I am reading the book "Notes on logic and set-theory" from P.T Johnstone.
It gives the following definition:
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:
$F_\Omega(X) $ is the free $\Omega$-structure generated by $X$, i.e., given any $\Omega$-structure $A$ and any function $f:X \rightarrow A$, there exists a unique homomorphism $\bar{f}:F_\Omega(X) \rightarrow A$ extending $f$
About the preceding definition of "free $\Omega$-structure generated by $X$" I have the following questions:
- What does it mean that an homomorphism extends a function?
- What is the general meaning of that definition?
Concerning your second question, I am not quite sure what you would like to hear, but in some sense $F_{\Omega}(X)$ is the "biggest" $\Omega$-structure containing the given set of variables $X$: for every $\Omega$-structure $A$, which contains the variables $X$ you can find a morphism of $F_{\Omega}(X)$ into $A$. Also, the theorem in your question tells you that whenever you want to construct a morphism $F_{\Omega}(X) \rightarrow A$ of $\Omega$-structures, you only need to define it on $X$, for it will then extend uniquely to all of $F_{\Omega}(X)$. That is one nice thing about free $\Omega$-structures.
In a more general vein, "free" objects appear in many different contexts in mathematics (free groups, free vector spaces, free monoids etc.), and they can be defined in category-theoretic terms as certain functors, which are (typically left-) adjoint to so-called forgetful functors (you may ignore this last sentence on a first reading, if it confuses you). You maybe want to check out the wikipedia entry to learn more about it.
By the way, just in order to test your understanding of $F_{\Omega}(X)$, here is a little question.