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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?

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    $\begingroup$ A less elementary, but highly important, situation in which these maps must be injective is when the adjunction is monadic, as in the case of free algebraic structures and in almost all other situations in which someone would be talking about "free" objects. $\endgroup$ Mar 22, 2020 at 23:01

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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 $U(g')(f(x))=U(g')(f(y))$.

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  • $\begingroup$ Really helpful! Thanks very much! $\endgroup$
    – user762130
    Mar 22, 2020 at 18:58

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