In group theory, a sequence $$ G_0\stackrel{f_1}{\longrightarrow} G_1 \stackrel{f_2}{\longrightarrow} G_2 \stackrel{f_3}{\longrightarrow}\; . . .\, \stackrel{f_n}{\longrightarrow} G_n $$ of groups and homomorphisms is called exact if the image of each homomorphism is the kernel of the subsequent one: $$ \operatorname{Im}f_i=\,\operatorname{Ker}f_i\quad,\qquad\mbox{for}\quad\mbox 1\leq i < n\;. $$ Also see this picture borrowed from Wikipedia. (Here the sequence is set infinite in both directions, though.)
If some morphism $f_{i+1}$ in this figure is a monomorphism, then its Ker is $e_i$.
After this figure, Wikipedia provides three examples:
Example (a).
In $\,e_0 \longrightarrow X \longrightarrow Y\,$, $\;$the image of $\,e_{\textstyle{_0}}$ is $\,e_{\textstyle{_X}}\,$, $\;$wherefore $\, X \longrightarrow Y\, $ is a monomorphism.
QUESTION:
What is the meaning of the left arrow, $\,e_{\textstyle{_0}} \longrightarrow X\,$?
I thought, there should be some initial group $G_0$ on the left (so we would get $\,G_0 \longrightarrow X\,$).
Or is it implied that this $G_0$ is trivial, i.e., consists of the neutral element $e_{\textstyle{_0}}$ solely?
Example (b).
In $\,X \longrightarrow Y \longrightarrow e\,$, $\;$the kernel of the rightmost map is $Y$. $\;$Hence the image of the map $X\longrightarrow Y$ is all of $Y$. So this is an epimorphism.
Example (c) is a combination of the former and latter ones.
The sequence $0 \longrightarrow X \longrightarrow Y \longrightarrow 0\,$ is exact iff the map $X \longrightarrow Y$ is both a monomorphism and epimorphism (i.e., is an isomorphism).
Wikipedia also defined a short exact sequence as
$$
0\longrightarrow A \stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C \longrightarrow 0
$$
where $f$ is a monomorphism and $g$ is an epimorphism.
I guess, "$0$" on the left is the neutral element $\,e_{\textstyle{_0}}$ of the initial group, while "$0$" on the right is the neutral element $e$ of the final group -- is that right?
Based on Example (b) above, $g$ is epimorphic (surjective) in this short exact sequence.
Based on Example (a), $f$ is monomorphic (injective), assuming that $\,e_{\textstyle{_0}}$ (denoted with "$0$" on the left) is mapped to $\,e_{\textstyle{_A}}$.
But then I again fail to grasp the meaning of the leftmost arrow: why is the preimage of $A$ just a neutral element of the initial group? And why is it mapped exactly to the neutral element of $A$? I am missing something big here.
Here the symbols $e_0, e, 0$ all mean the trivial group. Since a group homomorphism must map identity to identity, in this case there is only one possible homomorphism $e_0\to X$, and so we don't write it above the arrow.