Why is $\mathbb{S}^1$ a cogroup object in $\mathbf{Top.}$? $\require{AMScd}$
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 $\mathbf{HTop.}$.
 A: 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 $\mathbb{Z}[x,x^{-1}]$
A: 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.
