The following is taken from linear algebra and geometry by Leung
$\color{Green}{Background:}$
The $\textit{cokernel}$ of a linear transformation $c:X\to Y$ is defined as
$$\text{Coker }c=Y/\text{Im }c$$
$(a)$ Show that if the diagram of linear spaces and linear transformations $a$ and $d$ define linear transformations
$$\begin{array}{ccccccccc} X' & \xrightarrow{a} & X \\ b\big\downarrow & & \big\downarrow c \\ Y' & \xrightarrow{d} & Y \\ \end{array}$$
is commutative, then the linear transformations $\text{Ker }b \to \text{Ker } c,$ $\text{Im }b \to \text{Im } c,$ and $\text{Coker }b \to \text{Coker } c.$
$\color{Red}{Questions:}$
For the exercise above, I am having trouble completing it due to not certain if my commutative diagrams are correct and also making use of the transformations given.
For the case of: $\text{Coker }b \to \text{Coker } c$
I think I have the following commutative diagram.
$$\begin{array}{ccccccccc} X' & \xrightarrow{a} & X \\ b\big\downarrow & & \big\downarrow c \\ Y' & \xrightarrow{d} & Y \\ q'\big\downarrow & & \big\downarrow q \\ Y'/\text{Im }b & \xrightarrow{r} & Y/\text{Im }c \\ \end{array}\quad\quad (1)$$
I need to show the existence of $r:Y'/\text{Im }b\to Y/\text{Im }c$ as a linear transformation, and show that $c$ is linear.
For the existence part, I think I need to show that the explicit map for $r$ is $r(y'+\text{Im } b)=y+\text{Im } c$ exists. I am having a difficult time doing so.
$r$ being well defined is pretty easy to show after showing its existence. From commutative diagram $(1)$ we know that $c\cdot a = d\cdot b,$ we also have the following facts from the diagram:
$(q\cdot d)(y')=q(d(y'))=y+\text{Im }c$ and $(r\cdot q')(y')=r(q'(y'))=r(y'+\text{Im } b)$
I don't know how to make use of $c\cdot a = d\cdot b,$
For the case of: $\text{Ker }b \to \text{Ker } c$
I am not sure if the commutative diagram should either be
$$(a)\quad\begin{array}{ccccccccc} \text{Ker }b & \xrightarrow{w} & \text{Ker }c\\ m\big\downarrow & & \big\downarrow n \\ X' & \xrightarrow{a} & X \\ b\big\downarrow & & \big\downarrow c \\ Y' & \xrightarrow{d} & Y \\ \end{array} \text{or }\quad (b) \quad\begin{array}{ccccccccc} X' & \xrightarrow{a} & X\\ b\big\downarrow & & \big\downarrow c \\ Y' & \xrightarrow{d} & Y \\ b^{-1}\big\downarrow & & \big\downarrow c^{-1} \\ \text{Ker }b & \xrightarrow{w} & \text{Ker }c \\ \end{array}\quad (2)$$
where $b^{-1}, c^{-1}$ both denote inverse image maps.
For the case of: $\text{Im }b \to \text{Im } c$
I think the commutative diagram might be:
$$\begin{array}{ccccccccc} X' & \xrightarrow{a} & X \\ b\big\downarrow & & \big\downarrow c \\ X'/\text{Ker }b & \xrightarrow{r} & X/\text{Ker }c \\ t\big\downarrow & & \big\downarrow v \\ Y' & \xrightarrow{z} & Y \\ \end{array}\quad (3)$$
where the linear transformations $t, v$ are suppose to be bijections. I am trying to make use of the first isomorphism theorem for linear transformation for vector spaces. But I think it is suppose to be for the direct image of $b, c$ and not $t,v.$
Thank you in advance
Nota that for the kernel, you do not have an inverse map $b^{-1}$! The defining maps for kernel and cokernel are as follows (note the "dual" nature of theit respective arrows, which is also why the cokernel is called cokernel in the first place). $$ \ker b\to X'\stackrel b\to Y'\to \operatorname{coker} b $$ The arrow on the left is the canonical inclusion (viewing the kernel as a subspace of $X'$), and the the arrow on the right is the canonical projection to the quotioent $Y'/\operatorname{im} b$. Since these are the maps that you are granted automatically and without any choice or arbitrariness (you may learn more on that in the context of so-called universality), you should also try to construct the desired maps only from these.
So, combined you have $$ \begin{matrix}\ker b&\to& X'&\to& Y'&\to& \operatorname{coker} b \\&&\downarrow&&\downarrow\\ \ker c&\to& X&\to& Y&\to& \operatorname{coker} c \end{matrix}$$ You already found your way to define the maps between the cokernels: From $\operatorname{coker} b$ go left to $Y'$, then down to $Y$, then right to $\operatorname{coker} c$. In the first step, you made a choice, but can show that the final result is well-defined nevertheless.
For the kernels, you can in fact do the exact same thing (more precisely: the dual thing). Go right, down, left and give a conclusive argument why the go-left step is possible.