Cokernels in Co-Cartesian Diagram are isomorphic.

Question

We assume the category is abelian. Let the below be a co-cartesian diagram. A co-cartesian diagram is just the dual of a cartesian diagram. \begin{array}{ccc} A& \xrightarrow{f} & B \\[3pt] \downarrow {a} & & \downarrow{b} \\ C& \xrightarrow{g} & D \end{array}

I want to prove that $cok(a) \simeq cok(b)$. This is a part of a proposition from the book cohomology of sheaves by Birger Iversen. I could prove it for the case of abelian groups but somehow i couldn't transform the procedure into language of arrows. Thank you.

Answer 1

Hint: Use Mitchell's theorem: an Abelian category is equivalent to a full subcategory of a category of R-modules, where R is a ring.

https://en.wikipedia.org/wiki/Mitchell's_embedding_theorem

Answer 2

Edit: this proof is wrong, $d$ cannot be defined as written.

We prove the statement for $f$ and $g$, being analogous in the other one. Let us name $p:B\to coker (f)$ and $q:D \to coker (g)$.

First of all, notice that we can complete our diagram with a new map $\varphi:coker (f) \to coker (g)$. Indeed, since $qbf=qga=(qg)a=0$, $qb:B \to coker(g)$ can be factorized via $coker(f)$. We then consider $d:D\to B$ determined by $1:B\to B$ and $0:C\to B$. $pdg=0$ since $dg=0$ by requirement, and so $pd$ factorizes through $q$: there exists $\psi: coker (g) \to coker (f)$ making $$ \begin{array}{ccc} D& \xrightarrow{f} & coker(g) \\[3pt] \downarrow {d} & & \downarrow{\psi} \\ B& \xrightarrow{g} & coker(f) \end{array}$$ commutative.

Next, it suffices to show that they are one the inverse of the other. Since $db=1$, we easily obtain $\psi\varphi p = p$ and because $p$ is an epimorphism $\psi\varphi=1$. In particular, notice that $\ker (\varphi) = 0 = coker( \psi)$. The other part is done by checking that $coker (\varphi)=0$. Let $h: coker (g) \to X$ be such that $h\varphi=0$. We consider the map from $D\to X$ given by $0=h\varphi p =hqb : B \to X$ and $0:C\to X$. The obtained map is trivially the zero one. On the other side, also $hq$ satisfies the required properties: $hqg=0:C \to X$ and $hqb=0:B\to X$. Therefore $hq=0$ because the morphism $D\to X$ is unique by definition. Since $q$ is an epimorphism, $h=0$ and so we have obtained that $h\varphi=0$ implies $h=0$. So, $coker(\varphi)=0$ and $\varphi$ is an epimorphism. Finally, $\varphi =\varphi 1 = \varphi \psi \varphi$ and by epimorphism property $\varphi\psi=1$.

In fact, the last part is too formal: in an abelian category every monomorphism is a kernel and every epimorphism is a cokernel, therefore it is easy to see that monomorphism and epimorphism implies isomorphism.

Sorry for being 2 years late.