The following retyping of four slides are taken from: Richard Crew's Homological Algebra Lecture 2 and is a continuation of this post Meaning of: "$M'$ is the kernel of the canonical surjective morphism" and "$\text{Im} f$ is the kernel of $\text{Coker }f$?"
$\color{Green}{Background:}$
[Slide 3]
It is $\textit{cocartesian}$ if it makes $X$ a pushout of $Y'\to X'$ and $Y'\to Y. Consider now a commutative diagram
$$\begin{array}{ccccccccc} Y' & \xrightarrow{f'} & X' \\ g'\big\downarrow & & \big\downarrow g \\ Y' & \xrightarrow{f} & X \\ \end{array}\quad (1)$$
in the preabelian category $A.$ If $i':K'\to Y'$ is a kernel of $f', f'i=0$ and thus $fg'i'=gf'i'=0.$ If $i':K\to Y$ is a kernel of $f$ this says that $fg'$ factors uniquely through $i,$ whence a commutative diagram
$$\begin{array}{ccccccccc} K' & \xrightarrow{i'} & Y' & \xrightarrow{f'} & X' \\ h\downarrow & & g'\big\downarrow & & \downarrow g& \\ K & \xrightarrow{i}& Y & \xrightarrow{f} & X \end{array}$$
[Slide 5]
Let's reconsider the diagram
$$\begin{array}{ccccccccc} K' & \xrightarrow{i'} & Y' & \xrightarrow{f'} & X' \\ h\downarrow & & g'\big\downarrow & & \downarrow g& \\ K & \xrightarrow{i}& Y & \xrightarrow{f} & X \end{array}$$
where $K'\to Y'$ is a kernel of $f'$ and $K\to Y$ is a kernel of $f$. We do not assume the right hand square is cartesian; but we can apply the preceding construction to get a larger diagram
$$\begin{array}{ccccccccc} \text{Ker }(h) & \xrightarrow{} & \text{Ker }(g') & \xrightarrow{} & \text{Ker }(g) \\ \big\downarrow & & \big\downarrow & & \big\downarrow & \\ K' & \xrightarrow{i'} & Y' & \xrightarrow{f'} & X' \\ h\big\downarrow & & g'\big\downarrow & & \big\downarrow g& \\ K & \xrightarrow{i}& Y & \xrightarrow{f} & X \\ \end{array} (2)$$
in which all squares are commutative.
[Slide 8]
Suppose $A$ is a preabelian category and $f:X\to Y$ is a morphism in $A.$ Since kernels and cokernels exist in $A$ we get a diagram
$$\text{Ker }(f)\xrightarrow{i}X \text{}\xrightarrow{f}Y\text{}\xrightarrow{p}\text{Coker }(f)$$
in $A.$ By construction pf=0, so there is a unique $q:X\to \text{Ker }(p)$ such that $f=jq,$ where $j:\text{Ker }(p)\to Y$ is a kernel of $p:$
$$\text{Ker }(f)\xrightarrow{i}X \text{}\xrightarrow{q}\text{Ker }(p)\text{}\xrightarrow{j}Y\xrightarrow{p}\text{Coker }(f)$$
Here $j$ is a monomorphism, i.e.$j:\text{ker }(p)\to Y$ is a subobject of $Y$ and is unique up to unique isomorphism. It is called the $\textit{image}$ of $f$ and we denote it by $j:\text{Im }f:\to Y:$
$$\text{Ker }(f)\xrightarrow{i}X \text{}\xrightarrow{q}\text{Im }(f)\text{}\xrightarrow{j}Y\xrightarrow{p}\text{Coker }(f)$$
[Slide 9]
On the other hand, $fi=0$ implies that $f$ $\textit{also}$ factors through this cokernel of $i:\text{ker }f\to X.$ We denote (any chocie of a) cokernel by $k:X\to \text{Coim} f. $ So now we have a picture
$$\text{Ker }(f)\xrightarrow{i}X \text{}\xrightarrow{k}\text{Coim }(f)\text{}\xrightarrow{f''}Y\xrightarrow{p}\text{Coker }(f)$$
in which $f''k=f.$ The cokernel $k:X\to\text{Coim }(f)$ is called $\textit{coimage}$ of $f.$
The next step is to observe that we can $\textit{combine}$ these two pictures. The cokernel $k:X\to \text{Coim}(f)$ is an epimorphism, so $pf=pg''k=0$ implies $pf''=0$ (since $pg''k=0k)$. It follows that $f'':\text{Coim }(f)\to Y$ factors through the kernel $j:\text{Im }(f)\to Y$ of the cokernel $p:Y\to \text{Coker }(f).$ The picture is now
$$\text{Ker }(f)\xrightarrow{i}X\xrightarrow{q} \text{Coim }(f)\xrightarrow{f}\text{Im }(f)\xrightarrow{j}Y\xrightarrow{p}\text{Coker }(f)$$
$\color{Red}{Questions:}$
The definition of a kernel, cokernel, image, coimage for a linear/homormophic map $A\xrightarrow{f}B$ between vector spaces, abelian groups and modules are defined as follows: $\text{ker }f=\{a\in A:f(a)=0\},$ and $\text{coker }f=B/\text{im }f,$ \text{}=,$ $\text{}=.$
But according to Richard Crew's lecture, in various places within the four slides listed above, they are defined for whole morphisms:
In slide 3, we have,
$i':K'\to Y'$ is a kernel of $f', f'i=0,$
$i':K\to Y$ is a kernel of $f$
In slide 5,
$K'\to Y'$ is a kernel of $f'$
$K\to Y$ is a kernel of $f$
In slide 8,
$j:\text{Ker }(p)\to Y$ is a kernel of $p:$
$j:\text{Im }f:\to Y:$ is called the $\textit{image}$ of $f$
In slide 9,
$k:X\to \text{Coim} f. $ is consider a cokernel
In the above four slides what does it mean to refer to entire map/morphism as a cokernel or kernel?
Also in slide 9 when it says,
"cokernel of $i:\text{ker }f\to X.$ "
cokernel $k:X\to\text{Coim }(f)$ is called $\textit{coimage}$
"kernel $j:\text{Im }(f)\to Y$"
"cokernel $p:Y\to \text{Coker }(f).$"
In all three cases, is it talking about a cokernel, kernel coimage of a map or is cokernel, kernel, of the map that it is referring to, meaning:
"cokernel $k:X\to \text{Coim}(f)$," refers to $k:X\to \text{Coim}(f)$ is a cokernel of the map $f$ or does it mean $\text{coker }k?$ or "cokernel $k:X\to\text{Coim }(f)$ is called $\textit{coimage}$", is it $k$ a cokernel for map $f$ or is a comimage for map $f$ or does it denote $\text{coker }k$/$\text{coim }k?$