# why split epi and mono implies iso?

I was doing some exercises on the definitions of epics, monos, split monos, etc..., and I asked myself that if you could take, for instance an epi which is mono, and then deduce it is an iso, which is of course false; trying to recover the fact that in $Set$ epi and mono imply iso. So I wanna ask, if mono and coequalizer imply iso, or mono and split epi imply iso, or if both are false. I just wondered, since both coequalizer and split epi are stronger than epi; I've tried to show either, with no success.

Thanks

-

If $f$ is a coequalizer of maps $g, h$ then $fg = fh$. If $f$ is also mono then $g = h$. The coequalizer of the pair $(g, g)$ is the identity map which is indeed an isomorphism. So mono + coequalizer does mean isomorphism.
For mono and split epi assume $fg = \mathrm{id}$. Then $fgf = f$ so if $f$ is mono $gf = \mathrm{id}$, hence $f$ is iso. Alternatively, observe that $f$ is a coequalizer for $(gf, \mathrm{id})$.
in the first paragraph I don't understand why $f$ is an iso –  rush fan Feb 17 '13 at 22:52
Coequalizer is defined via a universal property, so when it exists it is unique up to isomorphism, hence $f$ is isomorphic to the identity map because they are both coequalizers of the pair $(g, g)$. Being isomorphic to the identity map implies that you are yourself an isomorphism. –  Jim Feb 17 '13 at 22:57