K-free algebra over $\overline{X}$

Question

This is Definition 10.9 of the book "A Course in Universal Algebra" by Burris and Sankappanavar (page 73, Millennium Edition).

http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra.pdf

Definition 10.9 Let $K$ be a family of algebras of type $\mathfrak{F}$. Given a set $X$ of variables define the congruence $\theta_K(X)$ on $T(X)$ by

$\theta_K(X)=\bigcap\Phi_K(X)$,

where $\Phi_K(X)=\{\phi\in \textrm{Con } T(X):T(X)/\phi \in IS(K)\}$; and then define $F_K(\overline{X})$, the $K$-free algebra over $\overline{X}$, by $F_K(\overline{X})=T(X)/\theta_K(X)$, where $\overline{X}=X/\theta_K(X)$.

In the following Remarks (Remark (5)), the book says that "If $K$ has a nontrivial algebra $A$ and $T(X)$ exists, then $X \cap (x /\theta_K(X))=\{x\}$as distinct memebers $x, y$ of $X$ can be separated by some homomorphism $\alpha:T(X) \rightarrow A$."

I am having hard time understanding the above remark. In particular, I have not found an example of $\phi\in \textrm{Con } T(X)$, where $T(X)/\phi \in IS(K)$ for the Remark (5).

Any help will be appreciated.

Answer 1

Hint: Let $a, b \in A \colon a \neq b$. Fix some $x \in X$ and consider the homomorphism $\alpha \colon T(X) \to A$ which extends this mapping $x \mapsto a, y \mapsto b, \forall y \in X \setminus \{x\}$. It exists because $T(X)$ is the absolutely free algebra over $X$ (or you can prove directly by the induction on the construction of a term that such an extension exists).

Consider the congruence $\phi = \ker \alpha$.

1. Prove that $T(X)/\phi \in IS(K)$ (use the homomorphism theorem and the fact that the homomorphic image is the subalgebra of the codomain of this homomorphism).
2. Prove that $\lnot x \phi y, \forall y \in X \setminus \{x\}$ (use the definition of the kernel congruence).

From 1 it follows that $\phi \in \Phi_{K}(X)$ and from 2 it follows that $\alpha$ separates $x$ from all elements of $X \setminus \{x\}$.