Prove that $f$ is monomorphism

Question

I have one exercise related to category theory and I don't know how to do it.

On the following commutative square, prove that if $g$ is monomorphism, then $f$

$$\require{AMScd} \begin{CD} A @>f>> C \\ @V\alpha_1VV @VV\alpha_2V \\ B @>g>> D \end{CD} $$

I've done the following: Let $s,t:Z \to A $ two morphism such that $fs=ft$. Then $\alpha_2fs=\alpha_2ft$, as the square is commutative, we have that $g\alpha_1s=g\alpha_1t$. As $g$ is monomorphism, we have that $\alpha_1s=\alpha_1t$. And I don't know how to continue...

Thanks for your help

Answer 1

It is wrong, for example:

$$\require{AMScd} \begin{CD} \mathbb{Z} @>\pi>> \mathbb{Z}_2 \\ @V0VV @VV0V \\ 0 @>1_0>> 0 \end{CD} $$

We have that $1_00=0\pi$, $1_0$ is monomorphism but $\pi$ is not.