Let $f: G \rightarrow G'$ be a group homomorphism with kernel $H$. Then we know as an elementary fact of abstract algebra that there is an injective group homomorphism $f^*:G/H \rightarrow G'$ such that $f^*(x+H) = f(x)$. More generally, let $\psi: G \rightarrow G/H$ be a surjective homomorphism with kernel $H$. Then again we have an injective homomorphism $f_{\psi}^* : G/H \rightarrow G'$ defined by $f^*_{\psi}(x+H)=f(y)$ where $\psi(y) = x+H$.
Could you verify the validity of this reasoning? Thanks.
Edited:
To give you an example about the motivation of this question, we know that given a homomorphism $f:G \rightarrow G'$ with kernel $H$ there exists always a homomorphism $f^*: G/H \rightarrow G'$ such that $f = f^* \circ \phi$ where $\phi:G \rightarrow G/H$ is the canonical mapping.
Now, what if instead of the canonical mapping $\phi$ we have another surjective homomorphism $\psi:G \rightarrow G/H$ with kernel $H$? Does then exist a homomorphism $f^*_{\psi} : G/ H \rightarrow G'$ such that $f = f^*_{\psi} \circ \psi$?