Show $(a,b)$ is in the equivalence relation $L_s$ iff $(ap,bp)$ is in $L_{s/p}$ where p is a congruence on semigroup S So the question that I'm struggling with is as follows: 
Let S be an arbitrary semigroup, and define an equivalence relation $L_S$ on S by
$(a, b) ∈ L_S$ iff $S^1 a = S^1b$.
Let $ρ$ be a congruence on S such that $ρ ⊆ L_S$.
Show that $(a, b) ∈ L_S$ in S iff $(aρ, bρ) ∈ L_{S/ρ}$ in
$S/ρ$.
I believe that I have the forward direction but I'm not sure so firstly is this correct?
$(a, b) ∈ L_S$ and $ρ ⊆ L_S$ implies that $aρb$
$\Rightarrow aρ=bρ$
$\Rightarrow (S/ρ)(aρ)=(S/ρ)(bρ)$
$\Rightarrow (aρ, bρ) ∈ L_{S/ρ}$
For the reverse proof I am completely stumped! I have tried plugging away at a few things and proof by contradiction but nothing seemed to lead the the answer I was hoping for.
 A: Your argument is incorrect. Indeed $\rho \subseteq \mathcal{L}_S$ means $(a,b) \in \rho$ implies $(a,b) \in \mathcal{L}_S$ (and not the other way around, as you wrote).
Anyway, the implication
$$
(a, b) \in \mathcal{L}_S \implies (a\rho, b\rho) \in \mathcal{L}_{S/\rho} 
$$
is true without any assumption on $\rho$. Indeed let $\rho: S \to S/\rho$ be the quotient map. Then $(a, b) \in \mathcal{L}_S$ means that $S^1a = S^1b$. This implies that $(S\rho)^1(a\rho) = (S^1a)\rho = (S^1b)\rho = (S\rho)^1 (b\rho)$, that is $(a\rho, b\rho) \in \mathcal{L}_{S/\rho}$.
Now, for the opposite implication
$$
(a\rho, b\rho) \in \mathcal{L}_{S/\rho} \implies (a, b) \in \mathcal{L}_S  
$$
your hypothesis is required. If $(a\rho, b\rho) \in \mathcal{L}_{S/\rho}$ there exists some $c, d$ in $S^1$ such that $(c\rho)(a\rho) = b\rho$ and $(d\rho)(b\rho) = a\rho$. It follows that $(ca)\rho = b\rho$ and $(db)\rho = a\rho$. Therefore $(ca, b) \in \rho$ and $(db, a) \in \rho$. Since $\rho \subseteq \mathcal{L}_S$, one gets $(ca, b) \in \mathcal{L}_S$ and $(db, a) \in \mathcal{L}_S$. It follows that $S^1b = S^1ca \subseteq S^1a$ and $S^1a = S^1db \subseteq S^1b$, whence $S^1a = S^1b$ as required.
