Reversal of a Finite Sum

Let $K$ be a field. Let $\mathbb{N}$ be the set of natural numbers, with addition $+$ as in that definition and with the order $\le$; let $N\in\mathbb{N}$, and let $[N]$ be the initial segment determined by $N$. Sums below are the finite sums of $K$; the field $K$ plays no role in claim 1. Then the following hold.

1. (The reversing map)(i) For every $a\in[N]$ there is exactly one $k\in\mathbb{N}$ with $a+k=N+1$, and this $k$ lies in $[N]$. Writing $\sigma_{N}(a)$ for it, the map $\sigma_{N}:[N]\to[N]$ satisfies $\sigma_{N}(\sigma_{N}(a))=a$ for every $a\in[N]$ and is a bijection. Moreover, if $F$ is an ordered field, with multiplicative identity $1$ and with $x-y$ abbreviating $x+(-y)$, and $\iota_{F}:\mathbb{N}\to F$ is its canonical map, then

$$\iota_{F}\bigl(\sigma_{N}(a)\bigr)=\iota_{F}(N)+1-\iota_{F}(a)\qquad\text{for every }a\in[N].$$

2. (Reversal)(ii) Let $h:[N]\to K$ be a map with values written $h_{a}$. Then

$$\sum_{a=1}^{N}h_{\sigma_{N}(a)}=\sum_{a=1}^{N}h_{a} .$$