The following is based on an exercise from the book $linear algebra and geometry" by Leung.
$\color{Green}{Background:}$
Given the linear transformation $c:X\to Y$ and the commutative square diagram below also with linear transformations $b,c,d.$
$$\begin{array}{ccccccccc} X' & \xrightarrow{a} & X \\ b\big\downarrow & & \big\downarrow c \\ Y' & \xrightarrow{d} & Y \\ \end{array},\quad (1)$$
Show that the linear transformations $a,d$ define a linear transformation $\text{Coker }b \to \text{Coker }c$
Attempted solution:
From the diagram above, I am given $a\cdot c=b\cdot d,$ I then extend the square diagram like below.
$$\begin{array}{ccccccccc} X' & \xrightarrow{a} & X \\ b\big\downarrow & & \big\downarrow c \\ Y' & \xrightarrow{d} & Y \\ q'\big\downarrow & & \big\downarrow q \\ Y'/\text{Im }b & \xrightarrow{r} & Y/\text{Im }c \\ \end{array}\quad\quad (2)$$
I think I need to show $r(\text{Coker }b) \subset \text{Coker }c,$ in order to show $\text{Coker }b \to \text{Coker }c.$ Elements of $\text{Coker }b$ and $\text{Coker }c$ are respectively $y'+ \text{Im }b$ and $y+\text{Im }c.$ It is easy to show that the linear transformation $r(\text{Coker }b)$ is well defined. So let $x'\in X',$ then $(c\cdot a)(x')=(d\cdot b)(x')$
$\Longrightarrow c(a(x'))=d(b(x')),$
$\Longrightarrow c(x)=d(y'),$
But $(r\cdot q')(y')=(q\cdot d)(y')$
$\Longrightarrow r(q'(y'))=q(d(y')),$
and $q'(y')=y'+\text{Im }b\Longrightarrow r(y'+\text{Im }b)=y+\text{Im }c$
Also, $q(d(y'))=q(c(x))=q(y)=y+\text{Im }c$
Hence $r(y'+\text{Im }b)=q(y)=y+\text{Im }c,$ which implies $r(\text{Coker }b) \subset \text{Coker }c.$ So the linear transformation we are looking for is $r:\text{Coker }b\to \text{Coker }c.$
$\color{Red}{Questions:}$
I posted the above exercise here and here. In both posts, I don't understand from the commenters how to go about fixing the solution to my post. I think from the second post, Osama Ghani adviced me that I can define $$ by looking at a preimage of something in coker $$ in $′$ and then applying $\cdot.$ Is this what he meant?
Let $^{−1}(\{y+\text{Im }b\})=′,$ be the inverse image of $q',$ then $(q'\cdot z^{-1})(\{y+\text{Im }b\})=(q'(z^{-1}(\{y'+\text{Im }b\}))=q'(y').$ Hence $r\cdot (q'\cdot z^{-1})(\{y+\text{Im }b\})=r(q'(y'))=q(d(y'))\Longrightarrow r(y+\text{Im }b)=q(d(y'))=q(c(x))=q(y)=y+\text{Im }c.$
If the above is not what he meant, can someone showm me how to correctly go about showing that the commutative square above can induce a linear transformation $r:\text{Coker }b\to \text{Coker }c.$
Thank you in advance