If $\psi:G \to H$ is a surjective homomorphism, then $|\{g \in G: \psi(g)=h_1\}| = |\{g \in G: \psi(g)=h_2\}|, \forall h_1,h_2 \in H.$ If $\psi:G \to H$ is a surjective homomorphism, then $|\{g \in G: \psi(g)=h_1\}| = |\{g \in G: \psi(g)=h_2\}|, \forall h_1,h_2 \in H.$ 
Could anyone advise on the proof? If $\psi$ is injective, then the result follows. So, what happens if $\psi$ is not injective? By 2nd isomorphism thm, $G/Ker(\psi) \cong H.$ Is this the correct start? Thank you. 
 A: Hint: if $\psi(g_1)=\psi(g_2)$, then $\psi(g_1g_2^{-1})=1$, so $g_1g_2^{-1}\in\ker\psi$. If $K=\ker\psi$, then $g_1K=g_2K$. What about the converse?

Since DonAntonio spoiled the fun, here's the complete answer.
Consider $K=\ker\psi$; then we can factor $\psi$ as $\tilde\psi\circ\pi$, where $\pi\colon G\to G/K$ is the projection and $\tilde\psi$ is injective; it's also surjective, because $\psi$ is.
Thus, for $h\in H$, $\{g\in G:\psi(g)=h\}=\pi^{-1}(\tilde\psi^{-1}(h))=\pi^{-1}(g)=gK$ where $g$ is any element such that $\psi(g)=h$. So the cardinality is just $|gK|=|K|$.

A less “theoretical” way to look at the business is to observe that $\psi(g_1)=\psi(g_2)$ if and only if $g_1g_2^{-1}\in K=\ker\psi$, so if and only if $g_1K=g_2K$ (this is implicit in the above computation). Thus the inverse image  of $h\in H$ is just a coset $gK$, where $\psi(g)=h$. Since all cosets have the same cardinality, the result follows.
A: An idea following egreg's answer: 
$$\psi(g)=h_1\in H\;\implies\;\psi(gn)=h_1\;\;\forall\,n\in N:=\ker\psi\implies \psi(gN)=h_1$$
and the other way around:
$$\psi(x)=h_1\implies \psi(g^{-1}x)=\psi(g)^{-1}\psi(x)=h_1^{-1}h_1=1\implies g^{-1}x\in N\iff xN=gN$$
and thus we see that 
$$\{g\in G\;;\;\phi(g)=h_1\in H\}=|gN|=|N|$$
and we're done.
