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

Outline of Proof
We reduce the case of arbitrary sets to Indexed Summation does not Depend on Permutation.

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

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

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

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


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

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

Also see

 * Change of Variables in Summation over Finite Set