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In Abstract and Concrete Category of Adamek, Herrlich and Stretcker, I'm dealing with exercise 7Dd. It says that a strict monomorphism $f:A\to B$ followed by a section $g:B\to C$ give rise to a a strict monomorphism $fg:A\to C$.

I try to prove it: Let $f'$ such that, for any $r,s:C\to\cdot$,we have $fgr=fgs$ implies $f'r=f's$. We have to show that $f'=af$ for a unique $a$. Let $h$ such that $gh=1$. Then $fu=fv$ implies $fghu=fghv$ which implies $f'hu=f'hv$. Since $f'$ is strict, by assumption, we obtain $f'h=af$ for a unique $f$. I have difficulties to deduce $f'=afg$ from it.

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Let $E$ denote the domain of $f'$. The trick is to factor $f': E \to C$ through $g$ first, using the fact that $g$ is the equalizer of the pair $hg, 1_C: C \to C$. Indeed, since $fg(hg) = fg(1_C)$, we have $f'(hg) = f' 1_C$ by hypothesis. Because of the observation of $g$ being the equalizer, we can therefore write $f' = bg$ for some (unique) $b: E \to B$.

Again by hypothesis, we have that $f' = bg$ equalizes every pair $r, s: C \to D$ that is equalized by $fg$, i.e.,

$$fgr = fgs \Rightarrow bgr = bgs.$$

In particular, using $gh = 1_B$, we now have a sequence of implications

$$fr' = fs' \Rightarrow fg(hr') = fg(hs') \Rightarrow bg(hr') = bg(hs') \Rightarrow br' = bs'$$

so that by strictness of $f$, we may write $b = af$ for some (unique) $a$. We conclude $f' = bg = afg$.

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