In Linear Algebra by Georgi E. Shilov the author gives the following theorem:
The operator $A:X\to Y$ has a left inverse$\iff$$A$ is a monomorphism. The operator $B:Y\to X$ has a right inverse$\iff$$B$ is an epimorphism.
I will skip the details of the proof as it is rather long, but I will include an image of it for context. The part which has confused me is the following statement:
"However, if $A$ is not a monomorphism, there exists a nonzero vector $x\in X$ such that $Ax=0$".
For example, I found the following that supports this statement.
$$A=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}, \ \ Ax=0 \ \ \ \forall x=\begin{pmatrix}0\\0\\n\end{pmatrix} \ \ \text{where} \ n \in K$$ It is clear to me there are countless examples of this, which begs the question; why does non-monomorphism imply that there always exists some vector to satisfy this relation?
If $A$ is not a monomorphism, there exist $x_1,x_2 \in X$, $x_1 \ne x_2$ such that $Ax_1 = Ax_2$. Then $x_1-x_2 \ne 0$ but $$A(x_1-x_2) = Ax_1 - Ax_2 = 0.$$ Hence $x_1-x_2$ is our desired vector.