Let $A$ and $B$ be finite groups with subgroups $C$ and $D$, respectively. Further, there is a surjective group homomorphism $\pi:A\to B$ which induces a surjective homomorphism $\pi{^\prime}: C\to D$. Then can we say that $\pi=\pi'$ or at least that $C\leq\pi^{-1}(D)$?
I don't know if it would help, but if both $C$ and $\pi^{-1}(D)$ has index $2$ in $A$ then would either of these results mentioned above hold? Or, can we say more that $C\cong\pi^{-1}(D)$?
Or, in generality, nothing can be concluded?
I would really appreciate it if somebody could provide some clarification on this. Also, please elaborate on your opinion.
You cannot say that $\pi=\pi'$ unless $A=C$. Your definition says that $\pi'$ is induced by $\pi$, meaning that $\pi'=\pi|_C$, the restriction of $\pi$ to $C$. Two functions are equal if and only if they have the same domain, the same value at every point in the domain, and (depending on your precise definition of function the following may be moot or inapplicable) the same codomain. So unless $A=C$, you don't have $\pi=\pi'$, because they have different domains.
By definition, if $\pi$ sends $C$ into $D$, then $C\subseteq\pi^{-1}(D)$, regardless of whether $\pi$ is surjective or not. The symbol $\leq$ here represents subgroups, so you would need to prove that $\pi^{-1}(D)$ is a subgroup of $A$ (it is, but it needs to be proven).
In general, it need not be the case that $C\cong \pi^{-1}(D)$. For example, consider $G=C_2\times C_2$, the Klein $4$-group, let $B=C_2$, the cyclic group of order $2$, let $\pi\colon A\to B$ be the projection onto the first coordinate, let $C$ be the diagonal subgroup, consisting of $(1,1)$ and $(x,x)$ and let $D=B$. Then the restriction of $\pi$ to $C$ is a surjection from $C$ onto $D$, but $\pi^{-1}(D)$ is all of $A$, and not isomorphic to $C$.
If $C$ has finite index in $A$ and $\pi^{-1}(D)$ has the same index, then necessarily we have $C=\pi^{-1}(D)$, since $[G:C]=[G:\pi^{-1}(D)][\pi^{-1}(D):C]$, and so $C$ and $\pi^{-1}(D)$ are equal (under the assumption that $\pi(C)\subseteq D$), not merely isomorphic.