Subobjects in Set are subsets

A monomorphism $m:A\to X$ in $\mathbf{S}\mathbf{e}\mathbf{t}$ is injective: if $m(a)=m(a^{\mathrm{\prime }})$, the two maps $g,h:\{\ast \}\to A$ with $g(\ast )=a$ and $h(\ast )=a^{\mathrm{\prime }}$ satisfy $m\circ g=m\circ h$, so $g=h$ and $a=a^{\mathrm{\prime }}$. Conversely an injection is monic by the same cancellation. For such an $m$, the corestriction $\bar{m}:A\to m[A]$ is a bijection and satisfies $m={\iota }_{m[A]}\circ \bar{m}$, while ${\iota }_{m[A]}=m\circ {\bar{m}}^{-1}$. Thus $m$ mutually factors with the inclusion of its image and represents that subset by [L1] and [L2].

If the inclusions of subsets $S,T\subseteq X$ mutually factor, their image sets in $X$ coincide, hence $S=T$. Conversely equal subsets give the same inclusion. Therefore taking the image and taking the inclusion are inverse assignments between subobjects and subsets.

The argument also applies to the unique injection $\mathrm{\varnothing }\hookrightarrow X$, so the empty subset supplies the least subobject rather than an exceptional case.