Prove $\mathcal T∘\mathcal R=\mathcal T∘\mathcal S$ for any $\mathcal T$ iff $\mathcal R∘\mathcal T=\mathcal S∘\mathcal T$ for any $\mathcal T$.

Question

I am trying to understand if for two binary relations $\mathcal R$ and $\mathcal S$ the following statements are equivalent.

1. $\mathcal R=\mathcal S$
2. $\mathcal R\circ\mathcal T=\mathcal S\circ\mathcal T$ for any (possible) relation $\mathcal T$
3. $\mathcal R[X]=\mathcal S[X]$ for any (possible) $X$
4. $\mathcal R^{-1}=\mathcal S^{-1}$
5. $\mathcal T\circ\mathcal R=\mathcal T\circ\mathcal S$ for any (possible) relation $\mathcal T$

So if $1$ holds then trivially $2$ holds.

Now if $2$ holds then for any $X$ we can consider the identity relation $\operatorname{id}_X$ so that the equality $$ \tag{2.1}\label{2.1}\mathcal R\circ\operatorname{id}_X=\mathcal S\circ\operatorname{id}_X $$ holds. So by \eqref{2.1} we argue that the equality $$ \mathcal R[X]=\mathcal R\big[\operatorname{id}_X[X]\big]=(\mathcal R\circ\operatorname{id}_X)[X]=(\mathcal S\circ\operatorname{id}_X)[X]=\mathcal S\big[\operatorname{id}_X[X]\big]=\mathcal S[X] $$ holds.

Now if $(y,x)$ is in $\mathcal R^{-1}$ then $(x,y)$ is in $\mathcal R$ and so $y$ is in $\mathcal R[x]$; however, if by hypothesis the equality $$ \mathcal R[x]=\mathcal S[x] $$ holds then $y$ is in $\mathcal S[x]$ and thus $(x,y)$ is in $\mathcal S$: we conclude that $(y,x)$ is in $\mathcal S^{-1}$ which is in particular not empty. Moreover by analogous arguments we can see that if $\mathcal S^{-1}$ is not empty then even $\mathcal R^{-1}$ is not empty and in particular it contains $\mathcal S^{-1}$. So by extensionality we conclude that $\mathcal R^{-1}$ and $\mathcal S^{-1}$ are equals.

Now if $\mathcal R^{-1}$ and $\mathcal S^{-1}$ are equals then the equality $$ \mathcal R^{-1}\circ\mathcal T^{-1}=\mathcal S^{-1}\circ\mathcal T^{-1} $$ holds so that even the equality $$ \mathcal T\circ\mathcal R=\big((\mathcal T\circ\mathcal R)^{-1}\big)^{-1}=(\mathcal R^{-1}\circ\mathcal T^{-1})^{-1}=(\mathcal S^{-1}\circ\mathcal T^{-1})^{-1}=\big((\mathcal T\circ\mathcal S)^{-1}\big)^{-1}=\mathcal T\circ\mathcal S $$ holds.

Finally if 5. holds then the equality $$ \mathcal T^{-1}\circ\mathcal R=\mathcal T^{-1}\circ\mathcal S $$ holds so that as above it is not complicate to conclude that the equality $$ \mathcal S^{-1}\circ\mathcal T=\mathcal R^{-1}\circ\mathcal T $$ holds. So as in 2. we conclude that $$ \mathcal R^{-1}[X]=\mathcal S^{-1}[X] $$ for any set $X$. Finally as in $3$ we conclude that $$ \mathcal R=\mathcal S $$

So first of all I ask if actually statements 1-5 are equivalent and so I ask if I well proved it. Specifically I ask to prove it more using a more clear argument since it seems to me it would be better to prove $1\to 5\to 2\to 3\to 4\to 1$ but unfortunately I am not able to prove $5\to 2$. So couls someone help me, please?

Answer 1

A possible solution could be the following which use the following theorem.

Theorem 1

If $\mathcal R$ and $\mathcal S$ are two binary relations then the following statements are equivalent.

1. $\mathcal R\subseteq\mathcal S$;
2. $\mathcal R\circ\mathcal T\subseteq\mathcal S\circ\mathcal T$ for any binary relation $\mathcal T$;
3. $\mathcal R[X]\subseteq\mathcal S[X]$ for any binary relation $\mathcal T$;
4. $\mathcal R^{-1}\subseteq\mathcal S^{-1}$.

Proof.

