On page 1 of the article Congruence classes in regular varieties (Behlolavek & Chajda) is the the following statement:
Let $\mathcal{A} = (A,F)$ denote an algebra. Then a non-empty subset $C \subseteq A$ is a class of a congruence relation on $\mathcal{A}$ if and only if for every unary polynomial $\tau$ for $\mathcal{A}$, we have: $$\tau(C) \cap C = \emptyset \mbox{ or } \tau(C) \subseteq C.$$
This doesn't seem to be true. For example, consider the cross-shaped set $C=(\mathbb{R} \times \{0\}) \cup (\{0\} \times \mathbb{R})$ viewed as a subset of $\mathbb{R}^2$, which is viewed as a vector space over the real line. The unary polynomials $\tau$ of $\mathbb{R}^2$ are of the form $\tau(x)=ax,$ so we always have $\tau(C) \subseteq C$. But $C$ is not the class of a congruence relation on $\mathbb{R}^2$.
Question. Does anyone know what the correct version of this statement is?
The statement is correct. "Polynomials" are allowed to refer to individual elements of your algebra as constants, so the unary polynomials for a vector space are not just $ax$ for $a$ a scalar but also $ax+v$ where $a$ is a scalar and $v$ is an element of the vector space. Your example $C$ does not satisfy $\tau(C)\cap C=\emptyset$ or $\tau(C)\subseteq C$ for all polynomials $\tau$ of this form.