Let $\alpha : G → H$ be a homomorphism, and suppose that $\alpha (g) = h$. For any $a \in G$, show that $\alpha(a) = h$ if and only if $a = gk$ for some $k \in \text{ker} (\alpha)$.
My attempt: Suppose that $a=gk, k \in \text{ker} (\alpha)$. Then
$\alpha(a) =\alpha(gk) = \alpha(g) \alpha(k) =\alpha(g) e=\alpha(g) =h$
Conversely, suppose that $\alpha$ is a homomorphism. If $\alpha(a) =\alpha(g) $, then $\alpha(a)(\alpha(g)) ^{-1} =e$. So
$h=\alpha(a) =\alpha(g)\alpha(a)( \alpha(g)) ^{-1}$
Let $k=ag^{-1}$ then
$h=\alpha(a) =\alpha(gk) $.
Thus $a=gk, k\in \text{ker} (\alpha)$.
Am I right?
Your proof is fine.
You could improve it by making it a little clearer how you conclude that $k\in\ker(\alpha)$.