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Suppose $f\colon A\longrightarrow B$ is a cover in a category (i.e. a morphism which does not factor through any proper subobject) and let $g\colon B\longrightarrow C$ a morphism with image $D\rightarrowtail C$. I have been asked to prove or disprove that then the image of $g\circ f$ coincides with the image of $g$. I think this should be true but I am not able to prove it.

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  • $\begingroup$ Which notion of image is meant here? $\endgroup$ – Qiaochu Yuan Jan 17 '18 at 11:35
  • $\begingroup$ By image of $g\colon B\rightarrow C$ I mean a subobject $m\colon D\rightarrowtail C$ (more precisely a class of isomorphism of monomorphisms with codomain $C$) satisfying that $g$ factors through $m$ and if $g$ factors through any other mono $n$ then there is a morphism $k$ such that $ k m= n$ (i.e. $m$ is the smallest one wrt to the partial order of subobjects). $\endgroup$ – J. Karen Jan 17 '18 at 11:52
  • $\begingroup$ Don't you mean $nk=m$ in your comment above? $\endgroup$ – Arnaud D. Jan 17 '18 at 11:56
  • $\begingroup$ Yes, of course, sorry! $\endgroup$ – J. Karen Jan 17 '18 at 12:04
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    $\begingroup$ @Orat It's more commonly called an extremal epimorphism, and sometimes a surjection. $\endgroup$ – Arnaud D. Jan 17 '18 at 20:29
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Consider the category with two objects $I$ and $T,$ with all hom-sets singletons except for $T$ which has three self-maps $1_T,f,f^2$ obeying $f^2=f^3.$ The composite $T\to I\to T$ is $f^2:T\to T.$ (If it helps: this category is the Karoubi envelope of the monoid $T=\langle f\mid f^2=f^3\rangle.$) Take $f=g.$ They're both extremal epi, but the composite $f^2$ factors through $I\rightarrowtail T.$

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