The problem says : (in a category with epi-equalizer factorizations)
if $gf$ is an equalizer show that $f$ is an equalizer , using this result, also show that if $f$ and $g$ are equalizers then $gf$ is an equalizer.
What I've found so far:
- In this question there is a counterexample in a category without (epi-equalizer) factorizations.
My attempt of a solution ($gf$ equalizer $\rightarrow$ $f$ equalizer ):
Since $gf$ is an equalizer it has to be the equalizer for some arrows $\alpha$ ,$\beta$ , I hope to show that $f$ is the equalizer for $\alpha g$ , $\beta g$
We have $\alpha(gf) = \beta(gf)$ , then $(\alpha g) f = (\beta g) f$.
Let $\phi$ be another candidate for equalizer so $(\alpha g) \phi = (\beta g) \phi$.
Since $gf$ is an equalizer and $(\alpha g) \phi = (\beta g) \phi $ , there exists an unique arrow $\psi$ which satisfies $gf\psi = g\phi$
Let $e,i$ be a (epi-equalizer) factorization of $g$ .
We have $ief\psi = ie\phi$ , since $i$ is an equalizer it's also mono, so $ef\psi = e\phi$.
The picture would look like this :
This is where I can't continue , if I could make that $e$ go away from the last equation I'd get the existence of the arrow ($f\psi = \phi$) , the uniqueness would follow from factoring f. I have tried to factor $\phi$ or $gf$ but I couldn't see how it would help me.
So far , I haven't used that $e$ is epi or that $i$ is an equalizer (only that it's mono)