$\color{Green}{Background:}$
In the context of the topic of exact sequences, I often see the following:
suppose I have a homomorphic map between groups, rings, modules, etc or a linear transformations between vector spaces: $\varphi:A\to B,$ and let $\text{ker }\varphi,$ denote the kernel of $\varphi,$ $\text{coim }\varphi=A/ \text{ker }\varphi,$ $\text{Im }\varphi,$ to denote the image of $\varphi,$ $\text{coker }\varphi=B/\text{Im }\varphi.$ In terms of elements, we have if $a\in A,$ then $\varphi(a)=b$ for some $b\in B.$ (Depending on context in which $\varphi$ is defined, it might not make sense to discuss $B\to \text{coker }\varphi,$ etc.) Also, from the map $\varphi,$ I can create the following list of different maps:
$f_1:\text{ker }\varphi\to A,\quad (1)$
$f_2:A\to A/\text{ker }\varphi,\quad (2)$
$f_3:\text{Im }\varphi\to B,\quad (3)$
$f_4:B\to B/\text{Im }\varphi,\quad (4)$
$f_5:A/\text{ker }\varphi\to \text{Im }\varphi,\quad (5)$
$f_6:\text{ker }\varphi\to A/\text{Im }\varphi\quad (6).$
Equivalently, $(4)$ can be written as $f_4:B\to \text{coker }\varphi,$ $(5)$ can be written as $f_5:\text{coim }\varphi\to \text{Im }\varphi,$ and $(6)$ can be written as $f_6:\text{ker }\varphi\to \text{coker }\varphi.$
$\color{Red}{Questions:}$
For describing the above five maps in terms of elements for the above six maps, are the following correct?
for $(1)$, if $a\in \text{ker }\varphi\subset A,$ then is $f_1$ can be described as $f_1(a)=a,$ and $(f_1\circ \varphi)(a)=f_1(\varphi(a))=f(0)=0.$
for $(2)$ if $a\in A,$ then $f_2$ can be defined as $f_2(a)=a+\text{ker }\varphi.$
for $(3)$ if $b\in \text{Im }\varphi \subset B,\exists a\in A,$ so that $\varphi(a)=b.$ But $\text{Im }\varphi\subset B,$ so $f_5(\varphi(a))=f_5(b)=b$.
for $(4)$ if $a\in A, \exists b\in B, \varphi(a)=b,$ then we have $A\xrightarrow{\varphi}B\xrightarrow{f_4}B/\text{Im }\varphi,$ defined as $f_4(\varphi(a))=f_4(b)=b+\text{Im }\varphi.$
for $(5)$ if $a\in A,$ then $f_5$ can be described as $f_5(a+\text{ker }\varphi)=\varphi(a).$
for $(6)$ if $a\in \text{ker }\varphi\subset A,$ then is $f_6$ can be described in terms of $f_6(a)=a+\text{Im }\varphi,$ and $f_6$ is surjective since it sends elements to the equivalence class of $A/\text{Im }\varphi.$
Thank you in advance.
This is all about notation, definitions and keeping your wits about you (good practice for diagram chases).
For example:
(1) $ker\space\varphi$ is a subgroup of $A$ so $f_1$ is the inclusion: $f_1(a)=a$. For the second statement here, draw a diagram
$$ker\space\varphi\xrightarrow{f_1}A\xrightarrow{\varphi} B$$
So $\varphi(f_1(b))=0$ where $b\in ker\space\varphi$ [order matters; noted in comments]
(2) $f_2(a)= a + ker\space\varphi$; you are just sending $a$ to its equivalence class under $\varphi$. Note that if $\varphi(a)=\varphi(b)$ then $f_2(a)=f_2(b)$.
The others are SIMILAR; try to work it out slowly and carefully just pay attention to where elements live and what the terms mean.
Added to finish (6):
The map must land in $coker\space\varphi=B/Im\space\varphi$
$f_6:ker\space\varphi\rightarrow B/Im\space\varphi$
note that it is the composition
$ker\space\varphi\rightarrow A\rightarrow B\rightarrow B/Im\space\varphi$
where the first map is the inclusion $f_1$, the second is $\varphi$ and the third is $f_4$. So to describe it, starting with some $a\in ker\space\varphi$, $\varphi(a)=0$ ultimately this composite is $0$. You may have intended just
$A\rightarrow coker\space\varphi$ which takes $a$ to the equivalence class of $\varphi(a)$.