I've just started reading about universal algebra, and have already hit a problem (see the two bullet points at the bottom).
My book gives the following definitions (paraphrased):
An operational type is a pair $ (\Omega,\alpha) $, where $ \Omega $ is a set of operational symbols and $\alpha$ is an 'arity' function which assigns to each operational symbol a natural number (including 0).
Given an operational type $ (\Omega, \alpha) $ and a set $A$, an $ \Omega$-structure on $A$ is a family of functions $ (\omega_A | \omega \in \Omega) $, where $ \omega_A : A^{\alpha(\omega)} \to A $ for each $\omega \in \Omega$ is the interpretation of the symbol $\omega$ in the structure $A$.
A homomorphism $f : A \to B $ of $\Omega$-structures is a function such that $f(\omega_A(a_1,...,a_{\alpha(\omega)})) = \omega_B(f(a_1),..., f(a_{\alpha(\omega)}) $
Given a set $X$ of variables and operational type $\Omega$ ($X \cap \Omega = \emptyset $), the set $ F_\Omega (X) $ of $\Omega$-terms is defined inductively:
a) If $x \in X$, then $x \in F_\Omega (X) $
b) If $\omega \in \Omega$, with $\alpha(\omega) = n$ and $t_1, t_2, ..., t_n \in F_\Omega(X) $, then $\omega t_1 t_2 ... t_n \in F_\Omega(X) $.
It then states the following theorem:
i) $F_\Omega(X) $ has an $\Omega$-structure
ii) $F_\Omega(X) $ is the free $\Omega$-structure generated by $X$
Now, my problem is with the second part of this theorem. Suppose we have a function $f: X \to A$, where $A$ is some $\Omega$-structure. I can see that we can define a homomorphism $ \bar{f} $ such that
if $ t = x \in X $, $\bar{f}(t) = f(x)$,
and if $ t = \omega t_1 t_1 ... t_n $, then $ \bar{f}(t) = \omega_A(\bar{f}(t_1), ... , \bar{f}(t_n)) $.
I think, but am not 100% sure, that because we've defined the set $F_\Omega(X) $ inductively, all $\Omega$-terms are either in $X$ or are in the form $ \omega t_1 ... t_n $ for some $\omega, t_i$. Is this correct?
I can see that this $\bar{f}$ is a homomorphism, but why is it the unique homormorphism extending $f$? I tried constructing a uniqueness argument, but it didn't really get anywhere: suppose we have some homomorphism $g : F_\Omega (X) \to A $ such that $g(x) = f(x) \ \forall x \in X $. If $ t = \omega t_1 ... t_n $ for some $ \omega \in \Omega $ and $ t_i \in F_\Omega (X) $, then $ g(t) = g(\omega t_1 ... t_n) = g( \omega_{F_\Omega(X)} (t_1, ... , t_n) ) = \omega_A ( g(t_1), ... , g(t_n)) $. Then I'm unsure how to finish the argument. I've thought about continuing this idea on $g(t_1), ... g(t_n) $ inductively until the arguments are contained in $X$, and then use the fact that $ \bar{f}(x) = g(x) \ \forall x \in X$. I feel it should be easier than this and so think I'm missing something obvious.
Thanks a lot!