By concrete category we will understand pair $(\mathcal{C},U)$ , where $C$ is a category and $U:\mathcal{C}\to\mathcal{SET}$ is a faithful (covariant) functor. Assume that $(\mathcal{C},U)$ is a concrete category.

Definition 1. Let $X$ be an object in $\mathcal{C}$ . Subset $B\subseteq U(X)$ (possibly empty) is called a basis of $X$ if for any object $Y$ in $\mathcal{C}$ and any function $g:B\to U(Y)$ there exists exactly one morphism $\alpha:X\to Y$ such that $U(\alpha)(x)=g(x)$ for any $x\in B$ . In this case we will say that $g$ lifts to $\alpha$ .

Definition 2. Object $X$ will be called free if there exists basis of $X$ .

Free objects generalize the notion of free modules over a ring. Some of the properties of free modules can be easily generalized to free objects in arbitrary concrete category. For example:

Proposition. Let $X$ and $Y$ be free objects with bases $B$ and $B'$ respectively and let $f:B\to B'$ be a function. The following statements hold:
$\mathrm{i)}$ If $f$ is an injection, then there exists a section $\alpha:X\to Y$ in $\mathcal{C}$ ;
$\mathrm{ii)}$ If $f$ is a surjection, then there exists a retraction $\beta:X\to Y$ in $\mathcal{C}$ ;
$\mathrm{iii)}$ If $f$ is a bijection, then $X$ and $Y$ are isomorphic.

Proof. $\mathrm{i)}$ Assume that $f:B\to B'$ is an injection. Let $f_1:B\to U(Y)$ be defined as $$f_1(x)=f(x)$$ for all $x\in B$ . Now, since $f:B\to B'$ is an injection, then there exists a surjection $f':B'\to B$ such that $$f'(f(x))=x$$ for all $x\in B$ . Let $f_2:B'\to U(X)$ be defined by $$f_2(y)=f'(y)$$ for all $y\in B'$ . Now both $X$ and $Y$ are free and thus there are morphism $\alpha: X\to Y$ and $\beta:Y\to X$ such that $$U(\alpha)(x)=f_1(x)\mbox{ and }U(\beta)(y)=f_2(y)$$ for all $x\in B$ and $y\in B'$ . It is easy to check, that this implies that $$U(\beta\circ\alpha)(x)=x$$ for all $x\in B$ . But $U(\mathrm{id}_{X})(x)=x$ for all $x\in B$ and thus canonical injection $i:B\to U(X)$ lifts to both $\beta\circ\alpha$ and $\mathrm{id}_{X}$ . Since lift is unique, then $\beta\circ\alpha=\mathrm{id}_{X}$ , so $\alpha$ is a section.

$\mathrm{ii)}$ Note that if $f:B\to B'$ is a surjection, then there exists an injection $g:B'\to B$ such that $f(g(y))=y$ for all $y\in B'$ . Thus, from $\mathrm{i)}$ we obtain that $\beta\circ\alpha=\mathrm{id}_{Y}$ for $\alpha:Y\to X$ and $\beta:X\to Y$ constructed as in $\mathrm{i)}$ . Therefore $\beta:X\to Y$ is a retraction.

$\mathrm{iii)}$ If $f$ is a bijection, then proof of $\mathrm{i)}$ and $\mathrm{ii)}$ shows that there are two morphisms $\alpha:X\to Y$ and $\beta:Y\to X$ such that $\beta\circ\alpha=\mathrm{id}_{X}$ and $\alpha\circ\beta=\mathrm{id}_{Y}$ . Thus $X$ and $Y$ are isomorphic. $\square$

Remark 1. Free objects does not have to exist. For example, the category of finite groups (without the trivial group) and group homomorphisms (where $U$ is a forgetful functor) does not have free objects (this is because there are no nontrivial group homomorphisms between groups with relatively prime orders).

Remark 2. Note that, if there is a free object $X$ in a concrete category $(\mathcal{C},U)$ such that $\emptyset$ is a basis of $X$ , then $X$ is an initial object. This follows directly from the definition, since any morphism $\alpha:X\to Y$ is a lift of $f:\emptyset\to U(Y)$ , thus it has to be unique. Conversly one can easily show, that initial object is always free with $\emptyset$ as a basis.