I am trying to understand exact sequences.
Suppose that we have an exact sequence $$0\rightarrow A\xrightarrow{f}B\xrightarrow{g}C\rightarrow0$$ and a morphism $B\xrightarrow{x}D$ such that $x\circ f=0$.
I do not understand why that implies that $x=y\circ g$ for some morphism $y$.
Since $x\circ f=0$, then $x= h\circ \text{coker}f$ for some morphism $h$. By the same reason, $g=t\circ \text{coker}f$.
I also know that $\text{ker}(\text{coker}f)=\text{ker}g$ because the complex is exact.
However, I do not know how to conclude from here that $x=y\circ g$.
Can anyone help me, please? Thank you very much in advance.
Since in abelian categories all epimorphisms and monomorphisms are normal, $g$ and $f$ are normal. Hence, $$g \cong \text{coker}(\text{ker}g).$$ Moreover, by exactness, $\text{coker}f=\text{coim}g$. That is, $$\text{coker}f\cong \text{coker}(\text{ker}g).$$ Hence, $g\cong \text{coker}f.$