Prove the endomorphism with surjection is invertible

Question

Let $A$ be a commutative ring with unity.

Let $\pi_a:A \rightarrow A$ be an endomorphism defined as:

$\pi_a(x) = ax$ where $a,x \in A$ (a is fixed)

Prove that the function $\pi_a(x)$ is surjective if and only if $a$ is invertible

I have seen similar questions on this topic, but all the ones I have found deal with topics/ideas (such as modules/metric spaces) I have not studied yet. Also, I am only asking for help on the forward direction.

My attempt (forward direction)

Assume $\pi_a(x)$ is surjective.

Since $\pi_a(x)$ is an endomorphism and surjective, it is also injective (I haven't proved that yet)

Thus, $\pi_a(x)$ is an isomorphism $\Rightarrow$ has an inverse function (we are allowed to use this result of isomorphisms).

Thus, $\pi_a(x)$ is invertible

Side question: does this even answer the original question since we were asked to prove $a$ was invertible (I am confused here)

Answer 1

$\Rightarrow$

Let $A$ be a commutative ring with unity.

Let $\pi_a:A\rightarrow A$ such that $\pi_a(x) = ax$ where $a$ is fixed.

Assume $\pi_a(x)$ is surjective.

Then, $\exists x \in A$ such that $\pi_a(x) = ax = 1_A$ (unity in $A$)

Since $ax=1_A$, $x$ is the inverse of $a,$ thus $a$ is inveritble.

$\Leftarrow$

Let $A$ be a commutative ring with unity.

Let $\pi_a:A\rightarrow A$ such that $\pi_a(x) = ax$ where $a$ is fixed.

Assume $a$ is invertible.

Then, $\forall y \in A$, $\pi_a(x) = y$ when $x = a^{-1}y$ and such an $x$ exists since $y \in A$ and $a$ is invertible.