4. Existence of free algebras in arbitrary varieties

Proposition. For every variety $V$ and every $n$ there exists a free algebra in $V$ with $n$ generators. In other words, $F _ V (n)$ always exists.

Outline of proof #1: The method of saving term relations in common.

This is a generalization of the ``table'' method (above) for a single algebra: We start by considering all functions $\delta: \{1,\dots, n\} \rightarrow A$ where $A$ runs through all algebras in $V$. Since $V$ is too large to be a set, there are also too many $\delta$'s, so we restrict our attention to cases where the image of $\delta$ generates $A$, and we remark that up to isomorphism there is only a set (rather than a class) of ways in which an image of such a $\delta$ can sit inside the $A$ it generates. Let $\Delta$ consist of one $\delta$ from each isomorphism class. Then inside $A ^ \Delta$, for $i=1,\dots n$ let $g _ i$ be the element whose $\delta$-th coordinate is $\delta(i)$, and let $F$ be the subalgebra of $A ^ \Delta$ generated by $g _ 1,\dots, g _ n$. Then we remark that $F$ has the Universal Mapping Property (UMP), so is free. I call this the method of ``saving relations in common'', because the only relations $t=u$ between the $g _ i$ are those true in every factor, and the factors account for all ways that $n$ elements of an algebra in $V$ can be related. As you see, there are two elements in this proof: choosing the isomorphism types and taking the subalgebra of a product.

Outline of proof #2: The method of overshooting.

For $T = T _ \tau (n)$ (the algebra of all terms in $n$ variables), let $F _ V (n) = T/\theta _ 0$, where $\theta _ 0 = \cap \{\theta \in$   Con$(T): T / \theta \in V\}$. Here $\theta _ 0$ is the least congruence relation $\theta$ on $t$ such that $T/\theta \in V$. One can show that $F _ V (n)$ inherits the UMP from $T$, which is free in the variety of all algebras of type $\tau$. I call this the ``overshooting'' method: Since $T$ is free but much too big, you have overshot, and you must trim $T$ down to where it fits in $V$, by taking $T$ modulo a congruence relation.