If a variety of algebras V has the CEP then $HS(K) = SH(K), K \subseteq V$

Question

CEP := congruence extension property

$SH(K) \le HS(K)$ is always true; I have a lemma about it, not going to list it here.

Just need to get $HS(K) \le SH(K)$

Let $A \in HS(K)$ then $A$ is the homomorphic image of a subalgebra of some $B \in K$. So namely, $\varphi : B \to C$ is a homomorphism with $B' \le B$ and $A \le C$ we get that $A = \varphi[B']$

And

$B \in K$ so $B$ has the CEP. So for $\theta \in Con(B')$ there is $\gamma \in Con(B)$ such that $\theta = \gamma \cap (B')^2$. Need to get $A$ to be a subalgebra of a homomorphic image of $B$.

I suspect the correspondence theorem is the key here. since $[\gamma, \nabla] \simeq B'/\theta$ under this theorem.

You don't need the $C$ above. Actually, you just can start saying "namely, $\varphi:B'\to A$ is an onto homomorphism." — Pedro Sánchez Terraf, Oct 17 '16 at 20:12

score 2 · Accepted Answer · answered Oct 17 '16 at 20:54

Let $\mathbf{A} \in HS(K)$.
Then $\mathbf{A} = \mathbf{B}/\theta$, where $\mathbf{B} \in S(K)$ and $\theta$ is a congruence on $\mathbf{B}$.
From $\mathbf{B} \in S(K)$ it follows that $\mathbf{B} \leq \mathbf{C}$, for some $\mathbf{C} \in K$.
By CEP, there is a congruence $\sigma$ on $\mathbf{C}$ such that $\sigma \cap B^2 = \theta$.

Thus $\mathbf{A} = \mathbf{B}/\theta \leq \mathbf{C}/\sigma$, that is $\mathbf{A} \in SH(K)$.

If a variety of algebras V has the CEP then $HS(K) = SH(K), K \subseteq V$

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