1

From set theory we know $\text{Im}(gf)=g(\text{Im}(f))$. I was wondering how to make this work with an arbitrary abelian category. I know that every morphism $f:X\to Y$ in any abelian category has a unique (up to unique isomorphism) decomposition through the image object $\text{Im}(f)=\text{ker}\text{coker}(f)\cong \text{coker}\text{ker}(f)$. So, $f=ip$ where $p:X\twoheadrightarrow \text{Im}(f)$ and $i:\text{Im}(f)\hookrightarrow Y$ the canonical maps.

For a composable pair $g$, $f$, I guess one way is to define $\text{Im}(gf)$ as $\text{Im}(g\circ i)$. I made this choice because,

  1. This is all I need for my purpose (I am trying to show that if $C$ is a split exact chain complex over an arbitrary abelian category, then the identity map on $C$ is null homotopic. See here. I prefer not to work with elements, not to use Mitchell Embedding theorem and I haven't yet learnt chasing members trick from Mac Lane).

  2. I think this is at least true because we get two factorizations of $gf$ one through Im$(gi)$, and one through Im$(gf)$. Since image factorizations are unique up to unique isomorphisms in abelian categories, Im$(gf)\cong$ Im$(gi)$. Since I am not sure how to draw commutative diagrams without the help of quiver, I give a link to my diagram over at quiver here.

Does this seem right, are there any other ways to interpret/work with image of composition of morphisms? Any suggestions will be greatly helpful.

Lastly, how to generalize this to an arbitrary category?

frelg
  • 363
  • 2
    I think you have answered your question. $Im(g\circ f)$ and $Im(g\circ i)$ are a priori defined in the abelian catogory, but your second point shows they are in fact uniquely isomorphic, which is the analogue of $Im(gf) = g(Im(f))$. For more general than abelian categories, you can still define images, but they don't necessarily exist. https://ncatlab.org/nlab/show/image – MPos Feb 15 '24 at 13:14
  • 1
    But due to the uniqueness of factorization property, if both images exist, then they would be isomorphic as well. – MPos Feb 15 '24 at 13:15
  • Thank you very much :)) – frelg Feb 15 '24 at 13:30

0 Answers0