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Let $\mathcal{A}$ be an abelian category. For a morphism $f:A\rightarrow B$, its image is usually defined in the following way: a map $i:I\rightarrow B$ is called an image of $f$ if it is a kernel of a cokernel of $f$.

I was trying to make an equivalent definition for the image of $f$ similar to the usual definitions of kernel and cokernels. I arrived to the following: a map $i:I\rightarrow B$ is called an image of $f$ if it is a monic satisfying the following two conditions:

  1. There exists a map $g:A\rightarrow I$ such that $ig=f$;
  2. If $j:J\rightarrow B$ is another map such that there exists $h:A\rightarrow J$ with $jh=f$ then there exits a unique map $k:I\rightarrow J$ such that $jk=i$.

I was trying to see if the condition for $i$ to be a monic is unnecessary but I coudn't proved it. So the question is, is a map $i:I\rightarrow B$ satisfying conditions 1 and 2 a monic?, or how I should modify 1 or 2 in order to eliminate the condition for $i$ to be a monic?.

Maybe it is better to put this in the category $R-mod$ with $R$ an associative ring. I would appreciate your suggestions in both cases.

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You may be interested in this notion of image – Hurkyl Dec 27 '12 at 2:14
See also Freyd's Abelian Categories, page 42 ( There he defines the image of $f:A \to B$ to be the smallest subobject of $B$ through which $f$ factors. The image, representing a subobject, is monic by definition. Essentially, you should take condition (2) and say "If $j:J\to B$ is another subobject through which $f$ factors, then there exists a unique..." The "kernel of the cokernel" characterization of the image of a map in an abelian category can then be proved as a proposition (which he does, Theorem 2.16). – John Myers Dec 27 '12 at 6:36
up vote 2 down vote accepted

The regular image doesn't satisfy your universal property (when $f$ isn't monic). For example, suppose we're just working with a commutative ring $A$, two ideals $K$ and $L$ of $A$, and the quotient map $A \rightarrow A/(K, L) =: B$. The image is just $A/(K,L) =: I$, and the map $A \rightarrow A/(K,L)$ factors $A \rightarrow A/K =: J\rightarrow A/(K,L)$, but you can't lift always $A/(K,L) \rightarrow A/K$.

However, $A \xrightarrow{id} A$ does satisfy your universal property!

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Is it me, or are you using $I$ to denote two different things here? – Ben Millwood Dec 27 '12 at 1:56
I made an edit that I hope is the same as what you meant, but clearer. Please tell me if not. – Ben Millwood Dec 27 '12 at 2:04
Absolutely spot on! Many thanks :) – uncookedfalcon Dec 27 '12 at 4:02

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