# Category Theory: Free Abelian Groups and Coproducts

This is probably obvious to someone familiar with Category Theory (I'm just starting) but why are the following two statements true?

$$\operatorname{Hom}(\operatorname{F}(S),-)\cong \operatorname{Map}(S,-)\cong \operatorname{Prod}(-;S)$$ and $$\operatorname{Hom}(\operatorname{Copr}(\mathbb{Z};S),-)\cong \operatorname{Prod}(\operatorname{Hom}(\mathbb{Z},-);S) \cong \operatorname{Prod}(-;S),$$ Where, $\operatorname{F}(S)$ is the free abelian group on the set $S$, $\operatorname{Map}$ denotes morphisms in the category of sets, $\operatorname{Copr}(\mathbb{Z};S)$ is the coproduct of the family of copies of the integers $\mathbb{Z}$ indexed by the set $S$ and $\operatorname{Prod}$ is the product functor.

For the first 2 isomorphisms: The first one is the definition of the free Abelian group. For the second, take an Abelian group $A$. Consider the homomorphism $$f_A:Map(S,A) \to Prod(A;S): u \mapsto (u(s))_{s \in S}$$ and show that is an isomorphism.

The other isomorphisms are proved similarly.

• It's clear how a map $u:S\to A$ is isomorphic to evaluating $u$ at every $s\in S$. So that means I have an incorrect understanding of the product functor. How do you get $(u(s))_{s \in S}$ as an element of $Prod(A;S)$? – Bob Jan 20 '16 at 4:33
• Well the product in $\mathbf{Ab}$ is just the good old cartesian product. – Nitrogen Jan 20 '16 at 4:43
• Ah I get it, thanks! The translation of category theory from the greek is, "so simple, yet so confusing". – Bob Jan 20 '16 at 4:51