The following is taken from Modules an approach to linear algebra by Blyth
$\color{Green}{Background:}$
$\textbf{Example:}$ If $f:A\to B$ is a morphism of abelian groups then we have the exact sequence
$0\xrightarrow{}\text{Ker }f\xrightarrow{i}A\xrightarrow{\sharp}A/\text{Ker }f\xrightarrow{}0\quad (*)$
in which $i$ is the canonical incolusion and $\sharp$ is the canonical epimorphsm. Likewise, we havve an exact sequence
$0\xrightarrow{}\text{Im }f\xrightarrow{} B\xrightarrow{}B/\text{Im }f\xrightarrow{}0\quad (**)$
$\color{Red}{Questions:}$
I don't understand the notation for the maps $0\xrightarrow{}\text{Ker }f$ and $0\xrightarrow{}\text{Im }f.$ Normally, if I have a morphism $0\to M,$ where $M$ is some algebraic object, then $0\to M,$ would mean $0\mapsto 0\in M.$ But for the case of $0\xrightarrow{}\text{Ker }f\xrightarrow{i}A,$ do I interpret as follow:
Givevn $0\to \text{Ker }f $ to mean $0\mapsto 0\in\text{Ker }f,$ and $f$ being a morphism of abelian groups from $A$ to $B$ gives $f(0)=0,$ where $0$ denotes the identity element. Then $\text{Im }(0\to \text{Ker }f)=\text{Ker }f=\{0\}.$ But $i:\text{Ker }f=0\to A:=i:0\mapsto A,$ and this implies $\text{Ker }i=\text{Ker }f=\{0\}$ hence $\text{Im}(0\to \text{Ker } f)=\text{Ker }i.$
Similarly, for $0\xrightarrow{}\text{Im }f\xrightarrow{}B.$ in $(**).$ $0\xrightarrow{}\text{Im }f$ means $0\mapsto 0\in \text{Im }f,$ and $f$ being a morphism of abelian groups from $A$ to $B,$ then $f(0)=0$ where $0$ denotes the identity element. So $\text{Im }(0 \to \text{Im }f)=\text{Ker }f=\{0\}.$ But $\text{Ker }(\text{Im }f\to B)=\{0\}.$ Hence $\text{Im }(0 \to \text{Im }f)=\text{Ker }f=\text{Ker }(\text{Im }f\to B)=\{0\}.$ But what happens if $\text{Ker}f\neq \{0\}?$
Thank you in advance.
I believe your second SES is $$0 \rightarrow \text{Im}(f) \rightarrow B \rightarrow B / \text{Im}(f) \rightarrow 0$$
Since Ker(f) $\hookrightarrow A $ is an inclusion, it is injective. This is how we should read the sequence $$ 0 \rightarrow \text{Ker}(f) \stackrel{i}{ \rightarrow }A $$
It says that $i$ is an injective map. As you say, $0$ maps to $0$.
Similarly for the second sequence.