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From Rotman's Algebraic Topology:

When we say that $\text{Hom}(\space , G)$ takes values in Groups, then it follows, of course, that $\text{Hom}(X, G)$ is a group for every object $X$ and $g^*$ is a homomorphism for every morphism $g$; a similar remark holds if $\text{Hom}(G,\space )$ takes values in Groups.

What does "$\text{Hom}(, G)$takes values in Groups" mean? The text seems to indicate that $\text{Hom}(X, G)$ is a group for every object $X$ follows from assuming this, but I don't know what this means.

Anyone know what this means?

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    $\begingroup$ "$\operatorname{Hom}(X, G)$ is a group for every object $X$" is exactly what it means. The "$g^*$ is a homomorphism for every morphism $g$" part is better described as something that follows. $\endgroup$ – Arthur Sep 25 '17 at 14:13
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    $\begingroup$ "takes values in" is precisely what the author explains in that sentence. Maybe instead of "then it follows, that", they should have written "we mean by this, that". In other words, it means that $\operatorname{Hom}(\cdot, G)$ is a functor from whatever category to Grp (or possibly to some other category with an obvious forgetful functor to Grp) $\endgroup$ – Hagen von Eitzen Sep 25 '17 at 14:13
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    $\begingroup$ Could mean that there is a factorization $C \to (Grp) \to (Set)$ where the second functor is the forgetful one and the composition is your $\mathrm{Hom}(-,G)$. That's equivalent to what he says afterwards (but don't drop the requirement of functoriality!). $\endgroup$ – Johann Haas Sep 25 '17 at 14:20
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It is stronger than saying that $\text{Hom}(X, G)$ is a group for every $X$; it is also saying that this group structure is natural in $X$.

More formally, for any object $G$ the functor $\text{Hom}(-, G)$ takes values in sets, meaning that it is a functor $C^{op} \to \text{Set}$. What it means for this functor to take values in groups is that it factors through a functor $C^{op} \to \text{Grp}$. And in fact you can show that such a factorization is equivalent to a choice of group object structure on $G$.

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