# Isomorphism from R^n via n-Term Sequence

## Theorem

Let $G$ be a unitary $R$-module.

Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $G$.

Let $R^n$ be the $R$-module $R^n$.

Let $\psi: R^n \to G$ be defined as:

$\displaystyle \map \psi {\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n} } = \sum_{k \mathop = 1}^n \lambda_k a_k$

Then $\psi$ is an isomorphism.

## Proof

By Unique Representation by Ordered Basis, $\psi$ is a bijection.

We have:

 $\displaystyle \sum_{k \mathop = 1}^n \lambda_k a_k + \sum_{k \mathop = 1}^n \mu_k a_k$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \paren {\lambda_k a_k + \mu_k a_k}$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \paren {\lambda_k + \mu_k} a_k$

and we have:

 $\displaystyle \beta \sum_{k \mathop = 1}^n \lambda_k a_k$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \beta \paren {\lambda_k a_k}$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \paren {\beta \lambda_k} a_k$

thus proving that $\psi$ is also a homomorphism.

$\blacksquare$