I am trying to solve the following problem:
Let $M$ be an $A$-module. Prove that $\operatorname{Hom}_A(A,M) \cong M$.
I am wondering if the homomorphism that I should show is a group homomorphism or module homomorphism? Also, is my proof of injectivity and surjectivity correct?
Let $f \colon \operatorname{Hom}_A(A,M) \rightarrow M$ be given by $f(\phi) = \phi(1)$.
First. Showing that $f$ is a module homomorphism.
Let $\varphi, \psi $ be $A$-module homomorphisms from $A \to M$, i.e., $\varphi, \psi \in \operatorname{Hom}_A(A,M),$ then $$f(\varphi + \psi) = (\varphi + \psi)(1) = \varphi (1) + \psi(1) = f(\varphi) + f(\psi) \,.$$
Let $r \in A$, and let $\psi \in \operatorname{Hom}_A(A,M)$, then $$f(r \psi ) = r \psi(1) = r f(\psi) \,.$$
Second. Showing that $f$ is injective.
Assume that $f(\varphi) = f(\psi)$ then $\varphi(1) = \psi(1)$ and knowing that a module homomorphism from $A$ to $M$ is completely determined by the image of $1$ then we can deduce that $\varphi = \psi$ as required.
Third. Showing that $f$ is surjective.
Let $m \in M$, define an $A$-module homomorphism $\varphi \colon A \to M$ by $\varphi (1) = m$, we can do this because a module homomorphism from $A$ to $M$ is completely determined by the image of $1$. Therefore, $f(\varphi) = \varphi (1) = m$ as required.