$f$ : $X\to$ $Y$ and $g$ : $Y\to$$Z$ a sequence in abelian categories. Show that if $gf$=$0$ if and only if exist a monomorphism $h$:$Im(f)$ $\to$ $Ker(g)$ such $kh$=$j$, where $j$:$Im(f)$$\to$ $Y$ and $k$: $Ker(g)$ $\to$ $Y$ are the canonical morphisms.
To prove the statement above, my idea to prove the "forward"is that since $gf$=$0$ then $Im(f)$=$Ker(g)$ ? then if this is true I can see $h$ as the identity morphism so the diagram conmute.Can anyone help me please end the proof this statement? Thanks!
Suppose $g\circ f = 0$ and that $h$ exists. Then t.f.d.c. Maps $\overline f$ and $\overline g$ are induced by the universal properties of $\ker g$ and cokernel $f$, respectively.
Recall $f$ factors as $A\overset e\to \textrm{im}\, f\overset j\to B$ with $e$ epi. Then since $0 = c_f\circ f = (c_f\circ j)\circ e$, we see that $c_f\circ j = 0$. Then $g\circ f = \overline g\circ c_f\circ j\circ e = 0$.
Next suppose such an $h$ exists. Let $Y\overset c\to \mathrm{coker}\, f$ be the cokernel of $f$, so $0 = c\circ f = c\circ j\circ e$. Since $e$ is epi, $c\circ j = 0$. Then $g\circ f = g\circ j\circ e = g\circ k_g\circ h\circ e = 0$.