In Mac Lane's Categories for the Working Mathematician, page 110, second edition, he states that, in the category of groups $Grp$, $F\_n$ being groups and $x\_n, y\_n$ being two cones on these groups (thus a family of group morphisms from X (resp. Y) to $F\_n$), these cones form a group by multiplication $x\_n y\_n$.


As I understand it, it just seems false. Let be two group homomorphisms $g, f$, there is no reason why $x \mapsto g(x) f(x)$ should also be a group morphism, as we cannot conclude whether $g(x)f(x)g(x)^{-1}f(x)^{-1} = e$.


So I probably missed something... I could use a hint!


Edit: as hinted in the comments, it is probably that Mac Lane was reasoning from inside Sets category, instead of Grp. But in this case it looks like it defeats the purpose of the example, as it comes just after a theorem stating that Sets has all small limits, and it would have been a good occasion to show how it works elsewhere than in Sets.



Product of two group morphisms not a group morphism
From the comments:


The *cones* are in $\mathrm{Set}$, but the set of *cones*, under pointwise multiplication is the limit in $\mathrm{Grp}$. In fancier language, MacLane's saying that limits in $\mathrm{Grp}$ are created in $\mathrm{Set}$.


