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Let $F$ be the following covariant functor from the category of sets to the category of left module over a ring $R$ with identity. For each set $X$, $F\left(X\right)$ is the free $R$-module on $X$. If $f : X\rightarrow X'$ is a function, let $F\left(f\right) : F\left(X\right)\rightarrow F\left(X'\right)$ be the unique module homomorphism $\overline f : F\left(X\right)\rightarrow F\left(X'\right)$ such that $\overline fi=f$, where $i$ is the inclusion map $X\rightarrow F\left(X\right)$.

This is an example in the Hungerford. I think that the given functor $F$ is not well-defined. For an object $X$ in the category of sets there are many free $R$-module on $X$. So it seems to consider a category of isomorphic classes of left module over a ring $R$ with identity and a covariant functor from the category of sets to this category. Is it right?

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« the Hungerford» = «the algebra book authored by Hungerford»? – Mariano Suárez-Alvarez Feb 3 '14 at 5:51
up vote 6 down vote accepted

No, he has in mind a specific construction of the free module on a set, so that fixes the choice.

For example, you could take the set of all functions $X\to R$ with finite support with its obvious structure of $R$-module as $F(X)$, or you could take some other fixed construction.

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There is no sensible category of «isomorphism classes of left mdules over a ring $R$». You could use instead a skeleton of the category of $R$-modules, but that is not a great idea for several reasons (see here for some discussion on this) – Mariano Suárez-Alvarez Feb 3 '14 at 5:53
(With a sufficiently strong form of the axiom of choice, you could instead simply pick one possible free module $F(X)$ for each $X$. You need a very strong form of AC because you need to make a nonsetful of choices for this, and that makes some people cringe) – Mariano Suárez-Alvarez Feb 3 '14 at 5:56

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