TITLE: Difference between 'free algebra' and 'absolutely free algebra'
QUESTION [5 upvotes]: What is the difference between a free algebra and an absolutely free algebra? Wikipedia and Encyclopaedia of Maths are not very clear on the subject...

REPLY [5 votes]: Absolutely free $\Omega$-algebra is the same thing as the term-algebra , where $\Omega$ is a given signature (in the sense of universal algebra). 
We have the  category $\textbf{$\Omega$-Alg}$, where objects are $\Omega$-algebras and arrows are $\Omega$-algebra homomorphisms. There is the canonical forgetful functor $U:\textbf{$\Omega$-Alg} \rightarrow \textbf{Set}$, which maps $\Omega$-algebras to their underlying set. The absolutely-free algebra is simply a functor $T:\textbf{Set} \rightarrow \textbf{$\Omega$-Alg}$ which is left-adjoint to $U$.
A variety  $\textbf{V}$ of $\Omega$-algebras is a full-subcategory of $\textbf{$\Omega$-Alg}$ which is closed under products, subalgebras and homomorphic images (in $\textbf{$\Omega$-Alg}$). There is also a canonical forgetful funtor $U_V:\textbf{V} \rightarrow \textbf{Set}$, which is just the restriction of $U$ to  $\textbf{V}$. The free-algebra in $\textbf{V}$, is simply a functor $F_V:\textbf{Set} \rightarrow \textbf{V}$, which is left-adjoint to $U_V$. 
Various explicit constructions of term-algebras and free algebras exist, however, they all are left adjoints to their respective forgetful functors, and are thus unique up to (unique) isomorphism. 
So, to summarize: 

Absolutely-free $\Omega$-algebra = left adjoint to $U:\textbf{$\Omega$-Alg}\rightarrow \textbf{Set}$

and

Free-algebra in $\mathbf{V}$ =  left adjoint to $U_V:\textbf{V} \rightarrow \textbf{Set}$. 

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