Suppose $f: A \longrightarrow B$ is a morphism in an abelian category $\mathcal{C}$.

What I consider an abelian category:

  1. $\mathcal{C}$ is additive.
  2. Every morphism has a kernel and a cokernel.
  3. Every monomorphism is a kernel and every epimorphism is a cokernel.

With that, we can define:

$Im(f)= kernel(cokernel(f))$


where $k: K \longrightarrow A$ is a kernel of $f$ if $ k \circ f = 0_{K,B}$ and whenever $h \circ f = 0$, $h$ factors uniquely through $k$. (i.e. $h= k \circ h'$). And $q:B \longrightarrow C$ is a cokernel of $f$ if $ f \circ q = 0_{A,C}$ and whenever $f \circ h = 0$, $h$ factors uniquely through $q$ (i.e. $h = h' \circ q$).

Notation: $0_{A,B}$ is the zero morphism obtained composing $A \longrightarrow 0$ and $0 \longrightarrow B$.

Once I have defined $Im(f)$ and $Coim(f)$, I want to check that there exists a natural map between them, called $\overline{f}$ which is isomorphism.

I have been working with epimorphisms and monomorphisms notions but I am a little bit lost. Any help/hint?

Related but do not understand: Equivalent conditions for a preabelian category to be abelian

  • $\begingroup$ It depends on the definition of abealian category you have in mind: there are very different ones. $\endgroup$ – Tommaso Scognamiglio Dec 4 '18 at 11:02
  • $\begingroup$ My bad, already edited. Thank you. $\endgroup$ – idriskameni Dec 4 '18 at 11:06
  • $\begingroup$ I do not understand you. Could you be more specific? $\endgroup$ – idriskameni Dec 4 '18 at 11:10
  • $\begingroup$ your definition just states that monics and epics are kernels and cokernels, but they need to be that also in a natural way, hence you actually want that your epimorphisms are the cokernels of their kernels! and monomorphisms dually $\endgroup$ – Enkidu Dec 4 '18 at 11:11
  • $\begingroup$ also, the fact that this is an isomorphism is in fact equivalent to 3) $\endgroup$ – Enkidu Dec 4 '18 at 11:29

the natural map comes from the following construction:

consider: $A \xrightarrow{f} B$ , denote by $K \hookrightarrow A$ and $B \twoheadrightarrow C$ the kernel and cokernel of $f$. Furthermore denote by $K_C \hookrightarrow B$ the kernel of the cokernel of f (the image) and by $A \twoheadrightarrow C_K $ the cokernel of the kernel (the coimage).

Then we know that $K \hookrightarrow A \xrightarrow{f} B$ is $0$ hence this factors over $C_K$ as $C_K \xrightarrow{\alpha} B$, now since $A \twoheadrightarrow C_K$ is an epic, we can deduce, since $A \xrightarrow{f} B \twoheadrightarrow C$ is $0$ that $\alpha$ factors over $K_C$ as $\beta: C_K \to K_C$ and this is your desired map. Remark, kernels are in general monics and cokernels epics, but not generally the other way around, hence this comes into play for showing that this is an iso.

  • $\begingroup$ I am a little bit lost when you say that it factors over $C_K$ as... or it factors over $K_C$ as... Could you explain a bit more? Thank you very much. $\endgroup$ – idriskameni Dec 4 '18 at 16:26
  • 1
    $\begingroup$ the universal property of the kernel $K \xrightarrow{\iota} X$ of $f:A \to B$ states that every morphism $g$ such that $f\circ g = 0$ factors uniquely over $K \xrightarrow{\iota} X$, i.e. there is a unique $\widehat{g}$ such that $\iota \circ \widehat{g}=g$ (this uniqueness by the way gives that every kernel is a monic). And for the cokernel dually. I hope that makes the factoring clearer. In the end you use the univ property of the cokernel of the kernel first to factor the morphism $f$ over the coimage and then you do the same thing dually for the image using the univ property of the kernel $\endgroup$ – Enkidu Dec 5 '18 at 10:11

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