This only holds if $ψ^{-1}(H_2)=H_1$, then use the isomorphism theorem:

$G_1 \to G_2 \to G_2/H_2$ is surjective. The kernel of the composition is then $H_1$. Now apply the isomorphism theorem and you're done.

Otherwise you could take for example $Z\times Z \to Z$ by $(1,1)\mapsto 1$ then the kernel is generated by $(1,-1)$ and you could take $H_1 = 0 \times Z$, but $Z \neq 0$.

Isomorphism between quotient groups with normal subgroups
Similar question: Using Silverlight for an entire website?

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