The following is taken from pg 37-38 of: Further Algebra with Applications by: P Cohn. It is also a continuation of Meaning of: "$M'$ is the kernel of the canonical surjective morphism" and "$\text{Im} f$ is the kernel of $\text{Coker }f$?"
$\color{Green}{Background:}$
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 $X$.
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 $\text{ker }\alpha.\quad (1)$
Dually the cokernel of $\alpha:X\to Y$ is an initial object in the category of all maps $\mu:Y\to C$ such that $\alpha\mu=0$. This is a quotient object of $Y$, unique up to equivalence if it exists; it (or also the map from $Y$) will be denoted by $\text{coker }\alpha$. Given any map $\alpha:X\to Y,$ assume that $\text{ker }\alpha$, $\text{coker }\alpha$ exist. Then we can define two further objects, the image of $\alpha$, a subobject of $Y,$ and the $\textit{coimage}$ of $\alpha$, a quotient object of $X$:
$$\text{im }\alpha =\text{ker } \text{coker }\alpha, \text{coim }\alpha =\text{coker } \text{ker }\alpha\quad (2)$$
Again they need not exist, but if they do, they are unique up to equivalence. Further we have the following diagram:
Here $\text{coim }\alpha$ is the largest quotient of $X$ kiling $\text{ker }\alpha$, hence there is a map $\kappa:\text{coim }\alpha\to Y$ such that $\alpha=(\text{coim }\alpha)\kappa$. It follows that ($\text{coim }\alpha)\kappa(\text{coker }\alpha)=0;$ but $\text{coim }$ is epic, so $\kappa(\text{coker }\alpha)=0,\quad (3)$
and since $\text{im }\alpha$ is the largest subobject of $Y$ killed by $\text{coker }\alpha$, there is a unique map $\alpha:\text{coim }\alpha\to \text{im }\alpha$ to make the diagram commute. If we had or proceeded in dual fashion, starting from $\text{im }\alpha$ and going via a map $X\to \text{im }\alpha$, we would have obtained another map $\alpha';:\text{coim }\alpha\to \text{im }\alpha$ to make the square commute,
Now $\alpha=(\text{coim }\alpha)\alpha'(\text{im }\alpha)=(\text{coim }\alpha)\alpha''(\text{im }\alpha);\quad (4)$
since $\text{coim }\alpha$ is epic and $\text{im }\alpha$ is monic, it follows that $\alpha'=\alpha''$, so the maps coincide and there is complete symmetry. In important cases $\alpha'$ is an isomorphism.
$\color{Red}{Questions:}$
I have some questios about different places of the quoted passages above. I have indented the relevant passage and numbered them for ease of references. In the $(1)$ above, where did the map $f$ come from? Also when it says: the map $\lambda$ to $X,$ will be denoted by $\text{ker }\alpha$. What does Cohn mean by the map $\lambda$ to $X$? The map $\lambda$ should be a map from $A\to X$.
In $(2)$, the notations: $\text{ker } \text{coker }\alpha, \text{coker } \text{ker }\alpha$, what does it mean in terms of maps. So if I have a map $\alpha$, do I take the cokernel of $alpha$, and then create another map out of that, then take the kernel of that resulting map. SImilarly for $\text{coker } \text{ker }\alpha$.
For $(3)$ and $(4),$ the notations: $(\text{coim }\alpha)\kappa$, $\text{coim }\alpha)\kappa(\text{coker }\alpha)$, $\kappa(\text{coker }\alpha)$, $(\text{coim }\alpha)\alpha'(\text{im }\alpha),(\text{coim }\alpha)\alpha''(\text{im }\alpha)$, are they composition of maps, sets, or both? Say $(\text{coim }\alpha)\kappa$, is that a compositoin of a direct image of a map with another map? or is it a multiplication of a set with a map?
Thank you in advance