Definition of $\Omega$-algebra I'm studying universal algebra. I have this definiton: given a signature $\Omega$, an $\Omega$-algebra is comprised of a "carrier" for the algebra and an "interpretation" for every operation symbol. The carrier of $A$ is a set (written $|A|$). The interpretation of $\omega\in \Omega$ is a function $a_{\omega}:|A|^{ar(\omega)}\longmapsto |A|$, where ar is the function associating to each operation symbol its arity.
Now, I can't understand the following "equivalent" definition: an $\Omega$-algebra is essentially a carrier $|A|$ and a single function
$$a:\displaystyle\sum_{\omega\in\Omega}|A|^{ar(\omega)}\longmapsto |A|$$
in which $\sum$ is the disjoint union operator. The domain is the set of terms which consist of a single operation applied to elements of $|A|$ and the function $a$ evaluates those in $|A|$.
This sounds obscure to me: what actually this functions $a$ do? If the domain is the disjoint union, how can 'a single operation applied to....' be elements of that domain?
 A: The function $a$ can receive arguments of several types (where each type is a term in the union). For consonance with the use of $\Sigma$, let's denote disjoint union as $+$.
Given a set $X$ and functions $a_1 \colon X \to X$ and $a_2 \colon X \times X \to X$, we may render the $a$s as a single function with signature
$$
  a \colon X + X \times X \to X,
$$
 where
$$
  a(y) = \left\{\begin{aligned}
      a_1(x) \quad &\text{iff}\ y = x \in X \\
      a_2(x_1, x_2) \quad &\text{iff}\ y = (x_1, x_2) \in X \times X
  \end{aligned}\right.
$$
So, given an $\Omega$-algebra $A$, with the operators $\omega$ and interpretations
$$
  a_\omega \colon \lvert A \rvert^{ar(\omega)} \longrightarrow \lvert A \rvert
$$
we have the equivalent but more compact representation as the single function
$$
  a:\displaystyle\sum_{\omega\in\Omega}|A|^{ar(\omega)} \longrightarrow |A|
$$
where
$$
  a(y) = \left\{\begin{aligned}
    a_{\omega_1}(…) \quad &\text{iff}\ …\\
    a_{\omega_2}(…) \quad &\text{iff}\ …\\
    &\vdots
  \end{aligned}\right.
$$
for all $\omega \in \Omega$, where I hope you will forgive my abuse of notation in enumerating the $\omega$s.
