Following [the case of groups](https://math.stackexchange.com/questions/541590/free-object-isomorphisms), I asked in [this MSE question](https://math.stackexchange.com/questions/1934096/quick-proof-that-free-objects-on-sets-of-different-cardinality-are-not-isomorphi) for a quick proof that given a free-forgetful adjunction $F\dashv U$ for some algebraic theory, we have $X\not\cong Y\implies FX\not\cong FY$. I was really surprised to find out this isn't true at all.


On the other hand, it seems that for the usual theories, i.e of (abelian) groups, (commutative) rings, modules, monoids, this is true, usually by finding a canonical way to vector spaces and using the dimension theorem. What is the underlying property of such theories that makes this theorem true? Can it be stated without reference to vector spaces, using only the axiom of choice? Is there a nice sufficient condition?



Free algebras on sets of different cardinality – for what theories are they non-isomorphic?
It is not true for "usual theories", at least if you accept that left $R$-modules for noncommutative $R$ is a usual theory. Take $R=\mathrm{End(}V)$ for some infinite-dimensional vector space $V$, then $R \cong R^2$ as left $R$-modules. Rings for which $\forall n,m : \mathbb{N}. R^n \cong R^m \Rightarrow n=m$ holds (the case of infinite bases follows by cardinality arguments) are called IBN-rings; they are studied for example in the beginning of Lam's *Lectures on modules and rings*.


As for the general question, the following criterion is very useful: Assume that there is some algebra $A$ with a finite underlying set $U(A)$ which has at least $2$ elements. Then, if $F(X) \cong F(Y)$ holds for two finite sets $X,Y$, we get
$$\hom(X,U(A)) \cong \hom(F(X),A) \cong \hom(F(X),A) \cong \hom(Y,U(A)).$$
Counting yields $X \cong Y$.


This was also remarked [here](https://mathoverflow.net/questions/126747/ibn-for-algebraic-theories), an almost identical question.


