In an Abelian category $\sf A$, suppose to have a diagram with exact rows, as in the Snake lemma: $\require{AMScd}$ $$\begin{CD} @.a'@>{u_a}>> a @>{v_a}>> a''@>>> 0\\ @. @VVf'V @VVfV @VVf''V @.\\ 0@>>>b'@>{u_b}>> b @>{v_b}>> b''@. \end{CD}$$
Construct the following diagram: $$\begin{CD} \operatorname {ker}f'@>i>> \operatorname {ker}f@>j>> \operatorname {ker}f''\\@VVk'V @VVkV @VVk''V\\ a'@>{u_a}>> a @>{v_a}>> a''\\ @VVf'V @VVfV @VVf''V \\ b'@>{u_b}>> b @>{v_b}>> b''\\@VVc'V @VVcV @VVc''V\\ \operatorname {cok}f'@>h>> \operatorname {cok}f@>l>> \operatorname {cok}f'' \end{CD}$$
Assume that $\sf A$ has enough projectives, so there is an epic $e:p\to \operatorname {ker}f''$, with $p$ projective. Then:
- $v_a$ is epic, so there is $\phi_1:p\to a$ such that $v_a\phi_1=k''e$.
- $v_bf\phi_1=0$, and $u_b$ is a kernel of $v_b$, so there is a unique $\phi_2:p\to b'$ such that $u_b\phi_2=f\phi_1$.
Let $\varphi_1:p\to a$ be another lift of $k''e$, and let $\varphi_2:p\to b'$ the unique morphism such that $u_b\varphi_2=f\varphi_1$. I have to show that $c'\phi_2=c'\varphi_2$.
I've tried in a few ways, but they all stopped early. I'm not asking for a solution; maybe an hint, but actually I'd just like to know if what I'm proving is really true at this point. (I found this task in Surowski's online paper where he proves the Snake lemma categorically). Thanks
What worked for me was setting $\psi_1:=\phi_1-\varphi_1:p\to a$, which factors through $\operatorname{ker}v_a\cong\operatorname{im}u_a$, and using that $a'\to \operatorname{im}u_a$ is epi.