The following is taken from pg 37 of: Further Algebra with Applications by: P Cohn.
$\color{Green}{Background:}$
Let $A$ be an additive category; given a map $\alpha:X\to Y,$ we shall define the kernel of $\alpha$ as a certain subobject of $X.$ Consider all maps $\lambda: A\to X$ such that $\lambda\alpha=0;$ for fixed $\alpha$ awe obtain a category by taking these maps as $\lambda$ as objects and as morphisms from $\lambda$ to $\lambda'$ maps $\varphi$ from the source of $\lambda$ to that of $\lambda'$ such that $\lambda=\varphi\lambda',$ with the obvious composition rule (obtained by composing maps in $A$). A final object in this category, if one exists, is called a kernel of $\alpha$. Thus a kernel of $\alpha$ is a map $\lambda:A\to X$ such that $\lambda\alpha=0$ and any other map $\lambda':A'\to X$ satisfying $\lambda'\alpha=0$ can be factored uniquely by $\lambda$, i.e. we have $\lambda'=\varphi\lambda$ for a unique map $\varphi$./ This can be expressed more briefly by saying that the kernel is the largest subobject 'killed' (i.e. mapped to $0$) by $\alpha$. The kernel need not exist, but if it does, it is unique up to equivalence, and in fact is a subobject of $X$.
$\lambda=\varphi\lambda' \quad\lambda'=\varphi\lambda\quad (1)$
For let $(A,\lambda)$ be the kernel and assume that $f\lambda=0;$ then by the uniqueness of the factorization, $f=0$. The kernel $A$ of $\alpha$ or also the map $\lambda$ to $X,$ will be denoted by $\text{ker }\alpha.\quad (2)$
$\color{Red}{Questions:}$
The two indented block above are points of confusion for me when reading the quoted passage above. In $(1)$, is $\varphi$ suppose to be an invertible map? Since $\lambda=\varphi\lambda',\Longrightarrow\varphi\lambda=\varphi(\varphi\lambda')\Longrightarrow\lambda'=\varphi\lambda.$ and $\lambda'=\varphi\lambda,\Longrightarrow\varphi\lambda'=\varphi(\varphi\lambda)\Longrightarrow\lambda=\varphi\lambda'.$
For $(2)$ we have earlier, the map $\lambda:A\to X,$ but what does it mean to say that "..also the map $\lambda$ to $X,$ will be denoted..."?
Thank you in advance.