Two group structures on Hom(C, G)

Question

Let $\mathbf{C}$ be a pointed category (i.e. a category with a zero object), and let $C$ and $G$ be respectively a cogroup and a group object in $\mathbf{C}$. In this situation, the set $Hom_{\mathbf{C}}(C, G)$ gets two different group structures, one because $C$ is a cogroup and the other because $G$ is a group. In problem 1.60 of Strom's Modern Classical Homotopy Theory is asked to show that these two structures coincide. How can I prove this? I have written down a couple of big diagrams (as hinted in the book), but I can't seem to find the answer.

Answer 1

You should understand how the operations interact.

Once I read this exercise in a category book: Consider a topological group G. Then the omotopy group of G has two operations: you can ordinarily compose two cycles ($*$), or multiply them pointwise ($\cdot$). Then prove that

1. $(\gamma * \gamma')(\delta * \delta') = (\gamma \delta) * (\gamma' \delta')$. This is quite straightforward if you think that pointwise multipliction is made (first cycle $\cdot$ first cycle) and (second cycle $\cdot$ second cycle).

2. There is an element $e$ which is an identity for both operations.

3. Show that these two properties imply that the operation coincide and are abelian.

I found this approach enlightening, because at first sight it does not seem obvious how to prove that the homotopy group of a topologicla group is abelian.

In our situation, let $\mu, \delta$ be respectively the multiplication and comultiplication, $\Delta: C \to C \times C$ the diagonal map, $\varepsilon, \eta$ the unit and the counit. Here you can't have the same equation, because that axioms also imply commutativity of operations. However, you can prove that:

1. Induced operations are defined by $(ff')= \mu(f,f')\Delta$, $(f*f')= \mu(f,f')\delta$.
2. $(f*f')(h*h') = (ff')*(hh')$
3. The constant function $u = \varepsilon \eta$ is an identity element for both operations.
4. These axioms imply the operations coincide.

Hope it helps, Andrea