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I cannot prove the `simple' statement. Any hint is appreciated. Below is what I tried.

Let $A\to C$ is a strong epi, and factors as $A\xrightarrow{g} B\xrightarrow{m} C $ with $m$ a mono. The clam is that $B\to C$ is an iso.

Recall that a strong epi has left lifting property w.r.t all monos. The only mono I have at hand is $B\to C$, and then I consider the natural commutative square $A\to C // B\to C $, then there is a lift $r:C\to B$, which should be the inverse of $B\to C$.

It is clear that $mr=1$. However I cannot show $rm=1$. What I obtain is $rmg=g$, but there is no assumption on $g$, I cannot deduce $rm=1$. It suffices to show $g$ is an epi.

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I'd like to know: What is a strong epi? –  Stefan Hamcke Oct 22 '13 at 21:10
If $m \circ r = \mathrm{id}$ and $m$ is a monomorphism, then $m$ is an isomorphism. Hint: consider $m \circ r \circ m$. –  Zhen Lin Oct 23 '13 at 8:02
@ZhenLin I got it, one should prove $m$ is a strong epi first. –  Ma Ming Oct 23 '13 at 8:39
It's really much easier than that. $m \circ r \circ m = m$, and $m$ is monic, so $r \circ m = \mathrm{id}$ as well. –  Zhen Lin Oct 23 '13 at 8:45
@ZhenLin Thanks a lot. Yes, the idea is the same. –  Ma Ming Oct 23 '13 at 17:52

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