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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.

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    $\begingroup$ MacLane writes "...the set L of all cones (all matching strings x)...". So, he seems to mean cones in the category of sets, not groups. $\endgroup$
    – Friedrich Knop
    Commented Apr 1, 2019 at 6:57
  • $\begingroup$ Ok, thank you. In this case, this is what lost me. I still don't see the point to be inside Sets, as we already know that Sets have all small limits. It would have been of much value to construct the limit in Grp, which would have been an example of how to do it without the use of the one-point set. $\endgroup$
    – Almeo Maus
    Commented Apr 1, 2019 at 7:02
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    $\begingroup$ "There's no reason why $x\mapsto g(x)f(x)$ should be a group homomorphism": you should easily find examples to confirm this expectation. $\endgroup$
    – YCor
    Commented Apr 1, 2019 at 7:32
  • $\begingroup$ @YCor Yes indeed, if the source group is $(R,+)$, if the destination group is the direct isometries of the $R^2$ plane, $E^+(2)$, then if we take $f(1)=rot([0,0],\pi/2)$ and $g(1)=trans(1*e_x)$, we get $g(1)f(1)g(1)^{-1}f(1)^{-1}\neq id$ $\endgroup$
    – Almeo Maus
    Commented Apr 1, 2019 at 8:01
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    $\begingroup$ The cones are in $\rm Set$, but the set of cones, under pointwise multiplication is the limit in $\rm Grp$. In fancier language, he's saying that limits in $\rm Grp$ are created in $\rm Set$. $\endgroup$
    – Mike Shulman
    Commented Apr 1, 2019 at 9:42

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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 $\mathrm{Set}$.

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