Summation over Finite Set is Well-Defined

Theorem
Let $G$ be an abelian group.

Let $S$ be a nonempty finite set.

Let $f : S \to G$ be a mapping.

Let $n$ be the cardinality of $S$.

let $\N_{<n}$ be an initial segement of the natural numbers.

Let $g,h : S \to \N_{<n}$ be bijections.

Then we have an equality of indexed summations of the compositions $f\circ g$ and $f\circ h$:


 * $\displaystyle\sum_{i \mathop= 0}^{n-1} f(g(i)) = \displaystyle\sum_{i \mathop= 0}^{n-1} f(h(i))$

That is, the definition of summation over finite set does not depend on the choice of the bijection $g : S \to \N_{<n}$.

Proof
Note that by Cardinality of Nonempty Finite Set is at Least One, $n\geq1$.

By Inverse of Bijection is Bijection, $g^{-1} : \N_{<n} \to S$ is a bijection.

By Composite of Bijections is Bijection, the composition $h\circ g^{-1}$ is a permutation of $\N_{<n}$.

By Indexed Summation does not Depend on Permutation, we have an equality of indexed summations:


 * $\displaystyle\sum_{i \mathop= 0}^{n-1} (f\circ g)(i) = \displaystyle\sum_{i \mathop= 0}^{n-1} (f\circ g)\circ(g^{-1}\circ h)(i)$

By Composition of Mappings is Associative and Bijection Composite with Inverse, the second indexed summations equals $\displaystyle\sum_{i \mathop= 0}^{n-1} f(h(i))$.

Also see

 * Definition:Summation over Finite Set