Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Take a morphism $f:X\to X$ in a category with the same domain and codomain. I want to test whether $f$ is a monomorphism. This means, taking arbitrary $g_1,g_2:Y \to X$ with $f\circ g_1=f\circ g_2$ it should follow that $g_1=g_2$.

Does it suffice for a retraction $f:X\to X$ to be a monomorphism that for all $g_1,g_2:X \to X$ with $f\circ g_1=f\circ g_2$ it follows that $g_1=g_2$?

share|cite|improve this question
The last sentence doesn't make any sense to me. The condition "for all $g_1,g_2:X\to X$ it follows that $g_1=g_2$" has nothing to do with $f$. – Alon Amit Mar 3 '12 at 9:25
Sorry, typo. Fixed. – roadrunner Mar 3 '12 at 10:40
If your category is additive (don't actually remember if additive is enough, but if you go with "abelian" then you're sure) it is enough that $\ker f =0$. – Andy Mar 3 '12 at 13:08

If $f: X\to Y$ is a retraction (even if domain and codomain are not the same) then it is indeed sufficient to consider maps $g_1,g_2: X\to X$ in the above: let $s: Y \to X$ be a map with $f\circ s = id_Y$, then you have $f\circ s\circ f = f = f\circ id_X$. Therefore $s\circ f = id_X$ and $f$ is in fact an isomorphism with inverse $s$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.