I need to make a statement which I think should be an elementary group theory fact. Here is the setup. I have two exact sequences of groups: $$ 1\longrightarrow \ker\varphi_1\longrightarrow G\overset{\varphi_1}{\longrightarrow} \mathbb{R}$$ $$ 1\longrightarrow \ker\varphi_2\longrightarrow G\overset{\varphi_2}{\longrightarrow} \mathbb{R}$$
The groups $G$ are the same (they are discrete countable groups, but not necessarily finitely generated). I can prove the kernels of the homomorphisms are equal. Moreover, the images are both discrete subsets of $\mathbb{R}$, and hence isomorphic to $\mathbb{Z}$. I would like to conclude that there is a homomorphism $\psi\colon\mathbb{R}\to \mathbb{R}$ (necessarily multiplication by some number) such that $\varphi_2=\psi\circ\varphi_1$.
Since the image of $\varphi_i$ is discrete in $\mathbb{R}$, we can just pretend the $\mathbb{R}$'s are $\mathbb{Z}$'s and rephrase the question as follows: I have two short exact sequences of groups, and I have maps in the first and second columns making the diagram commute. I want to conclude there is a map in the third column making the entire diagram commute.