$Ker(CoKer(f))$ satisfies the universal property of $im(f)$ for a morphism $f$ in Abelian category.

abelian-categoriescategory-theory

I'm new to Category theory and recently read about the "Abelian Categories". I've difficulties in establishing the existence of an object that satisfy the universal property of "Image" (the property is given here) in an Abelian Category, more precisely:

Problem. Suppose $\mathcal{C}$ be an Abelian category. Let $f\in Mor(A,B)$ (for two object $A, B$ of $\mathcal{C}$). Then $\text{Ker(Coker}(f))$ satisfies the universal property of $im(f)$.

Definitions and Notations (that I've read):

Suppose $f$ as above then-

Ker(f): Kernel of $f$ is an object $K$ together with a morphism $ker(f):K\to A$ satisfying the universal property given in the wiki page here.

CoKer(f): CoKernel of $f$ is an object $C$ together with a morphism $CoKer(f):B\to C$ satisfying the universal property given in the wiki page here.

Suppose $f$ is as in the problem. Let the object $K$ together with the morphism $Ker(CoKer(f)):K\to B$ is the kernel of $CoKer(f):B\to C$.

To prove: $K$ with $Ker(CoKer(f)):K\to B$ satisfies the universal property of $im(f)$.

First, since $CoKer(f)\circ f = 0$ then by universal property of $Ker(CoKer(f))$ there exists a unique morphism $e:A\to K$ such that $Ker(CoKer(f))\circ e=f$. So this establishes the first property of $im(f)$.

Now to prove the second property, let $K'$ be an object with a morphism $i:K'\to B$ and a morphism $e':A \to K'$ such that $i\circ e'=f$. We have to prove there exists an unique morphism $g:K \to K'$ such that $i\circ g=Ker(CoKer(f))$.

But I cannot see any way to claim such an unique $g$. Any help is appreciated. Thank you.

P.S.(Universal Property of Image): An object $I$ in $\mathcal{C}$ together with a morphism $i:I \to B$ is said to be image of $f$ if it satisfies the following universal property:

  • (1). $\exists e \in Mor(A, I):i\circ e=f$ and
  • (2). for any object $I'$ with a monomorphism $i' \in Mor(I', B)$ and a morphism $e' \in Mor(A, I')$ such that $i'\circ e'=f$, there exists a unique $g \in Mor(I, I')$ such that $i'\circ g=i$.

Best Answer

One of the main properties of abelian categories is that every monomorphism is a kernel, and more specifically, the kernel of its cokernel; and dually, every epimorphism is the cokernel of its kernel.

In particular, $f$ factors through a monomorphism $i$ if and only if $\operatorname{Coker}(i)\circ f=0$; by the universal property of the cokernel $\operatorname{Coker}(f)$, this is also equivalent to the condition that there exists a (unique) morphism $h$ such that $h\circ \operatorname{Coker}(f) =\operatorname{Coker}(i)$. But since $\operatorname{Coker}(f)$ is also the cokernel of its kernel, the existence of $h$ is also equivalent to $\operatorname{Coker}(i)\circ \operatorname{Ker}(\operatorname{Coker}(f))=0$, and by the universal property of the kernel $i$ of $\operatorname{Coker}(i)$ this is equivalent to the existence of a morphism $g$ such that $\operatorname{Ker}(\operatorname{Coker}(f))=i\circ g$.

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