Conclude $L_{[a]}$ is not injective if it is not surjective and use this to show the existence of $[b] \in \mathbb{Z_{n}}$ s.t. $[a][b] = [0]$

The full question is: Suppose that $[a]$ is not invertible in $\mathbb{Z_{n}}$. Consider the left multiplication map $L_{[a]} : \mathbb{Z_{n}} \rightarrow \mathbb{Z_{n}}$ defined by $L_{[a]}([b]) = [a][b]$. Since $[a]$ is not invertible, $[1]$ is not in the range of $L_{[a]}$, so it is not surjective.

1. Conclude that $L_{[a]}$ is not injective
2. Use this fact to show that there exists a $[b] \neq [0] \in \mathbb{Z_{n}}$ such that $[a][b] = [0]$

For this question I believe I have an answer for part (1.) but I need a little help for part (2.):

1. Assume $L_{[a]}$ is injective. If $\mathbb{Z_{n}}$ has $n$ elements, then by injectivity of $L_{[a]}$ we can say that the image of $L_{[a]}$ has at least $n$ elements. The image of $L_{[a]}$ is contained within $\mathbb{Z_{n}}$ so it has at most $n$ elements. Therefore the image of $L_{[a]}$ when $L_{[a]}$ is injective has exactly $n$ elements. However, we know that $[1]$ is not contained in the image of $L_{[a]}$ per the question statement, therefore $L_{[a]}$ cannot be injective.
2. For this part I know I am supposed to use the fact that $L_{[a]}$ need not obey the rules that:
   • if $f(x_{1}) = f(x_{2})$, then $x_{1} = x_{2}$
   • $\forall [y] \in \mathbb{Z_{n}}, \exists [x] \in \mathbb{Z_{n}}$ such that $L_{[a]}([x]) = [y]$
   • What this amounts to is the fact that not every element of the codomain needs to have an element in the domain associated to it and some elements of the image can have more than one element of the preimage associated to it.
     • We can see this with $[1]$ in the codomain having no elements of the domain which is left multiplied by $[a]$ to obtain it. So, what is it specifically about allowing 2 or more elements from the preimage to have the same image that implies the existence of $[0]$?

Any help or nudge in the right direction is greatly appreciated. Also, if my answer for part (1.) is incorrect any illumination or criticism is appreciated as well.

Your proof for 1 seems to be right

By the failure of injectivity, we have $[x_1] \neq [x_2]$ with $L_{[a]}([x_1]) = L_{[a]}([x_2])$ so we have $$[a][x_1] = [a][x_2]$$ $$[a][x_1] - [a][x_2] = 0$$ $$[a][x_1-x_2]=0$$ But as $[x_1] \neq [x_2]$, $[x_1-x_2] \neq 0$ so we have the result