Proof Verification: $\epsilon(\sigma)=\epsilon(\sigma^{-1}) \, \, \forall{}\sigma\in{}S_n$ 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?

When you say $n=m-2$, that should be $n\equiv m-2\mod 2$. The value of $\epsilon(\sigma)$ doesn't tell you what that exponent is, only that it's even or odd.
In fact, nothing about this depends on that "number of inversions" formula. It's entirely a consequence of $\epsilon$ being a homomorphism to the two-element group.
The way I would phrase it? $\epsilon(\sigma^{-1})=(\epsilon(\sigma))^{-1}$ since $\epsilon$ is a homomorphism. Then, in the two-element group $\{1,-1\}$, every element is its own inverse, so $(\epsilon(\sigma))^{-1}=\epsilon(\sigma)$. Done.
