is it a stronger property of equalizers?

Question

I understand that if $h$ is an eqalizer of $f,f:A\rightrightarrows B$ then $h$ is an isomorphism. Is it stronger to say that if an isomorphism $h$ is an equalizer of $f,f:A\rightrightarrows B$ then $id_A$ is also an equalizer? This apparently stronger property might imply some equations between compositions of arrows that do not hold in general, am I right ?

Answer 1

The seemingly stronger version is satisfied trivially because $id_A$ is always an equalizer of $f,f:A\to B$. One sees this by simply checking the definition directly: $f\circ id_A=f\circ id_A$, and given any $g:C\to A$ with $fg=fg$, there is exactly one arrow $x:C\to A$ with $id_A\circ x=g$.