Let $\mathbb{Z} = R$ be our base ring. I am trying to show for a countable direct product of $\mathbb{Z}$ modules there is an isomorphism between it and its dual. I am stuck on the part about surjectivity and I am a little confused because according to Dummite and Foote you can only get surjectivity of the map if the modules are projective and finitely generated. Let me explain the problem in detail:
Let $P = \oplus_{i \in \mathbb{N}} A_i$ where each $A_i = \mathbb{Z}$. How do we to show the map $c_P : P \rightarrow P^{**}$ given by $x \mapsto (y^{*} \mapsto \left< x, y^{*} \right>$ is surjective?
I know how to compute the dual of $\mathbb{Z}^{*} = Hom_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z})$ by showing the mapping of each $y* \in \mathbb{Z}^{*}$ given by $y^{*} \mapsto y^{*}(1)$ is an isomorphism so $\mathbb{Z}^{*} \cong \mathbb{Z}$. Now since the dual of a direct sum is the direct product of corresponding duals we have $P^{*} \cong \prod_{i \in \mathbb{N}} \mathbb{Z} \cong \mathbb{Z} \times \mathbb{Z} \times \ldots $
Form here I don't know what to do to prove the map $c_p$ is surjective. I am confused about the statements I have read saying we need the module to be projective and finitely generated. Is it just the fact that the dual of a direct product should be the direct sum of the duals?