FALSE: A subobject is a monomorphism rather than an equivalence class of representatives

Let $m:A\to X$ send $0$ to $0$, and let $n:B\to X$ send $\ast$ to $0$. Both maps are injective, and an injection $f$ in $\mathbf{S}\mathbf{e}\mathbf{t}$ is monic: if $f\circ g=f\circ h$ then $f(g(x))=f(h(x))$ for every $x$, so $g(x)=h(x)$ by injectivity and $g=h$. They are nonetheless different morphisms, because their domains differ.

The unique bijections $u:A\to B$ and $v:B\to A$ satisfy $m=n\circ u$ and $n=m\circ v$. Thus $m$ and $n$ mutually factor and [L2] makes their factor maps inverse isomorphisms.

Consequently the two different monomorphisms determine one equivalence class $[m]=[n]$, which is the subobject prescribed by [L1]. The individual monomorphisms are representatives, not the subobject itself.