I want to find $Ker(b)/Im(a)$ of the following maps:
$a: \mathbb Z \bigoplus \mathbb Z \rightarrow \mathbb Z \bigoplus \mathbb Z \bigoplus \mathbb Z$
$a(x,y)=(x-y,x-y,x-y) $
$b:\mathbb Z \bigoplus \mathbb Z \bigoplus \mathbb Z \rightarrow \mathbb Z \bigoplus \mathbb Z$
$b(x,y,z)=(z-x,0)$
I compute that $Ker(b)=\{(x,y,x) : x,y \in \mathbb Z\}$ and $Im(a)=\{(z,z,z) : z \in \mathbb Z \}$
Hence, $Ker(b)=<(1,0,1),(0,1,0)>$ and $Im(a)=<(1,1,1)>$
At this point, I define the function
$g(x,y,z)=(x-y,y-z,z-x)$ , $(x,y,z) \in Ker(b)$.
I notice that $Ker(g)=Im(a)$ and also I find that $Im(g)=\{(k,l,-k-l) : k,l \in \mathbb Z\} = <(1,0,-1),(0,1,-1)>$
and now using the First Isomorphism Theorem I find that $Ker(b)/Im(a) \simeq Im(g)$.
My question is, is there any other way of finding the quotient group, without thinking of a map $g$ (as here) and applying the First Isomorphism Theorem? I am asking because I dont know if in every situation it is possible to find directly that kind of map.
Thanks in advance.