\left(v_{i}\right)$ and so we can collapse this subsequence. Assume that $v_{0},\dots,v_{n}$ have minimal number of vertices in the homotopy class of $g$. Then the only sequence of $\left(g\left(v_{0}\right),g\left(v_{1}\right),\dots,g\left(v_{n}\right),g\left(v_{0}\right)\right)$ is given by one of the following:
constant map $\left(w_{0}\right)$,
around the circle $n$-times positively: \begin{align*} \left(w_{0},w_{1},w_{2},w_{0},w_{1},w_{2},w_{0},\dots,w_{0},w_{1},w_{2}\right), \end{align*} or
around the circle $m$-times negatively: \begin{align*} \left(w_{0},w_{2},w_{1},w_{0},w_{2},w_{1},w_{0},\dots,w_{0},w_{2},w_{1},w_{0}\right) \end{align*}
This shows that any pointed continuous map $f:S^{1}\longrightarrow S^{1}$ is homotopic to one of the maps \begin{align*} g_{n}:S^{1}\longrightarrow S^{1} \end{align*} defined by $z\mapsto z^{n}$ for $n\in\mathbb{Z}$, relative to the basepoint. This shows that the map $\Phi:\mathbb{Z}\longrightarrow\pi_{1}\left(S^{1}\right)$ is surjective. Note that $\pi_{1}\left(S^{1}\right)$ is generated by $\left[g_{1}\right]$. On the other hand, one can show that $g_{n}$ is not homotopic to $g_{m}$ relative to the basepoint if $n\neq m$, as $g_{n}$ is the continuous mapping that goes around $S^{1}$ for $n$ times. This shows that $\Phi$ is injective. This gives that $\pi_{1}\left(S^{1}\right)=\mathbb{Z}$.
My question:
How does one show that $\Phi$ is injective?
My attempt:
I started off naturally by assuming $m\neq n$ and $[g_{m}]=[g_{n}]$ in $\pi_{1}(S^{1})$ and so $[g_{n-m}]=c$, where $c$ is the constant map, mapping all $x\in X$ to the basepoint $v_{0}\in S^{1}$. But how does one obtain a contradiction here?
Any hint or suggestion is welcome.