Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual

Theorem
Let $\struct {R, +, \times}$ be a commutative ring with unity.

Let $G$ be an $R$-module.

Let $G^*$ be the algebraic dual of $G$.

Let $G^{**}$ be the double dual of $G$.

For each $x \in G$, let $x^\wedge: G^* \to R$ be defined as:
 * $\forall t \in G^*: \map {x^\wedge} t = \map t x$

Then:
 * $x^\wedge \in G^{**}$

Proof
We have that $x^\wedge$ is a mapping from $G^* \to R$.

It remains to be demonstrates that $x^\wedge$ is in fact a linear transformation.

Hence we need to show that:


 * $(1): \quad \forall u, v \in G^*: \map {x^\wedge} {u + v} = \map {x^\wedge} u + \map {x^\wedge} v$
 * $(2): \quad \forall u \in G^*: \forall \lambda \in R: \map {x^\wedge} {\lambda \times u} = \lambda \times \map {x^\wedge} u$

Hence:

and:

Hence the result.

Also see

 * Definition:Evaluation Linear Transformation/Module Theory, which uses $x^\wedge$ in its definition