Not certain whether my proof is right, would appreciate it if I could get some feedback on it. Also, the epsilon here is the sign function of the permutation so $\epsilon=sgn$
Proof:
Since the mapping $ \, \, \epsilon:S_n\rightarrow\{\pm{}1\}$ is a group homomorphism, I'll use the fact that
\begin{equation}
\epsilon(\sigma\tau)=\epsilon(\sigma)\cdot\epsilon(\tau)
\end{equation}
And let $\tau=\sigma^{-1}$ which will give us $\epsilon(\sigma\sigma^{-1})=\epsilon(e)$ with $e$ the identity permutation. And since the identity permutation has sign $1$ we have that $\epsilon(\sigma)\cdot\epsilon(\sigma^{-1})=1$.
Now since $\epsilon(\sigma)=(-1)^{\text{number of inversions of} \, \sigma}$ and $\epsilon(\sigma^{-1})=(-1)^{\text{number of inversions of} \, \sigma^{-1}}$ let $n$ and $m$ denote those powers respectively (to avoid cumbersome notation) we then have that
\begin{align}
&(-1)^n\cdot(-1)^m=1\\
&(-1)^{n+m}=(-1)^2\\
&n=m-2
\end{align}
And since $m-2$ doesn't alter the sign of $(-1)$ (there's a better way of saying this) we have that the number of both inversions is the same and hence the sign of both permutations is also the same.
How does this look?
Best Answer
You can't conclude that $n = m - 2$, only that $n+m$ is even, but that's enough to finish. (It's also just not true: $(12)^{-1} = (12)$, so in this case $n=m$, not $n=m-2$.)
You can also stop at $\epsilon(\sigma) \cdot \epsilon(\sigma^{-1}) = 1$. This already tells you that either $\epsilon(\sigma)=\epsilon(\sigma^{-1}) = 1$ or $\epsilon(\sigma)=\epsilon(\sigma^{-1}) = -1$.