I am trying to prove that the function $\bar g$ is well-defined. This is how I did it:
Define $\bar g: \ker\alpha \to \ker\beta$ by $\bar g(m) = g(m)$, for all $m$ in $\ker\alpha$. This means $m$ is in $A$. By commutativity of the square $ABED$, $\beta (g(m)) = j(\alpha(m)) = j(0) = 0$. Hence, $g(m)$ is in $\ker\beta$. Since $m$ is in $\ker\alpha$, then $\bar g(\alpha) = g(m)$. Hence, $\bar g(\alpha)$ is in $\ker\beta$. Hence, $\bar g$ is well-defined.
I want to know if this is a good way to go about it. I will also appreciate it if you could suggest a better approach.
Thanks in advance.
Yes, it works exactly like that. Since $ABED$ commutes, i.e. $j\alpha = \beta g$, the map $g$ sends every $m \in \ker \alpha$ inside $\ker \beta$. Define $\bar{g}$ as the restriction of $g$ to $\ker\alpha$ as you did.
Just be careful, you wrote $\bar{g}(\alpha)$ at some point, which doesn't make sense, I think you meant $\bar{g}(m)$.