Show that map from $\text{Hom$_A$($_AA,M$)} \to M$ is injective

Question

Let $A$ be a unital ring and $M\in\text{A-Mod}$

Then $\text{Hom$_A$($_AA,M$)} \cong M$ By $\phi \mapsto \phi(1)$

I want to show that this map is injective, without using Yoneda Lemma. I have a proof but I don't get one of the steps.

Let $a\in M$, of course $a=a\cdot1$

Take $\phi,\psi \in \text{Hom$_A$($_AA,M$)}$ and assume

$$\phi(1) = \psi(1)$$ Then $$a\phi(1) = a\psi(1), a\in A$$ And here comes the step I don't understand;

$$\implies \phi(a\cdot 1) = \psi(a\cdot1) \implies \phi(a) = \psi(a)$$

Why are we allowed to multiply in $a$ inside both functions in the above manner?

Answer 1

Because $\phi$ and $\psi$ are $A$-module-homomorphisms.