Are two substructures in a concrete category which share the same underlying set equal? Let us say that a substructure of an object $M$ of a concrete category $\mathbf{C}$ with forgetful functor $|.|$ is an objet $N$ such that $|N|\subset|M|$ and $i:|N|\to |M|$ which is the canonical injection is an embedding.
We also define an embedding as an injective morphism (meaning his image through $|.|$ is injective) $f:A\to B$ such that for every function $g:|X|\to|A|$, $g$ is a morphism if and only if $|f|\circ g$ is a morphism.
If, N and N' are two substructures of M such that $|N|=|N'|$, is it necessary that $N=N'$ or that $N\simeq N'$ ?
Edit n°1 : $i$ is supposed to lift through |.| to an element of $Hom(N,M)$.
 A: Forget injectivity, what's relevant is that an embedding $N\hookrightarrow M$ is a $|\cdot|$-cartesian morphism.
Explicitly, a morphism $N\xrightarrow{\phi} M$ in $\mathbf C$ is $|\cdot|$-cartesian if whenever $L\xrightarrow{\chi}M$ has a $|\cdot|$-factorization $|L|\xrightarrow{g}|N|\xrightarrow{|\phi|}|M|$, then there exists a unique morphism $L\xrightarrow{\psi}N$ so that $L\xrightarrow{\chi}M$ factors as $L\xrightarrow{\psi}N\xrightarrow{\phi}M$ with $|\psi|=g$.
In particular, this implies that $N\xrightarrow{\phi}M$ is a terminal object in the category with objects $L\xrightarrow{\chi}M$ where $|\chi|=|\phi|$ and morphisms are commutative triangles $L_1\xrightarrow{\psi}L_2$ with $\chi_2\circ\psi=\chi_1$ and $|N|\xrightarrow{|\psi|}|N|$ an identity morphism. Consequently, if $N_1\xrightarrow{\phi_1}M$ and $N_2\xrightarrow{\phi_2}M$ are $|\cdot|$-cartesian morphisms with $|\phi_1|=|\phi_2|$, then $N_1\cong N_2$.
In your special case of $|\cdot|$ being faithful (i.e. when $\mathbf C$ is concrete), we can identify morphisms in $\mathbf C$ with morphisms in the image of $|\cdot|$. Then $f$ being $|\cdot|$-cartesian says exactly that $|L|\xrightarrow{g}|M|$ is a morphism in $\mathbf C$ if (and only if) $|L|\xrightarrow{g}|M|\xrightarrow{f}|N|$ is a morphism in $\mathbf C$.
