Does the map from a set to the free object on the set have to be injective?

Question

I've seen the following definition of a free object in category theory.

Let $\mathcal C$ be a concrete category. Denote by $U\colon\mathcal C\to\mathrm{Set}$ the forgetful functor. Let $X$ be a set. Then an object $F(X)\in\mathcal C$ equipped with an arrow $f_X\colon X\to U(F(X))$ is called the free object of $\mathcal C$ on $X$ if: for all $A\in\mathcal C$ and any function $g\colon X\to U(A)$ in the category of sets, there exists a unique "extension" $g'\colon F(X)\to A$ in the category $\mathcal C$ such that $U(g')\circ f_X=g$.

I looked at some concrete examples of free objects in the category of groups and the category of modules. In each case, the arrow $f_X$ was injective. Does this follow from the definition? Why isn't it included in the definition?

Answer 1

No, it does not follow from the definition. Indeed, it does not even follow from the definition in the case of modules in general. Suppose $R$ is the zero ring (the ring with one element). Then every module over $R$ has one element, and this single module is free on every possible set via every possible map.

If you assume there exists an object $A$ of $\mathcal{C}$ such that $U(A)$ has more than one element, then it does follow that $f_X$ must be injective for any free object. Indeed, suppose $F$ is a free object on $X$ via a map $f_X:X\to U(F)$ and $f_X(x)=f_X(y)$ for some distinct $x,y\in X$. Since $U(A)$ has more than one element, there is a function $g:X\to U(A)$ such that $g(x)\neq g(y)$. Taking the unique $g':F\to A$ such that $U(g')\circ f_X=g$, we get a contradiction, since $U(g')(f(x))=U(g')(f(y))$.