Q:

$g \circ f$ trivial implies that $Im f \subseteq ker g$

If $f:X_1 \rightarrow  X_2$ and $g:X_2 \rightarrow  X_3$ are homomorphisms. 
If $g \circ f =0$ does it imply that $Im f \subseteq  ker g$? and how to show that? do you have an example?
thanks :)

A:

Take an element $y\in Im f$. Then $y$ is of the form $y=f(x)$, for some $x\in X_1$. Now apply $g$ to $y$. You get $g(y)=g(f(x))=(g\circ f)(x)$, which is zero by assumption. As an example, take $f$ arbitrary and $g(y)=0$ for every $y\in X_2$.

