Why is $\mathbb{S}^1$ a cogroup object in $\mathbf{Top.}$?

Question

$\require{AMScd}$

This is basic level question, but this kind of questions usually find no answer on stackexchange.

I am trying to introduce my self to categories theory and advanced algebraic topology. I have just learnt about group and cogroup structures. My question is about cogroup structures.

An object with a group structure $(G, \mu: G \times G \to G, u: * \to G, i: G \to G)$ satisfies among the other properties the following

\begin{CD} G \times * @<{id \times *_G}<< G\\ @V{id\times u}VV @VV{id}V \\ G \times G @>{\mu }>> G \end{CD}

Here $*$ is the terminal object and $*_G$ the unique morphism to from $G$ to $*$. This is the law of neutral element.

If I have understood properly, a cogroup object should sastify the reversed diagram, i.e.

$\require{AMScd}$ \begin{CD} G \coprod * @>{id \coprod *_G}>> G\\ @A{id\coprod u}AA @AA{id}A \\ G \coprod G @<{\mu }<< G \end{CD}

Now, choose as category $\mathbf{Top.}$ and $G = (\mathbb{S}^1, 1)$ with operation $\mu: \mathbb{S}^1 \to \mathbb{S}^1 \vee \mathbb{S}^1$ that wrap the circle "half-and-half". The claim is that this is a cogroup structure on $\mathbb{S}^1$, but to me seems not.

The above diagram becomes

\begin{CD} \mathbb{S}^1 \vee * @>{id \vee *}>> \mathbb{S}^1 \\ @A{id\vee u}AA @AA{id}A \\ \mathbb{S}^1 \vee \mathbb{S}^1 @<{\mu }<< \mathbb{S}^1 \end{CD}

But $*$ in $\mathbf{Top.}$ is both initial and terminal and so $\mathbb{S}^1 \vee \mathbb{S}^1 \to \mathbb{S}^1 \vee * \to \mathbb{S}^1$ would collpse the "right circle" of $\mathbb{S}^1 \vee \mathbb{S}^1$ to the basepoint of $\mathbb{S}^1$ and so $\mathbb{S}^1 \to \mathbb{S}^1 \vee \mathbb{S}^1 \to \mathbb{S}^1 \vee * \to \mathbb{S}^1$ would collpse the "right half-circle" of $\mathbb{S}^1$ to the basepoint of $\mathbb{S}^1$. Then it could not be the identity as claimed by the other arrow.

Now, I know for sure that I am wrong, but where?

EDIT.

Obviously the answer is: $\mathbb{S}^1$ is not a cogroup object in $\mathbf{Top.}$ but in $\mathbf{HTop.}$.

Answer 1

This is a statement about the homotopy category. Consider the following fact:

Let $X$ be an object in a category $C$ such that the corepresentable functor $h^X:C\to \operatorname{Set}$ factors through the forgetful functor $\operatorname{Grp}\to \operatorname{Set}$. Then these data precisely specify the diagrams you drew above but for natural transformations wrt the corepresentable functor $h^X$. By Yoneda, all of these natural transformations descend down to a cogroup structure on X in $C$ (so long as $C$ has enough coproducts).

So in particular, we know that in the homotopy category of based spaces $\operatorname{Ho}(\operatorname{Top}_\ast)$, the functor $\pi_1:\operatorname{Top}_\ast \to \operatorname{Grp}$ is corepresented by the corepresentable functor $h^{(S^1,\ast)}$ where $h^{(S^1,\ast)}(X,x)=[(S^1,\ast),(X,x)]=\pi_1(X,x)$, as desired.

Another example of a canonical cogroup that arises in this way is the group scheme $G_m$. This is the spectrum of the ring $\mathbb{Z}[x,x^{-1}].$

If you unwind what the corepresentable functor $h^{\mathbb{Z}[x,x^{-1}]}$ does on rings, you see that it sends a ring to its group of units. This then flips around to specify the cogroup structure by Yoneda on $\mathbb{Z}[x,x^{-1}]$

Answer 2

The circle is not a cogroup object in the category of based spaces. If $X$ is a cogroup in based spaces, then the composition $$ X \to X \vee X \to X\times X $$ would necessarily coincide with the diagonal $\Delta$, but the intersection of $X\vee X$ with the diagonal is the basepoint $(\ast,\ast)$. It then follows that $\Delta(X)$ is a point. This can only happen if $X$ itself is point.

In unbased spaces, a similar argument shows that $X =\emptyset$, the empty space.