Homotopy invariance in homology i have this from Hatcher's book "Algebric topology"

And i don't understand why we have $i-1$ in $(-1)^{i-1}$ and strict inequality in $P\partial(\sigma)$ ?
Please.
Thank you.
 A: Recall that if $\Delta^n = [v_0, v_1, \ldots, v_{n-1}, v_n]$ then for a map $\sigma\colon\Delta^n\to X$ we define the boundary operator $\partial\colon C_n(X)\to C_{n-1}(X)$ by
$$\partial(\sigma) = \sum_{j=0}^n (-1)^{j} \sigma \mid [v_0,v_1, \ldots, \hat{v}_j,\ldots, v_{n-1},v_n]$$
and if, for an arbitrary chain $\alpha$, we have
$$P(\alpha)=\sum_{i=0}^n(-1)^{i} F \circ (\alpha \times \mathbf{1}) \mid [v_0, \ldots, v_i, w_i, \ldots, w_n]$$ then $P\partial(\sigma)$ will be
$$P\left(\sum_{j=0}^n (-1)^{j} \sigma \mid [v_0,v_1, \ldots, \hat{v}_j,\ldots, v_{n-1},v_n]\right)$$ so we're taking the prism operator on an $(n-1)$-chain. When defining the prism operator we add in a $w_i$ next to $v_i$ as well as change the last $n-i$ vertices from $v_k$s to $w_k$s before restricting the function $F\circ(\sigma\times\mathbf{1})$ to this simplex. But remember if $k>j$ then $v_k$ is actually the $(k-1)$th vertex now because we no longer have $v_j$ in the order. If $k<j$ then $v_k$ is still the $k$th vertex as it's not shifted in its order. So we need to split these two cases up, and the shift accounts for the appearance of the $(-1)^{i-1}$ because odd and even vertex placements in the order have now swapped for all $i> j$. Also, $v_j$ does not appear in any chain for a fixed $j$ now because we've removed it, so the case $i=j$ in the sum doesn't make sense.
To make it a bit more clear, let's suppose that $n=5$, and we're looking at the term corresponding to $j=3$ with $i=1$. For ease of notation I'll ignore all of the function compositions and restrictions and just concentrate on the parity associated to the simplices. So, we start with the simplex $[v_0,v_1,v_2,v_3,v_4,v_5]$. The first thing we do is delete the vertex $v_j$ for $j=3$, and as $3$ is odd this means we have the modifier $-1$ associated to this simplex and we're left with $-[v_0,v_1,v_2,v_4,v_5]$. Now, in the case $i=1$ we will add in a $w_1$ and replace all the $v_k$s with $w_k$s for $k>i=1$. Here $v_i$ is the $1$st vertex in the order* (because $i<j$) which is odd so we add another $-1$ modifier leaving us with $(-1)(-1)[v_0,v_1,w_1,w_2,w_4,w_5]=+[v_0,v_1,w_1,w_2,w_4,w_5]$.
What if we take an example were $i>j$. Let's try $n=5$, $j=3$ but with $i=4$ this time. We again start with $[v_0,v_1,v_2,v_3,v_4,v_5]$ which becomes $-[v_0,v_1,v_2,v_4,v_5]$. Now, $i=4$ means we add a $w_4$ after $v_4$ and replace just $v_5$ with a $w_5$ this time. Also $v_4$ you will see is in the $3$rd place in the order now because we deleted $v_3$ which shifts everything on the right of $v_3$ one place to the left. So we associate another $-1$ modifier because $3$ is odd, giving $(-1)(-1)[v_0,v_1,v_2,v_4,w_4,w_5]=+[v_0,v_1,v_2,v_4,w_4,w_5]$.
Hopefully this highlights the necessity of splitting $i>j$ and $i<j$, and why $i=j$ doesn't actually make sense (because there's no $v_j$ to add a $w_j$ after).

*I'm concious of the fact that $1$st here doesn't actually mean there is nothing that comes before it, because the $0$th position holds that property... such is life when poorly defined English words are mixed with well defined mathematical notation.
