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In MacLane'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 MacLane 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.

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.

In MacLane'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!

In MacLane'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 MacLane 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.