Let $f : X \to Y$ and $g : Y \to Z$ be homomorphisms of left $R$-modules. Show that there exists an exact sequence of left $R$-modules $$ 0 \rightarrow \operatorname{ker}(f) \rightarrow \operatorname{ker}(g f) \rightarrow \operatorname{ker}(g) \rightarrow \operatorname{coker}(f) \rightarrow \operatorname{coker}(g f) \rightarrow \operatorname{coker}(g) \rightarrow 0. $$
My try: I want to construct a commutative diagram of abelian groups with exact rows and then use Snake lemma. I tried the following diagram $$ {\rm ker}(f)\stackrel{i}{\longrightarrow}X\stackrel{f}{\longrightarrow}f(X)\stackrel{}{\longrightarrow}0 $$ $$ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f \downarrow\quad\ \ \ \ gf\downarrow\ \quad\quad g\downarrow $$ $$ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!0\longrightarrow 0\ \ \longrightarrow\ \ \ \ Z\stackrel{{\rm id}}{\longrightarrow}\ \ \ \ \ Z $$
Is this diagram commutative? How to construct a correct diagram? Thanks a lot.
Use a diagram of the same shape as yours with $X, Y, \operatorname{Coker} f, 0$ in the top row and $0, Z, Z, 0$ in the bottom row with the obvious maps. Then apply the snake lemma.