How to prove no homotopy of paths in $\mathbb{R}^2\setminus\{(0,0)\}$? Consider the functions $f,g\colon[0,1] \to S^1 \subset \mathbb{R}^2 \setminus\{(0,0)\}$ such that
$$f(t) = (\cos(n\pi t),\sin(n\pi t)) \quad g(t)=(\cos(m\pi t),\sin(m\pi t))$$
for some integers $m \neq n$ where $f(0) = g(0)$ and $f(1) = g(1)$. How do you prove that the two paths $f$ and $g$ are NOT path homotopic, relative to their endpoints, for specific cases of $n,m$?

This question is the continuation of this discussion (Showing two paths in $\mathbb{R}^2∖(0,0)$ are not homotopic.). I have been reading Algebraic Topology by Hatcher and I am familiar with the Fundamental Group of the Circle, but not much more. In the linked post, the accepted answer uses $1$-forms, which I am not familiar with.

My attempt for $n=1,m=-1$: Let $f,g\colon[0,1] \to S^1 \subset \mathbb{R}^2\setminus\{(0,0)\}$ such that
$$f(t) = (\cos(\pi t),\sin(\pi t)) \quad g(t)=(\cos(\pi t),-\sin(\pi t)).$$
Assume that $f,g$ are path homotopic. Then there exists a homotopy $f_s:[0,1] \to \mathbb{R}^2-\{(0,0)\}$ such that
$$f_0(t)=f(t), \quad f_1(t)=g(t)$$
and for all $s \in I$,
$$f_s(0) = (1,0), \quad f_s(1)=(-1,0).$$
We can construct the path homotopy $g_s(t) = \frac{f_s(t)}{|f_s(t)|}$ and then use the quotient map $\pi:I \to I/\partial I \cong S^1$ and then use the map $\hat{g_s}:S^1 \to S^1$ where $\hat{g_s} \circ \pi = g_s$. Then... that's where I'm lost.
Any help is much appreciated!
 A: In the case that $n$ and $m$ are both even, this comes down to knowing that $\pi_1(S^1)\cong \mathbb{Z}$ is generated by the homotopy class of the path $\gamma(t) = (cos(2\pi t), sin(2\pi t))$. Based on this, a simple argument goes as follows:
Say $n = 2k$ and $m = 2l$. First, note that there are homotopies $f \sim k\gamma$ and $g \sim l \gamma$ (where $k\gamma$ is the concatenation of $k$ copies of $\gamma$). If $n\neq m$ then $k[\gamma]\neq l[\gamma]$ since $\pi_1(S^1)$ is freely generated by $[\gamma]$, so the result follows for $S^1$. The inclusion $S^1 \to \mathbb{R}^2\setminus\{0,0\}$ is a homotopy-equivalence (the homotopy-inverse sends $v$ to $\frac{v}{|v|}$) so it induces an isomorphism on $\pi_1$, therefore if $f\sim g$ in $\mathbb{R}^2\setminus\{0,0\}$ then they must be homotopic in $S^1$ as well, which is a contradiction.

The only other case is where $n$ and $m$ are both odd, so that $f$ and $g$ both start at $p = (1,0)$ but end at $q = (-1,0)$. This case is slightly weird since we're used to dealing with loops, but we will leverage the fact that there is a bijection between $\pi_1(S^1)$ and the set of homotopy classes of paths relative endpoints from $p$ to $q$, denoted $\pi_0(\Omega_{p,q}(S^1))$ (here $\Omega_{p,q}(S^1)$ denotes the space of paths from $p$ to $q$, similar to the notation for the loop space, and the path components of this space are the homotopy classes rel. endpoints). 
One such a bijection is given by taking the path $\alpha$ going counterclockwise from $q$ to $p$, and sending the class of a path $[\beta]\in \pi_0(\Omega_{p,q}(S^1))$ to the class of the loop $[\beta * \alpha]\in \pi_1(S^1)$; the inverse is given by sending a loop $[\delta]$ to $[\delta* \alpha^{-1}]$. 
Now we need to determine the loops given by concatenating $f$ and $g$ with $\alpha$ (I will give a sketch). If $n = 2k + 1 > 0$ then $f$ is a path which goes counterclockwise around the circle $k$ and a half times and so $f*\alpha$ goes around the circle $k + 1$ times, i.e. $[f * \alpha] = (k+1)[\gamma]$; if $n<0$ and hence $k = \frac{n-1}{2}< 0$ then $f$ goes clockwise around the circle $|k|$ minus a half times (draw some pictures for different values of $n$ and $k$), and contatenating with $\alpha$ backtracks a half circle and we get $[f*\alpha] = (|k| - 1)[\gamma^{-1}] =  -(|k| - 1)[\gamma] = (k+1)[\gamma]$. Similarly if $m= 2l + 1$ then $[g * \alpha] = (l+1)[\gamma]$. Again since $\pi_1(S^1)$ is generated by $[\gamma]$ and the map $[\_*\alpha]$ is a bijection it follows that $f$ and $g$ are homotopic relative endpoints iff $n = m$.

For your concrete example $n = 1$ and $m = -1$, then $k = 0$ and $l = -1$ and we get $[f * \alpha] = [\gamma]$ and $[g* \alpha] = 0$. In fact $g\sim \alpha^{-1}$ in $\Omega_{p,q}(S^1)$.
