The following is taken from pg 37-38 of: Further Algebra with Applications by: P Cohn. It is also a continuation of this post
$\color{Green}{Background:}$
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 $X$:
$$\text{im }\alpha =\text{ker } \text{coker }\alpha, \text{coim }\alpha =\text{coker } \text{ker }\alpha\quad (1)$$
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 $\kappa(\text{coker }\alpha)=0,\quad (2)$
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 $\alpha=(\text{coim }\alpha)\alpha'(\text{im }\alpha)=(\text{coim }\alpha)\alpha''(\text{im }\alpha);\quad (3)$
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 am having trouble deciphering notations from the above passage. For $(1)$ I understand that if we have a map $f:X\to Y,$ the the kernel of the map $f$ can be denoted using the inclusion map $i:\text{ker }f\to A,$ and the cokernel of the map $f$ can be denoted by $w:Y\to \text{coker }f,$ where $\text{coker }f=Y/\text{im }f.$
But then why is $\text{im }\alpha =\text{ker } \text{coker }\alpha$ and $\text{coim }\alpha =\text{coker } \text{ker }\alpha$? I look online for some explanations about this, apparently from beginning of section two of Jeffrey Carlson's notes, he cites Saunders Mclane's Category theory text and list the following two definitions:
The image $\text{im }f,$ is $\text{ker } \text{coker }f:\text{Im }f\to Y$
and
The coimage $\text{coim }f$ is $\text{coker } \text{ker }f:X\to \text{Coim }f.$
I understand that coimage of $f$ is defined as $A/\text{ker }f.$ and image of $f$ is just $f(A).$ What does or how does the notations $\text{coker } \text{ker }f, \text{ker } \text{coker }f.$
For $(2),(3)$ I have the same question about the notations: $\text{coim }\alpha,\text{coim }\alpha)\kappa(\text{coker }\alpha), \kappa(\text{coker }\alpha), (\text{coim }\alpha)\alpha'(\text{im }\alpha),$ and $(\text{coim }\alpha)\alpha''(\text{im }\alpha).$ Are all these compositions between maps? I understand that $\kappa,\alpha, \alpha''$ are maps. But for $\text{coim }\alpha, \text{coker }\alpha,$ etc. Are they each direct images or sets or something else?