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A while ago I was interested in categorical image of composition of morphisms in abelian categories, see this post. I learnt that $\text{Im}(gf)=\text{Im}(g i)$ where $i$ is the canonical monomorphism gotten by factorizing $f$ in the abelian category. Recently, I had to deal with the same thing but with kernels.

Context 1: Recall that abelian categories have all kernels. The categorical kernel of a morphism $f$ is the equalizer of $f$ and the zero morphism (which exists as abelian categories contain zero objects).

Context 2: In the category of sets, for a composable pair of morphisms $f,g$ we know $\ker(gf)=f^{-1}(\ker(g))$, ie the preimage of $\ker(g)$ along $f$.

Question: Is there a categorical version of this for arbitrary abelian categories?

My thoughts so far: Let $A\xrightarrow{f}B\xrightarrow{g}C$ be a composable pair of morphisms in an arbitrary abelian category. Let $k:\ker(g)\to B$ be the kernel map of $g$. I think I need to lift this map along $f$ to get a map from $\ker(g)\to A$. It may be (I am not at all sure) possible to take the kernel of this map to be $\ker(gf)$ using the universal property of kernels.

Any suggestions would be greatly helpful :))

frelg
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  • "In the category of sets [...]" what's the kernel of a morphism in $\mathrm{Set}$? $\mathrm{Set}$ is not abelian. – Ben Steffan Mar 02 '24 at 13:15
  • @BenSteffan take a full abelian subcategory of set I presume – Chris Mar 02 '24 at 13:26
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    You can easily see using the universal property that $\ker(g\circ f)=\ker(p\circ f)$ where $g=i\circ p$ is the canonical decomposition of $g$ in an abelian category (this is basically due to the fact that $i$ is a mono) – Temoi Mar 02 '24 at 13:50
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    @Temoi Thanks, this was exactly what I was looking for. If you were to make your comment an answer, I would accept it :)) – frelg Mar 02 '24 at 14:09

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You can easily see using the universal property that $\ker(g\circ f)=\ker(p\circ f)$ where $g=i\circ p$ is the canonical decomposition of $g$ in an abelian category (this is basically due to the fact that $i$ is a mono).

Temoi
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