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Let A be a pointed topological space. I want to show that A is a cogroup object in $Ho(Top_*)$ iff the functor $[A,\_] \colon Top_* \rightarrow Sets_*$ factors through the category of groups.

$\Leftarrow$: I want to find the multiplication and inverse map. Since $[A, A\vee A]$ is a group, it must contain an element (the identity). Let one of those representatives $m$ be the multiplication map.

One property the multiplication map must satisfy is for the composition $(*\vee id)\circ m$ to be homotopic to $id$. But already here I'm in trouble. The continuous function $* \vee id$ induces a group homomorphism $[A, A\vee A] \rightarrow [A,A]$. But since $m$ is the identity, this map must take $m$ to the identity of $A$. But it is not hard to see that the constant function $* \colon A \rightarrow A$ represents the identity in $[A,A]$. So if the property was true we would have a homotopy $id \sim *$. But this is only true when A is contractible.

Can someone help me finding the correct multiplication map (or point out my error above) and the correct inverse map (for the latter I only see two options, identity map or constant map).

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You've gone about it the wrong way. The map $A \to A \vee A$ you want is obtained by noting that $[A, A \vee A]$ is a group and that there is a natural bijection $[A, A \vee A] \times [A, A \vee A] \cong [A \vee A, A \vee A]$. (Here I am assuming that $A \vee A$ is the coproduct in $\textbf{Ho}(\textbf{Top}_*)$, which is probably not true for general $A$.) Now, $\textrm{id} : A \vee A \to A \vee A$ then corresponds to a pair of maps $f, g : A \to A \vee A$, and the cogroup operation of $A$ is the map $A \to A \vee A$ corresponding to $f \cdot g$, where $\cdot$ is the group operation of $[A, A \vee A]$. The counit of $A$ corresponds to the unit of the group $[A, 1]$, and the coinversion of $A$ corresponds to the inversion of the group $[A, A]$.

In general, an object $A$ in a category $\mathcal{C}$ is a cogroup object if and only if the representable functor $\mathcal{C}(A, -) : \mathcal{C} \to \textbf{Set}$ factors through $\textbf{Grp}$.

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Thanks Zhen. Could you clarify what you mean by "the inversion of the group [A, A]"? – Wing Nov 3 '12 at 23:15
$[A, A]$ is a group, so it has a map $[A, A] \to [A, A]$ sending an element $x$ to $x^{-1}$ (in the sense of the group operation, rather than composition). – Zhen Lin Nov 3 '12 at 23:17
Now I get it, thanks! – Wing Nov 3 '12 at 23:20
I'm having a hard time for instance to show the property mentioned in my first post. I have no idea how to show that the composition ($id\vee *)\circ m$ is homotopic to the identity map. Can you give any further hand? Thanks. – Wing Nov 4 '12 at 21:57
It's all pure abstract nonsense, as I said in the second paragraph. By the Yoneda lemma, two morphisms $f, g : B \to C$ in $\mathcal{C}$ are equal if and only if the maps $f^*, g^* : \mathcal{C}(C, X) \to \mathcal{C}(B, X)$ are equal for all objects $X$ in $\mathcal{C}$. Thus the equation you want is logically equivalent to the dual equation for the groups $[A, X]$. – Zhen Lin Nov 4 '12 at 22:00

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