How to show that the category of $\mathbb{Z}$-modules is isomorphic to the catefory of abelian groups. It seems obvious for me that a $\mathbb{Z}$-module is an abelian group and conversely. But I don't know how to write it using functors.
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1Well take the "identity functor" $Ab \to \Bbb Z-Mod$, sending an abelian group to the underlying $\Bbb Z$-modules and verify all the axioms ... – Nicolas Hemelsoet Oct 01 '18 at 12:39
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Yeah okay that what I thought just was weird to write because there is no proper definition of this functor. – Oct 01 '18 at 15:40
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1@Pierre21 The proper definition is to send a $\mathbb{Z}$-module to the underlying group gotten by forgetting the action of $\mathbb{Z}$, sending a map of $\mathbb{Z}$-modules to the identical map of abelian groups, and inversely, to equip an abelian group (morphism) with its unique strucutre of a $\mathbb{Z}$-module (morphism.) These are perfectly good functors even if they look a bit trivial, as they are. – Kevin Carlson Oct 02 '18 at 01:00
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The are many different foundations where mathematics (particularly, category theory, algebra) can be done. But usually the classical category theory (and algebra) is based on the set theory or one of its standard variations (for example, including classes or universes etc). So, by default, all questions on MathStackExchange (I guess) are perceived by the community within the framework of the set theory.
In a standard form of the set theory, by a standard definition, a $\mathbb{Z}$-module is not an abelian group (and, similarly, an abelian group is not a $\mathbb{Z}$-module). The reason is that an abelian group is a set with one operation and a $\mathbb{Z}$-module is a set with two operations (one of which is an action of $\mathbb{Z}$, i.e., an external operation). A set with one operation can't be a set with two operations, because the first is a pair, but the second is a triplet.
I do not want to seem an excessive formalist; on the contrary, I believe that it is one of the most important effects of the category theory, that it provides convinient tools for understanding when formally different objects are mathematically the same. It is the case when their categories are...equivalent. Isomorphism is even a stronger property.
So, by definition, to prove that abelian groups are mathematically the same as $\mathbb{Z}$-modules, it is sufficient to find the two functors $\mathcal{F}\colon\mathbf{Ab}\to(\mathbb{Z}-\mathbf{Mod})$ and $\mathcal{G}\colon(\mathbb{Z}-\mathbf{Mod})\to\mathbf{Ab}$, such that $\mathcal{G}\circ\mathcal{F}=\text{I}_{\mathbf{Ab}}$ and $\mathcal{F}\circ\mathcal{G}=\text{I}_{\mathbb{Z}-\mathbf{Mod}}$. In order to find such functors you should think how to construct in a natural way a $\mathbb{Z}$-module by an abelian group and vice versa. If you are not familiar with the categorical notions (functor, composition, identity functor, forgetful functor), then I recommend the textbook of Mac Lane "Categories for the working mathematician".
Oskar
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1As a bit of an aside, in some formulations a group has three operations (or two operations and a constant). Then again, maybe this isn't so much of an aside as yet another demonstration of the main point. – Oct 01 '18 at 16:14
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@Hurkyl I definitely agree. Also, to demonstrate this point via equivalent categories, we may define $\mathbf{Ab}$ as a category of abelian-group-objects (without fixed cartesian monoidal structure) in $\mathbf{Set}$ and $\mathbb{Z}-\mathbf{Mod}$ as a category of module-objects over $\mathbb{Z}$ in $\mathbf{Ab}$. – Oskar Oct 01 '18 at 16:24
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...or even as a category of commutative groupoids with one object. It's a good exercise to prove that all these categories are equivalent. – Oskar Oct 01 '18 at 16:33
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