($1\Rightarrow 2$) So if $(x,y)$ is in $\mathcal R\circ\mathcal T$ then there exists $a$ such that $(x,a)$ is in $\mathcal T$ and $(a,y)$ in $\mathcal R$ but $\mathcal R$ is contained into $\mathcal S$ and thus $(x,a)$ is in $\mathcal T$ and $(a,y)$ in $\mathcal S$: we conclude that $(x,y)$ is in $\mathcal S\circ\mathcal T$.

($2\Rightarrow 3$) So if $\mathcal R\circ\mathcal T$ is contained for any relation $\mathcal T$ into $\mathcal S\circ\mathcal T$ then $\mathcal R\circ\operatorname{id}_X$ is contained into $\mathcal S\circ\operatorname{id}_X$. Now if $y$ is in $\mathcal R[X]$ then there exists $x$ in $X$ such that $(x,y)$ is in $\mathcal R$ but $(x,x)$ is in $\operatorname{id}_X$ so that $(x,y)$ is in $\mathcal R\circ\operatorname{id}_X$ and so in $\mathcal S\circ\operatorname{id}_X$: we conclude that $(x,y)$ is in $\mathcal S$ and so $y$ is in $\mathcal S[X]$ since by hypothesis $x$ is in $X$.

($3\Rightarrow 4$) So if $(y,x)$ is in $\mathcal R^{-1}$ then $(x,y)$ is in $\mathcal R$ so that $y$ is in $\mathcal R\big[\{x\}\big]$ which by hypothesis is contained into $\mathcal S\big[\{x\}\big]$: we conclude that $y$ is in $\mathcal S\big[\{x\}\big]$ so that there exists $x_*$ in $\{x\}$ such that $(x_*,y)$ is in $\mathcal S$ but $x_*$ must be necessarly $x$ and so $(y,x)$ is in $\mathcal S^{-1}$ since $(x_*,y)=(x,y)$ is in $\mathcal S$.

($4\Rightarrow 1$) So if $(x,y)$ is in $\mathcal R$ then $(y,x)$ is in $\mathcal R^{-1}$ and so in $\mathcal S^{-1}$: we conclude that $(x,y)$ is in $\mathcal S$.

So let's we prove the following corollary

Corollary 2

If $\mathcal R$ and $\mathcal S$ are two binary relations then the inclusion $$ \mathcal R\subseteq\mathcal S $$ holds if and only if the inclusion $$ \mathcal T\circ\mathcal R\subseteq\mathcal T\circ\mathcal S $$ for any binary relation $\mathcal T$.

Proof.

If $\mathcal R$ is contained into $\mathcal S$ then $\mathcal R^{-1}$ is contained into $\mathcal S^{-1}$ so that by thm. 1 the inclusion $$ \mathcal R^{-1}\circ\mathcal T^{-1}\subseteq\mathcal S^{-1}\circ\mathcal T^{-1} $$ holds. Now if $(x,y)$ is in $\mathcal T\circ\mathcal R$ then $(y,x)$ is in $(\mathcal T\circ\mathcal R)^{-1}=(\mathcal R^{-1}\circ\mathcal T^{-1})$ so that it is in $(\mathcal S^{-1}\circ\mathcal T^{-1})=(\mathcal T\circ\mathcal S)^{-1}$ and so $(x,y)$ is in $\mathcal T\circ\mathcal S$.

Conversely if $\mathcal T\circ\mathcal R$ is contained into $\mathcal S\circ\mathcal R^{-1}$ then for any $\mathcal T$ then $\mathcal T^{-1}\circ\mathcal R$ is contained into $\mathcal T^{-1}\circ\mathcal R$ and so by thm. 1 the inclusion $$ \mathcal R^{-1}\circ\mathcal T=(\mathcal T^{-1}\circ\mathcal R)^{-1}\subseteq(\mathcal T^{-1}\circ\mathcal S)^{-1}=\mathcal S^{-1}\circ\mathcal T $$ holds: so by thm. 1 we conclude that $\mathcal R^{-1}$ is contained into $\mathcal S^{-1}$ and so $\mathcal R$ is contained into $\mathcal S$.

Now by extensionality a set $A$ is equal to a set $B$ iff $A$ is contained into $B$ and vice versa: so using thm. 1 and crl. 2 is possibile to prove the following theorem which actually is a corollary of thm. 1.

Theorem 3

If $\mathcal R$ and $\mathcal S$ are two binary relations then the following statements are equivalent:

1. $\mathcal R=\mathcal S$;
2. $\mathcal T\circ\mathcal R=\mathcal T\circ\mathcal S$;
3. $\mathcal R\circ\mathcal T=\mathcal S\circ\mathcal T$ for any relation $\mathcal T$
4. $\mathcal R[X]=\mathcal S[X]$ for any set $X$;
5. $\mathcal R^{-1}=\mathcal S^{-1}$.