why split epi and mono implies iso?
Solution 1:
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 $(gf, \mathrm{id})$.
The same holds for epi + equalizer and epi + split mono.