Transpose of Linear Transformation is a Linear Transformation

Theorem
Let $R$ be a commutative ring.

Let $G$ and $H$ be $R$-modules.

Let $G^*$ and $H^*$ be the algebraic duals of $G$ and $H$ respectively.

Let $\mathcal L_R \left({G, H}\right)$ be the set of all linear transformations from $G$ to $H$.

Let $u \in \mathcal L_R \left({G, H}\right)$.

Let $u^t: H^* \to G^*$ be the transpose of $u$.

Then $u^t: H^* \to G^*$ is itself a linear transformation.

Proof
By definition of Evaluation Linear Transformation, $\forall x \in G: y' \in H^*: \left \langle {x, u^t \left({y'}\right)} \right \rangle = \left \langle {u \left({x}\right), y'} \right \rangle$.

Since we have:

and:

it follows that $u^t: H^* \to G^*$ is a linear transformation.