All of the following are discussed in an abelian category $\mathcal{C}$. Given $X \stackrel{g}{\rightarrow} Y \stackrel{f}{\hookrightarrow}Z$. Consider the morphism $$\mathrm{coker}\ g \stackrel{k}{\rightarrow} \mathrm{coker}\ fg$$ given by $$\require{AMScd} \begin{CD} Y @>f>> Z\\ @VVV @VVV\\ \mathrm{coker}\ g @>k>> \mathrm{coker}\ fg \end{CD} $$ Is it a monomorphism? I know it's obvious in the category $\mathrm{Mod}(R)$ where R is a ring. But does this hold in general abelian categories and how to prove it?
Thank you.
As Daniel Schepler notes in the comments, the diagram you wrote down is a pushout diagram.
To see this, consider the universal morphisms $q_Y\colon Y\rightarrow \operatorname{coker}(g)$ and $q_Z\colon Z\rightarrow \operatorname{coker}(f\circ g)$. Let $j_1\colon\operatorname{coker}(g)\rightarrow Q$ and $j_2\colon Z\rightarrow Q$ be morphisms in $\mathcal{C}$ such that $j_1\circ q_Y=j_2\circ f$. By definition of the cokernel, precomposing with $g$ yields $j_2\circ f \circ g=j_1\circ q_Y \circ g= j_1\circ 0=0$. Since $q_Z$ is universal, there is thus a unique morphism $u\colon\operatorname{coker}(f\circ g)\rightarrow Q$ such that $u\circ q_Z=j_2$. We also have $u\circ k\circ q_Y=u\circ q_Z\circ f=j_2\circ f=j_1\circ q_Y.$ Since cokernels are epimorphisms, it follows that $u\circ k =j_1.$ This shows that the square is indeed a pushout diagram.
Since in an abelian category monomorphisms are stable under pushouts, it follows that $k$ is a monomorphism. The dual statement that epimorphisms are stable under pullbacks is, for instance, proved in Proposition 2 of Chapter VIII.4 in Mac Lane's Categories for the Working Mathematician.