Let $R$ be a commutative unital ring, $M$ and $N$ be $R$-modules, and $u:M\to N$ be an $R$-linear map. Let $i_M:M\to I_M$ and $i_N:N\to I_N$ be injective hulls for $M$ and $N$ and set $\Sigma M=\text{cok}(i_M)$, similarly for $\Sigma N$.
Since $I_N$ is an injective $R$-module, there exists an $R$-linear map $\alpha:I_M\to I_N$ with $\alpha i_M=i_Nu$. Then $$m+\text{im}(i_M)\mapsto\alpha(m)+\text{im}(i_N):\Sigma M\to\Sigma N$$ is a well-defined $R$-linear map $\alpha'$ with $\alpha'q_M=q_N\alpha$ (where $q_M$, $q_N$ are the quotient maps).
Now let $\beta:I_M\to I_N$ be another $R$-linear map with $\beta i_M=i_Nu$. Then $\alpha i_M=\beta i_M$ shows that $\alpha$ and $\beta$ agree on $\text{im}(i_M)$. In the same way as for $\alpha$, $\beta$ induces a map $\beta':\Sigma M\to\Sigma N$ with $\beta'q_M=q_N\beta$. I want to show that $\alpha'-\beta'$ factors through an injective module i.e. that there exists an injective $Q$-module $I$, a map $a:\Sigma M\to I$ and a map $b:I\to\Sigma N$ for which $ba=\alpha'-\beta'$.
The only injective modules involved in this problem are $i_M$ and $i_N$, and I cannot see how to factor through these. How can I factor $\alpha'-\beta'$ through an injective?
Context: I am going through the construction for the costable module category and trying to construct an endofunctor $\Sigma$ on it.
I think this should be equivalent to the case where $u=0$ as a map $M \to N$, in which you find a lift $\alpha: I_M \to I_N$ and then an induced map $\alpha' : \Sigma M \to \Sigma N$. The question is whether $\alpha'$ factors through an injective. It will factor through $I_N$. Consider the diagram $$ \require{AMScd} \begin{CD} 0 @>>> M @>{i_M}>> I_M @>{q_M}>> \Sigma M @>>> 0 \\ @. @VV{0}V @VV{\alpha}V @VV{\alpha'}V \\ 0 @>>> N @>{i_N}>> I_N @>{q_N}>> \Sigma N @>>> 0 \\ \end{CD} $$ Since $\alpha i_M = 0$, $\alpha$ induces a map $\Sigma M \to I_N$, and this will factor $\alpha'$.