Signature of permutations is a homomorphism Given the following definition of $signature$:
$\epsilon(\sigma)=(-1)^{n-k}$, where $k$ is the number of cycles (with disjoint supports, counting the 1-cycles) of the permutation, prove that $\epsilon$ is a homomorphism.
Now, in the preceding exercise, we are asked to show:
$(ab)(ax_1x_2...x_kby_1y_2...y_l)=(ax_1x_2...x_k)(by_1y_2....y_l)$
and
$(ab)(ax_1x_2...x_k)(by_1y_2...y_l)=(ax_1x_2...x_kby_1y_2...y_l)$
I was able to do this, but this only holds if everything is distinct, so I don't know how to apply this result to prove that $\epsilon$ is a homomorphism (and apparently this is what I'm supposed to do).
 A: Hint: For showing  $\epsilon(\sigma_1\sigma_2)\equiv\epsilon(\sigma_1)+\epsilon(\sigma_1)$ mod $2$ where $\sigma_i\in S_n$ for $i=1,2$
First, show that $\epsilon(\sigma\lambda)=\epsilon(\sigma)\pm1$ where $\lambda$ is transposition.
Second, A permutation cannot be written as a product of both an odd and an
even number of transpositions.
A: This is an expansion of the other answer, as I caught myself thinking about this again and would like to have a complete line of thought safeguarded.

Defining $\epsilon$ as in the post, we have that $\epsilon$ is well-defined, since the decomposition of a permutation $\sigma$ in disjoint cycles is unique (This can be seen by considering the group action of the group generated  by $\sigma$ on the set $\{1,\cdots, n\}$. The orbits of this action are representations of the cycles.)
Given a permutation $\sigma$, by the equalities shown and commutativity of disjoint cycles, it follows that, for any given permutation $\tau$, $\tau \sigma$ has its decomposition given by one more cycle or by one less cycle. Either way, it follows that $\epsilon(\tau \sigma)=-1$.
Since any cycle $(a_1a_2\cdots a_n)=(a_2a_1)(a_3a_2)\cdots(a_{n-1}a_{n-2})(a_n a_{n-1})$, we have that every permutation is a composition of transpositions. Now, let $\sigma=\Pi_{i=1}^n \tau_i$. We have by induction that $\epsilon(\sigma)=(-1)^n$. (Note that we don't need to prove that permutations can't be written as different parities of transpositions: this is a consequence, given our definitions).
From here, given two permutations $\sigma_1,\sigma_2$, decompose both in transpositions $\Pi_{i=1}^n \tau_i, \Pi_{j=1}^m \pi_j$. We have that $\sigma_1 \sigma_2$ is a product of $m+n$ transpositions. Therefore, $\epsilon(\sigma_1\sigma_2)=(-1)^{n+m}=(-1)^n(-1)^m=\epsilon(\sigma_1)\epsilon(\sigma_2)$. 
