Showing that sign function sends transposition to $-1$. Proposition: There exists a unique function $s:S_n \rightarrow \{-1,1\}$ such that if $\sigma = (ij)$ is a transposition then $s(\sigma) = -1$ and if $\sigma,\tau \in S_n$ then $s(\sigma \tau) = s(\sigma) s(\tau)$.
To do this we introduced the polynomial
$$
\Delta = \prod_{1 \leq i < j \leq n} (x_i-x_j)
$$
and defined
$$
\sigma(\Delta) = \prod_{1 \leq i < j \leq n} (x_{\sigma(i)}-x_{\sigma(j)}).
$$
We then define $s$ to be the constant by which $\sigma(\Delta) = s(\sigma)\Delta$, because the polynomials are just swapped by a sign. To show that $s$ takes transpositions to $-1$, we just counted the number of swaps in the product, but I got lost at this point. I don't understand the logic of how we break up the product. It goes as follows
Let $\sigma = (kl)$ with $k <l$. Then consider
$$
\prod_{1 \leq i < j \leq n} (x_{\sigma(i)}-x_{\sigma(j)}) = \prod_{i<j,i \neq k,  j \neq l} (x_{\sigma(i)}-x_{\sigma(j)}) \; \times (x_l-x_k) \; \times \prod_{j>k, j \neq l} (x_l-x_j) \; \times \prod_{i < l, i \neq k} (x_i-x_k).
$$
So we break this into 4 sub-products. The first product apparently has no sign changes, but why is that the case? If $i<k<j<l$ won't we have sign changes because $\sigma(k) = l$ and $\sigma(l)=k$? I know it isn't much to go on but is there any way to help me understand the notation or how we seem to be partitioning the product. Thanks in advance for any clarifications.
 A: It's easier to see if you write it out longhand. So, suppose $1\le k<l\le n$ and $\sigma$ switches $k$ and $l.$ Then
$(1).\ $ the products
$$(x_1-x_2)(x_1-x_2)\cdots (x_1-x_k),\ (x_2-x_3)(x_2-x_4)\cdots (x_2-x_k),\ (x_3-x_4)(x_3-x_5)\cdots (x_3-x_k),\cdots, \ (x_{k-1}-x_k)$$ and $(x_l-x_{l+1})(x_l-x_{l+2})\cdots (x_l-x_n)$ are left unchanged by $\sigma$ because $l>k.$
$(2). $ The products
$(x_{l+1}-x_{l+2})(x_{l+1}-x_{l+3})\cdots (x_{l+1}-x_n),\ (x_{l+2}-x_{l+3})(x_{l+2}-x_{l+3})\cdots (x_{l+3}-x_n),\cdots, \ (x_{n-1}-x_n)$
are left unchanged because they don't involve $k$ or $l$ at all.
$(3).\ $ The product $(x_k-x_{k+1})(x_k-x_{k+2})\cdots (x_k-x_l)$ is changed because each factor $f$ becomes $(-1)f$ on applying $\sigma$ to it. There are $l-k+1$ of these and so this product is multiplied by $(-1)^{l-k+1}$ in the new product.
$(4).\ $ Each of the products
$(x_{k+1}-x_{k+2})(x_{k+1}-x_{k+3})\cdots (x_{k+1}-x_l),\ (x_{k+2}-x_{k+3})(x_{k+2}-x_{k+4})\cdots (x_{k+2}-x_l),\ (x_{k+3}-x_{k+4})(x_{k+3}-x_{k+5})\cdots (x_{k+3}-x_l),\cdots, (x_{l-1}-x_l)$
is multiplied by $-1$ on applying $\sigma.$ As there are $l-1-(k+1)+1=l-k-1$ of these, $\sigma $ sends this (total) product to $(-1)^{l-k-1}$ times the original.
To finish, combine $(3)$ and $(4): \sigma \Delta = (-1)^{l-k+1}(-1)^{l-k-1}\Delta,$ which is evidently equal to $-\Delta.$
