I am trying to solve the following problem:
For $i = 1,2,$ let $G_i$ be a group and $H_i$ be a normal subgroup of $G_i.$ Let $\pi_i: G_i \to G_i/H_i$ be given by $\pi_i(a) = a H_i$ for all $a \in G_i.$ Let $f: G_1 \to G_2$ be a group homomorphism such that $f(H_1) \subset H_2.$ Let $f': H_1 \to H_2$ be the restriction of $f.$
$(a)$ Prove: there exist a unique group homomorphism $\bar{f}: G_1/H_1 \to G_2/H_2$ such that $\bar{f} \circ \pi_1 = \pi_2 \circ f.$
I understood that we are trying to prove that the following diagram commutes:
$$\require{AMScd} \begin{CD} G_1 @>{\pi_1}>> G_1/H_1\\ @VVV @VVV \\ G_2 @>{\pi_2}>> G_2/H_2 \end{CD}$$
But still I do not know how to define a function to help me in this proof, could someone help me please?