In $\mathbf{Set}$, monomorphisms are exactly injections and epimorphisms are exactly surjections

If $f$ is injective, $f\circ g=f\circ h$ implies $g=h$ pointwise, so $f$ is monic; if $f$ is not injective, choose $a_0\ne a_1$ with $f(a_0)=f(a_1)$, and the two maps from a singleton selecting $a_0,a_1$ show that $f$ is not monic.

If $f$ is surjective and $g\circ f=h\circ f$, then for each $b\in B$ choose an $a$ only for this fixed $b$ with $f(a)=b$, giving $g(b)=h(b)$; hence $g=h$ and $f$ is epic.

If $f$ is not surjective, let $g:B\to\{0,1\}$ be constantly $0$ and let $h$ be $0$ on $f[A]$ and $1$ outside $f[A]$; then $g\ne h$ but $g\circ f=h\circ f$, so $f$ is not epic.